🧠:<think>First, I need to calculate the total initial data load for the low-resolution images. Each low-resolution image has 25% of the original resolution in both dimensions, so the new dimensions are 750x500 pixels. With a color depth of 24 bits per pixel, each image has a size of 750 * 500 * 24 bits. Converting this to megabytes, I divide by 8 to get bytes and then by 1024^2 to convert to MB. This gives approximately 1.17 MB per low-resolution image. Multiplying by 120 images, the total initial data load is about 140.4 MB.Next, for the lossless compression, each original image is 18 MB, so the total original storage is 2160 MB. After applying a 30% reduction, each image is compressed to 70% of its original size, which is 12.6 MB per image. Multiplying by 120 images, the new total storage is 1512 MB. The percentage reduction in storage is calculated by comparing the difference between the original and new storage to the original storage, resulting in a 30% reduction.</think>

🧠:<think>Okay, so I have this problem about modeling the shift from physical to digital reading. There are two parts, and I need to solve both. Let me take it step by step.Starting with part 1: I need to find the time ( t ) when the number of digital readers equals the number of physical book readers. The functions given are:- ( P(t) = P_0 e^{-kt} ) for physical readers.- ( D(t) = D_0 ln(1 + at) ) for digital readers.So, I need to set these equal to each other and solve for ( t ). That means:( P_0 e^{-kt} = D_0 ln(1 + at) )Hmm, this looks like an equation where ( t ) is in both an exponential and a logarithmic function. I don't think I can solve this algebraically in a straightforward way. Maybe I need to use some kind of substitution or perhaps the Lambert W function? Wait, I remember that equations involving exponentials and logarithms sometimes can be transformed into a form that uses the Lambert W function, which is the inverse function of ( f(W) = W e^{W} ).Let me try to manipulate the equation:( P_0 e^{-kt} = D_0 ln(1 + at) )Let me divide both sides by ( D_0 ):( frac{P_0}{D_0} e^{-kt} = ln(1 + at) )Let me denote ( frac{P_0}{D_0} ) as a constant, say ( C ). So,( C e^{-kt} = ln(1 + at) )Hmm, now I need to solve for ( t ). Let me exponentiate both sides to get rid of the logarithm:( e^{C e^{-kt}} = 1 + at )But that seems more complicated. Maybe another approach. Let me set ( u = at ), so ( t = frac{u}{a} ). Then, the equation becomes:( C e^{-k frac{u}{a}} = ln(1 + u) )Which is:( C e^{-frac{k}{a} u} = ln(1 + u) )Still not sure. Maybe I can write it as:( ln(1 + u) = C e^{-frac{k}{a} u} )Exponentiating both sides:( 1 + u = e^{C e^{-frac{k}{a} u}} )This still looks complicated. Maybe I can take the natural logarithm again? Let me see:( ln(1 + u) = C e^{-frac{k}{a} u} )Wait, that's the same as before. Maybe I need to rearrange terms. Let me try to express this in terms of ( v = e^{-frac{k}{a} u} ). Then, ( u = -frac{a}{k} ln v ). Substituting back:( ln(1 - frac{a}{k} ln v) = C v )Hmm, this seems even more complicated. Maybe this approach isn't working. Perhaps I should consider that this equation might not have an analytical solution and instead needs to be solved numerically. But the question asks to express ( t ) in terms of ( P_0 ), ( D_0 ), ( k ), and ( a ). So maybe I need to express it implicitly or use the Lambert W function.Let me try another substitution. Let me set ( y = 1 + at ). Then, ( t = frac{y - 1}{a} ). Substituting into the equation:( C e^{-k frac{y - 1}{a}} = ln y )Which is:( C e^{-frac{k}{a} y + frac{k}{a}} = ln y )Factor out the constant:( C e^{frac{k}{a}} e^{-frac{k}{a} y} = ln y )Let me denote ( C e^{frac{k}{a}} ) as another constant, say ( K ). So,( K e^{-frac{k}{a} y} = ln y )Multiply both sides by ( -frac{k}{a} ):( -frac{k}{a} K e^{-frac{k}{a} y} = -frac{k}{a} ln y )Let me set ( z = -frac{k}{a} y ). Then, ( y = -frac{a}{k} z ). Substituting:( -frac{k}{a} K e^{z} = -frac{k}{a} ln(-frac{a}{k} z) )Simplify:( K e^{z} = ln(-frac{a}{k} z) )Hmm, this still doesn't look like the standard form for Lambert W. The standard form is ( z e^{z} = something ). Maybe I need to manipulate it differently.Alternatively, let's go back to the original equation:( P_0 e^{-kt} = D_0 ln(1 + at) )Let me divide both sides by ( P_0 ):( e^{-kt} = frac{D_0}{P_0} ln(1 + at) )Let me denote ( frac{D_0}{P_0} ) as another constant, say ( M ). So,( e^{-kt} = M ln(1 + at) )Take natural logarithm on both sides:( -kt = ln(M) + ln(ln(1 + at)) )Hmm, not helpful. Maybe I need to consider that this equation might not have a closed-form solution and instead needs to be expressed in terms of the Lambert W function. Let me try to rearrange terms to get it into a form where I can apply Lambert W.Let me set ( u = 1 + at ). Then, ( t = frac{u - 1}{a} ). Substitute into the equation:( e^{-k frac{u - 1}{a}} = M ln u )Which is:( e^{frac{k}{a} (1 - u)} = M ln u )Let me write this as:( e^{frac{k}{a}} e^{-frac{k}{a} u} = M ln u )Multiply both sides by ( e^{frac{k}{a} u} ):( e^{frac{k}{a}} = M ln u cdot e^{frac{k}{a} u} )Let me set ( v = frac{k}{a} u ). Then, ( u = frac{a}{k} v ). Substitute back:( e^{frac{k}{a}} = M ln(frac{a}{k} v) cdot e^{v} )Simplify:( e^{frac{k}{a}} = M left( ln a - ln k + ln v right) e^{v} )This is getting too complicated. Maybe I need to accept that this equation doesn't have a solution in terms of elementary functions and instead express ( t ) implicitly or use the Lambert W function.Wait, let me try another substitution. Let me set ( w = e^{-kt} ). Then, ( t = -frac{1}{k} ln w ). Substitute into the equation:( P_0 w = D_0 ln(1 + a (-frac{1}{k} ln w)) )Simplify:( P_0 w = D_0 lnleft(1 - frac{a}{k} ln wright) )This still looks messy. Maybe I need to rearrange terms differently.Alternatively, let me consider the equation:( P_0 e^{-kt} = D_0 ln(1 + at) )Let me divide both sides by ( D_0 ):( frac{P_0}{D_0} e^{-kt} = ln(1 + at) )Let me denote ( frac{P_0}{D_0} ) as ( C ) again. So,( C e^{-kt} = ln(1 + at) )Let me exponentiate both sides:( e^{C e^{-kt}} = 1 + at )Let me set ( z = e^{-kt} ). Then, ( t = -frac{1}{k} ln z ). Substitute back:( e^{C z} = 1 + a (-frac{1}{k} ln z) )Simplify:( e^{C z} = 1 - frac{a}{k} ln z )This still doesn't seem helpful. Maybe I need to consider that this equation might not have an analytical solution and instead needs to be solved numerically. But the question asks to express ( t ) in terms of the given constants, so perhaps I need to leave it in terms of the Lambert W function.Wait, let me try to rearrange the equation:( C e^{-kt} = ln(1 + at) )Let me set ( x = at ). Then, ( t = frac{x}{a} ). Substitute:( C e^{-k frac{x}{a}} = ln(1 + x) )Let me write this as:( C e^{-frac{k}{a} x} = ln(1 + x) )Let me set ( y = -frac{k}{a} x ). Then, ( x = -frac{a}{k} y ). Substitute:( C e^{y} = ln(1 - frac{a}{k} y) )Hmm, not helpful. Maybe I need to consider that this equation is transcendental and cannot be solved with elementary functions. Therefore, the solution for ( t ) must be expressed implicitly or using the Lambert W function.Wait, let me try to manipulate the equation again:( C e^{-kt} = ln(1 + at) )Let me set ( u = 1 + at ). Then, ( t = frac{u - 1}{a} ). Substitute:( C e^{-k frac{u - 1}{a}} = ln u )Which is:( C e^{frac{k}{a} (1 - u)} = ln u )Let me write this as:( C e^{frac{k}{a}} e^{-frac{k}{a} u} = ln u )Let me set ( v = frac{k}{a} u ). Then, ( u = frac{a}{k} v ). Substitute:( C e^{frac{k}{a}} e^{-v} = ln(frac{a}{k} v) )Simplify:( C e^{frac{k}{a}} e^{-v} = ln a - ln k + ln v )This is still complicated. Maybe I need to accept that this equation doesn't have a closed-form solution and instead express ( t ) in terms of the Lambert W function.Wait, let me try to rearrange the equation:( C e^{-kt} = ln(1 + at) )Let me set ( z = e^{-kt} ). Then, ( t = -frac{1}{k} ln z ). Substitute:( C z = ln(1 + a (-frac{1}{k} ln z)) )Simplify:( C z = lnleft(1 - frac{a}{k} ln zright) )Let me set ( w = ln z ). Then, ( z = e^{w} ). Substitute:( C e^{w} = lnleft(1 - frac{a}{k} wright) )This is still not helpful. Maybe I need to consider that this equation is too complex for an analytical solution and instead needs to be solved numerically. However, the question asks to express ( t ) in terms of the given constants, so perhaps I need to leave it in terms of the Lambert W function.Wait, let me try to rearrange the equation again:( C e^{-kt} = ln(1 + at) )Let me set ( u = at ). Then, ( t = frac{u}{a} ). Substitute:( C e^{-k frac{u}{a}} = ln(1 + u) )Let me write this as:( C e^{-frac{k}{a} u} = ln(1 + u) )Let me set ( v = ln(1 + u) ). Then, ( u = e^{v} - 1 ). Substitute:( C e^{-frac{k}{a} (e^{v} - 1)} = v )Simplify:( C e^{-frac{k}{a} e^{v} + frac{k}{a}} = v )Let me write this as:( C e^{frac{k}{a}} e^{-frac{k}{a} e^{v}} = v )Let me set ( w = e^{v} ). Then, ( v = ln w ). Substitute:( C e^{frac{k}{a}} e^{-frac{k}{a} w} = ln w )Multiply both sides by ( e^{frac{k}{a} w} ):( C e^{frac{k}{a}} = ln w cdot e^{frac{k}{a} w} )Let me set ( z = frac{k}{a} w ). Then, ( w = frac{a}{k} z ). Substitute:( C e^{frac{k}{a}} = ln(frac{a}{k} z) cdot e^{z} )Simplify:( C e^{frac{k}{a}} = (ln a - ln k + ln z) e^{z} )This is still complicated. I think I'm stuck here. Maybe I need to consider that this equation doesn't have a solution in terms of elementary functions and instead needs to be expressed using the Lambert W function. Let me try to rearrange terms to get it into a form where I can apply Lambert W.Let me go back to the equation:( C e^{-kt} = ln(1 + at) )Let me set ( u = 1 + at ). Then, ( t = frac{u - 1}{a} ). Substitute:( C e^{-k frac{u - 1}{a}} = ln u )Which is:( C e^{frac{k}{a} (1 - u)} = ln u )Let me write this as:( C e^{frac{k}{a}} e^{-frac{k}{a} u} = ln u )Let me set ( v = frac{k}{a} u ). Then, ( u = frac{a}{k} v ). Substitute:( C e^{frac{k}{a}} e^{-v} = ln(frac{a}{k} v) )Simplify:( C e^{frac{k}{a}} e^{-v} = ln a - ln k + ln v )Let me set ( w = ln v ). Then, ( v = e^{w} ). Substitute:( C e^{frac{k}{a}} e^{-e^{w}} = ln a - ln k + w )This is getting too convoluted. I think I need to accept that this equation might not have a closed-form solution and instead needs to be solved numerically. However, the question asks to express ( t ) in terms of the given constants, so perhaps I need to leave it in terms of the Lambert W function.Wait, let me try one more substitution. Let me set ( y = e^{-kt} ). Then, ( t = -frac{1}{k} ln y ). Substitute into the equation:( P_0 y = D_0 ln(1 + a (-frac{1}{k} ln y)) )Simplify:( P_0 y = D_0 lnleft(1 - frac{a}{k} ln yright) )Let me set ( z = ln y ). Then, ( y = e^{z} ). Substitute:( P_0 e^{z} = D_0 lnleft(1 - frac{a}{k} zright) )This still doesn't seem helpful. I think I'm stuck here. Maybe the equation is too complex for an analytical solution, and the answer is expressed implicitly or using the Lambert W function.Wait, perhaps I can rearrange the original equation:( P_0 e^{-kt} = D_0 ln(1 + at) )Let me divide both sides by ( D_0 ):( frac{P_0}{D_0} e^{-kt} = ln(1 + at) )Let me set ( u = at ). Then, ( t = frac{u}{a} ). Substitute:( frac{P_0}{D_0} e^{-k frac{u}{a}} = ln(1 + u) )Let me write this as:( frac{P_0}{D_0} e^{-frac{k}{a} u} = ln(1 + u) )Let me set ( v = ln(1 + u) ). Then, ( u = e^{v} - 1 ). Substitute:( frac{P_0}{D_0} e^{-frac{k}{a} (e^{v} - 1)} = v )Simplify:( frac{P_0}{D_0} e^{frac{k}{a}} e^{-frac{k}{a} e^{v}} = v )Let me set ( w = e^{v} ). Then, ( v = ln w ). Substitute:( frac{P_0}{D_0} e^{frac{k}{a}} e^{-frac{k}{a} w} = ln w )Multiply both sides by ( e^{frac{k}{a} w} ):( frac{P_0}{D_0} e^{frac{k}{a}} = ln w cdot e^{frac{k}{a} w} )Let me set ( z = frac{k}{a} w ). Then, ( w = frac{a}{k} z ). Substitute:( frac{P_0}{D_0} e^{frac{k}{a}} = ln(frac{a}{k} z) cdot e^{z} )Simplify:( frac{P_0}{D_0} e^{frac{k}{a}} = (ln a - ln k + ln z) e^{z} )This is still too complicated. I think I need to accept that this equation doesn't have a solution in terms of elementary functions and instead needs to be expressed using the Lambert W function. Let me try to rearrange terms to get it into a form where I can apply Lambert W.Let me consider the equation:( (ln a - ln k + ln z) e^{z} = frac{P_0}{D_0} e^{frac{k}{a}} )Let me set ( s = z + ln a - ln k ). Then, ( z = s - ln a + ln k ). Substitute:( s e^{s - ln a + ln k} = frac{P_0}{D_0} e^{frac{k}{a}} )Simplify the exponent:( s e^{s} e^{-ln a + ln k} = frac{P_0}{D_0} e^{frac{k}{a}} )Which is:( s e^{s} cdot frac{k}{a} = frac{P_0}{D_0} e^{frac{k}{a}} )Multiply both sides by ( frac{a}{k} ):( s e^{s} = frac{P_0}{D_0} e^{frac{k}{a}} cdot frac{a}{k} )So,( s e^{s} = frac{a P_0}{k D_0} e^{frac{k}{a}} )Now, this is in the form ( s e^{s} = C ), where ( C = frac{a P_0}{k D_0} e^{frac{k}{a}} ). Therefore, the solution is ( s = W(C) ), where ( W ) is the Lambert W function.So, ( s = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) )Recall that ( s = z + ln a - ln k ), and ( z = frac{k}{a} w ), ( w = e^{v} ), ( v = ln(1 + u) ), ( u = at ). This is getting too long, but let me try to backtrack.We have:( s = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) )But ( s = z + ln a - ln k ), and ( z = frac{k}{a} w ), ( w = e^{v} ), ( v = ln(1 + u) ), ( u = at ).Wait, this is too convoluted. Maybe I can express ( t ) in terms of ( s ).From earlier, ( s = z + ln a - ln k ), and ( z = frac{k}{a} w ), ( w = e^{v} ), ( v = ln(1 + u) ), ( u = at ).But this is too long. Maybe I can express ( t ) in terms of ( s ) directly.Wait, let me try to express ( t ) in terms of ( s ):From ( s = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) ), and ( s = z + ln a - ln k ), and ( z = frac{k}{a} w ), ( w = e^{v} ), ( v = ln(1 + u) ), ( u = at ).This is getting too long. Maybe I can express ( t ) as:( t = frac{1}{a} left( e^{v} - 1 right) )But ( v = ln(1 + u) ), which is ( v = ln(1 + at) ). This is circular.I think I need to accept that the solution for ( t ) is expressed in terms of the Lambert W function, and it's quite involved. Therefore, the answer is:( t = frac{1}{a} left( e^{Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right)} - 1 right) )But I'm not sure if this is correct. Let me check the steps again.Wait, from earlier, we had:( s e^{s} = frac{a P_0}{k D_0} e^{frac{k}{a}} )So, ( s = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) )Then, ( s = z + ln a - ln k )But ( z = frac{k}{a} w ), and ( w = e^{v} ), ( v = ln(1 + u) ), ( u = at ).This is too long. Maybe I can express ( t ) in terms of ( s ) as follows:From ( s = z + ln a - ln k ), and ( z = frac{k}{a} w ), ( w = e^{v} ), ( v = ln(1 + u) ), ( u = at ).So,( s = frac{k}{a} e^{v} + ln a - ln k )But ( v = ln(1 + u) = ln(1 + at) )This is still circular. I think I need to express ( t ) in terms of ( s ) as:( t = frac{1}{a} left( e^{v} - 1 right) )But ( v = ln(1 + at) ), so ( e^{v} = 1 + at ), which gives ( t = frac{1}{a} (e^{v} - 1) ). But ( v = ln(1 + at) ), so this is consistent.But from earlier, ( s = z + ln a - ln k ), and ( z = frac{k}{a} w ), ( w = e^{v} ). So,( s = frac{k}{a} e^{v} + ln a - ln k )But ( e^{v} = 1 + at ), so:( s = frac{k}{a} (1 + at) + ln a - ln k )Simplify:( s = frac{k}{a} + k t + ln a - ln k )But ( s = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) ), so:( Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) = frac{k}{a} + k t + ln a - ln k )Solve for ( t ):( k t = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{k}{a} - ln a + ln k )Therefore,( t = frac{1}{k} left[ Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{k}{a} - ln a + ln k right] )Simplify:( t = frac{1}{k} Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{1}{a} - frac{ln a}{k} + frac{ln k}{k} )This seems complicated, but it's an expression in terms of the Lambert W function. Therefore, the time ( t ) when the number of digital readers equals the number of physical book readers is:( t = frac{1}{k} Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{1}{a} - frac{ln a}{k} + frac{ln k}{k} )But I'm not sure if this is the simplest form or if I made a mistake in the substitutions. Maybe I should check with a simpler case or see if there's a different approach.Alternatively, perhaps the equation can be rearranged differently. Let me try to write the original equation again:( P_0 e^{-kt} = D_0 ln(1 + at) )Let me set ( x = at ). Then, ( t = frac{x}{a} ). Substitute:( P_0 e^{-k frac{x}{a}} = D_0 ln(1 + x) )Let me write this as:( frac{P_0}{D_0} e^{-frac{k}{a} x} = ln(1 + x) )Let me set ( y = ln(1 + x) ). Then, ( x = e^{y} - 1 ). Substitute:( frac{P_0}{D_0} e^{-frac{k}{a} (e^{y} - 1)} = y )Simplify:( frac{P_0}{D_0} e^{frac{k}{a}} e^{-frac{k}{a} e^{y}} = y )Let me set ( z = e^{y} ). Then, ( y = ln z ). Substitute:( frac{P_0}{D_0} e^{frac{k}{a}} e^{-frac{k}{a} z} = ln z )Multiply both sides by ( e^{frac{k}{a} z} ):( frac{P_0}{D_0} e^{frac{k}{a}} = ln z cdot e^{frac{k}{a} z} )Let me set ( w = frac{k}{a} z ). Then, ( z = frac{a}{k} w ). Substitute:( frac{P_0}{D_0} e^{frac{k}{a}} = ln(frac{a}{k} w) cdot e^{w} )Simplify:( frac{P_0}{D_0} e^{frac{k}{a}} = (ln a - ln k + ln w) e^{w} )This is still complicated. Maybe I can set ( u = w + ln a - ln k ). Then, ( w = u - ln a + ln k ). Substitute:( frac{P_0}{D_0} e^{frac{k}{a}} = u e^{u - ln a + ln k} )Simplify the exponent:( frac{P_0}{D_0} e^{frac{k}{a}} = u e^{u} e^{-ln a + ln k} )Which is:( frac{P_0}{D_0} e^{frac{k}{a}} = u e^{u} cdot frac{k}{a} )Multiply both sides by ( frac{a}{k} ):( frac{a P_0}{k D_0} e^{frac{k}{a}} = u e^{u} )So, ( u e^{u} = frac{a P_0}{k D_0} e^{frac{k}{a}} )Therefore, ( u = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) )Recall that ( u = w + ln a - ln k ), and ( w = frac{k}{a} z ), ( z = e^{y} ), ( y = ln(1 + x) ), ( x = at ).So, ( u = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) )But ( u = w + ln a - ln k ), and ( w = frac{k}{a} z ), ( z = e^{y} ), ( y = ln(1 + x) ), ( x = at ).This is too long, but let me try to express ( t ) in terms of ( u ):From ( u = w + ln a - ln k ), and ( w = frac{k}{a} z ), ( z = e^{y} ), ( y = ln(1 + x) ), ( x = at ).So,( u = frac{k}{a} e^{y} + ln a - ln k )But ( y = ln(1 + x) = ln(1 + at) ), so ( e^{y} = 1 + at ).Therefore,( u = frac{k}{a} (1 + at) + ln a - ln k )Simplify:( u = frac{k}{a} + k t + ln a - ln k )But ( u = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) ), so:( Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) = frac{k}{a} + k t + ln a - ln k )Solve for ( t ):( k t = Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{k}{a} - ln a + ln k )Therefore,( t = frac{1}{k} left[ Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{k}{a} - ln a + ln k right] )Simplify:( t = frac{1}{k} Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{1}{a} - frac{ln a}{k} + frac{ln k}{k} )This seems to be the expression for ( t ) in terms of the Lambert W function. Therefore, the answer to part 1 is:( t = frac{1}{k} Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{1}{a} - frac{ln a}{k} + frac{ln k}{k} )Now, moving on to part 2: Calculate the time ( t ) at which the rate of decrease of physical book readers is exactly twice the rate of increase of digital readers.First, I need to find the derivatives of ( P(t) ) and ( D(t) ).Given:( P(t) = P_0 e^{-kt} )So, the rate of decrease is:( frac{dP}{dt} = -k P_0 e^{-kt} )Similarly, ( D(t) = D_0 ln(1 + at) )So, the rate of increase is:( frac{dD}{dt} = D_0 cdot frac{a}{1 + at} )The problem states that the rate of decrease of physical readers is exactly twice the rate of increase of digital readers. Therefore:( left| frac{dP}{dt} right| = 2 frac{dD}{dt} )Since ( frac{dP}{dt} ) is negative (decreasing), the absolute value is:( -frac{dP}{dt} = 2 frac{dD}{dt} )So,( k P_0 e^{-kt} = 2 cdot D_0 cdot frac{a}{1 + at} )Now, solve for ( t ).Let me write this equation:( k P_0 e^{-kt} = frac{2 a D_0}{1 + at} )Let me rearrange terms:( e^{-kt} (1 + at) = frac{2 a D_0}{k P_0} )Let me denote ( frac{2 a D_0}{k P_0} ) as a constant ( C ). So,( e^{-kt} (1 + at) = C )Let me set ( u = at ). Then, ( t = frac{u}{a} ). Substitute:( e^{-k frac{u}{a}} (1 + u) = C )Which is:( (1 + u) e^{-frac{k}{a} u} = C )Let me set ( v = frac{k}{a} u ). Then, ( u = frac{a}{k} v ). Substitute:( left(1 + frac{a}{k} vright) e^{-v} = C )Simplify:( left( frac{k + a v}{k} right) e^{-v} = C )Multiply both sides by ( k ):( (k + a v) e^{-v} = k C )Let me write this as:( (k + a v) e^{-v} = k C )Let me set ( w = v + frac{k}{a} ). Then, ( v = w - frac{k}{a} ). Substitute:( left( k + a left( w - frac{k}{a} right) right) e^{- (w - frac{k}{a})} = k C )Simplify inside the parentheses:( k + a w - k = a w )So,( a w e^{-w + frac{k}{a}} = k C )Factor out ( e^{frac{k}{a}} ):( a w e^{-w} e^{frac{k}{a}} = k C )So,( a w e^{-w} = k C e^{-frac{k}{a}} )Let me set ( z = -w ). Then, ( w = -z ). Substitute:( a (-z) e^{z} = k C e^{-frac{k}{a}} )Multiply both sides by -1:( a z e^{z} = -k C e^{-frac{k}{a}} )But ( z e^{z} ) is the form for the Lambert W function. So,( z e^{z} = frac{-k C e^{-frac{k}{a}}}{a} )Therefore,( z = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )Recall that ( z = -w ), ( w = v + frac{k}{a} ), ( v = frac{k}{a} u ), ( u = at ).So,( z = - (v + frac{k}{a}) = - left( frac{k}{a} u + frac{k}{a} right) = - frac{k}{a} (u + 1) )But ( u = at ), so:( z = - frac{k}{a} (at + 1) = -k t - frac{k}{a} )But from earlier, ( z = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) ), so:( -k t - frac{k}{a} = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )Solve for ( t ):( -k t = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) + frac{k}{a} )Multiply both sides by -1:( k t = - Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) - frac{k}{a} )Therefore,( t = - frac{1}{k} Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) - frac{1}{a} )Recall that ( C = frac{2 a D_0}{k P_0} ). Substitute back:( t = - frac{1}{k} Wleft( frac{-k cdot frac{2 a D_0}{k P_0} cdot e^{-frac{k}{a}}}{a} right) - frac{1}{a} )Simplify inside the Lambert W:( frac{-k cdot frac{2 a D_0}{k P_0} cdot e^{-frac{k}{a}}}{a} = frac{-2 a D_0}{P_0} cdot e^{-frac{k}{a}} cdot frac{1}{a} = frac{-2 D_0}{P_0} e^{-frac{k}{a}} )Therefore,( t = - frac{1}{k} Wleft( frac{-2 D_0}{P_0} e^{-frac{k}{a}} right) - frac{1}{a} )This is the expression for ( t ) in terms of the Lambert W function. Therefore, the answer to part 2 is:( t = - frac{1}{k} Wleft( frac{-2 D_0}{P_0} e^{-frac{k}{a}} right) - frac{1}{a} )I think this is the correct expression. Let me double-check the steps to ensure I didn't make a mistake.Starting from:( k P_0 e^{-kt} = frac{2 a D_0}{1 + at} )Then,( e^{-kt} (1 + at) = frac{2 a D_0}{k P_0} = C )Set ( u = at ), so ( t = frac{u}{a} ):( e^{-k frac{u}{a}} (1 + u) = C )Set ( v = frac{k}{a} u ), so ( u = frac{a}{k} v ):( e^{-v} left(1 + frac{a}{k} vright) = C )Multiply by ( k ):( (k + a v) e^{-v} = k C )Set ( w = v + frac{k}{a} ), so ( v = w - frac{k}{a} ):( (k + a (w - frac{k}{a})) e^{- (w - frac{k}{a})} = k C )Simplify:( (k + a w - k) e^{-w + frac{k}{a}} = k C )Which is:( a w e^{-w} e^{frac{k}{a}} = k C )Set ( z = -w ):( a (-z) e^{z} e^{frac{k}{a}} = k C )So,( -a z e^{z} = k C e^{-frac{k}{a}} )Multiply both sides by -1:( a z e^{z} = -k C e^{-frac{k}{a}} )Thus,( z e^{z} = frac{-k C e^{-frac{k}{a}}}{a} )So,( z = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )But ( z = -w = - (v + frac{k}{a}) = - ( frac{k}{a} u + frac{k}{a} ) = - frac{k}{a} (u + 1) = - frac{k}{a} (at + 1) )Therefore,( - frac{k}{a} (at + 1) = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )Solve for ( t ):( -k (at + 1) = a Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )Wait, I think I made a mistake here. Let me correct it.From ( z = -w = - (v + frac{k}{a}) ), and ( v = frac{k}{a} u ), ( u = at ):So,( z = - ( frac{k}{a} u + frac{k}{a} ) = - frac{k}{a} (u + 1) = - frac{k}{a} (at + 1) )Thus,( z = - frac{k}{a} (at + 1) )But ( z = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) ), so:( - frac{k}{a} (at + 1) = Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )Multiply both sides by -1:( frac{k}{a} (at + 1) = - Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )Divide both sides by ( frac{k}{a} ):( at + 1 = - frac{a}{k} Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) )Subtract 1:( at = - frac{a}{k} Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) - 1 )Divide by ( a ):( t = - frac{1}{k} Wleft( frac{-k C e^{-frac{k}{a}}}{a} right) - frac{1}{a} )Which is the same as before. Therefore, the expression is correct.So, summarizing:1. The time ( t ) when digital readers equal physical readers is:( t = frac{1}{k} Wleft( frac{a P_0}{k D_0} e^{frac{k}{a}} right) - frac{1}{a} - frac{ln a}{k} + frac{ln k}{k} )2. The time ( t ) when the rate of decrease of physical readers is twice the rate of increase of digital readers is:( t = - frac{1}{k} Wleft( frac{-2 D_0}{P_0} e^{-frac{k}{a}} right) - frac{1}{a} )I think these are the correct expressions, although they are quite involved and involve the Lambert W function, which is not an elementary function. Therefore, these are the solutions expressed in terms of the given constants.</think>

🧠:<think>Alright, so I have this problem about modeling the growth of fitness center memberships using a differential equation. It's been a while since I dealt with differential equations, but let me try to work through it step by step.First, the problem gives me the differential equation:[ frac{dN}{dt} = kNleft(1 - frac{N}{M}right) ]This looks familiar—it's the logistic growth model. I remember that the logistic equation models population growth where there's a carrying capacity, which is the maximum population that the environment can sustain. In this case, the "population" is the number of fitness center members, and the carrying capacity is ( M ).The first sub-problem asks me to find the values of ( k ) and ( M ) given that 20 years ago, the number of members was 500, and now it's 8000. So, ( N(0) = 500 ) and ( N(20) = 8000 ).I need to solve this differential equation to find ( N(t) ) in terms of ( t ), ( k ), and ( M ), and then use the given conditions to solve for ( k ) and ( M ).I recall that the solution to the logistic differential equation is:[ N(t) = frac{M}{1 + left(frac{M - N(0)}{N(0)}right)e^{-kt}} ]Let me verify that. If I plug in ( t = 0 ), I should get ( N(0) = 500 ). Plugging in:[ N(0) = frac{M}{1 + left(frac{M - 500}{500}right)e^{0}} = frac{M}{1 + frac{M - 500}{500}} ]Simplify the denominator:[ 1 + frac{M - 500}{500} = frac{500 + M - 500}{500} = frac{M}{500} ]So,[ N(0) = frac{M}{frac{M}{500}} = 500 ]Yep, that checks out. Good.Now, using the information that at ( t = 20 ), ( N(20) = 8000 ). So, plug that into the equation:[ 8000 = frac{M}{1 + left(frac{M - 500}{500}right)e^{-20k}} ]Let me denote ( frac{M - 500}{500} ) as a single term for simplicity. Let's call it ( C ). So, ( C = frac{M - 500}{500} ). Then the equation becomes:[ 8000 = frac{M}{1 + C e^{-20k}} ]But since ( C = frac{M - 500}{500} ), I can write:[ 8000 = frac{M}{1 + left(frac{M - 500}{500}right)e^{-20k}} ]This seems a bit complicated, but maybe I can rearrange it to solve for ( M ) and ( k ).First, let's rewrite the equation:[ 8000 left[1 + left(frac{M - 500}{500}right)e^{-20k}right] = M ]Expanding the left side:[ 8000 + 8000 left(frac{M - 500}{500}right)e^{-20k} = M ]Simplify ( 8000 times frac{M - 500}{500} ):[ 8000 div 500 = 16 ], so:[ 8000 + 16(M - 500)e^{-20k} = M ]Let me write that as:[ 8000 + 16(M - 500)e^{-20k} = M ]Now, let's bring the 8000 to the right side:[ 16(M - 500)e^{-20k} = M - 8000 ]Divide both sides by 16:[ (M - 500)e^{-20k} = frac{M - 8000}{16} ]Let me write this as:[ e^{-20k} = frac{M - 8000}{16(M - 500)} ]Take the natural logarithm of both sides:[ -20k = lnleft(frac{M - 8000}{16(M - 500)}right) ]So,[ k = -frac{1}{20} lnleft(frac{M - 8000}{16(M - 500)}right) ]Hmm, okay. So now I have an expression for ( k ) in terms of ( M ). But I need another equation to solve for both ( k ) and ( M ). Wait, but I only have two data points: ( N(0) = 500 ) and ( N(20) = 8000 ). So, I think this is the only equation I can get from the given information.Wait, is there another way? Maybe I can express ( k ) in terms of ( M ) and then substitute back into the equation for ( N(t) ). Alternatively, perhaps I can assume that 8000 is less than ( M ), which makes sense because the carrying capacity is the maximum, so ( M ) must be greater than 8000.Let me denote ( M ) as some value greater than 8000. Let me call ( M = 8000 + x ), where ( x > 0 ). Maybe that substitution can help.But before that, let me think about the behavior of the logistic function. The function approaches ( M ) as ( t ) approaches infinity. So, if ( N(20) = 8000 ), which is significantly larger than ( N(0) = 500 ), it suggests that the growth is still in the accelerating phase, but perhaps approaching the carrying capacity.Alternatively, maybe I can express ( e^{-20k} ) in terms of ( M ) and then see if I can find ( M ).Wait, let's go back to the equation:[ 8000 = frac{M}{1 + left(frac{M - 500}{500}right)e^{-20k}} ]Let me denote ( A = left(frac{M - 500}{500}right) ). Then the equation becomes:[ 8000 = frac{M}{1 + A e^{-20k}} ]So,[ 1 + A e^{-20k} = frac{M}{8000} ]Thus,[ A e^{-20k} = frac{M}{8000} - 1 ]But ( A = frac{M - 500}{500} ), so:[ frac{M - 500}{500} e^{-20k} = frac{M}{8000} - 1 ]Multiply both sides by 500:[ (M - 500) e^{-20k} = frac{500M}{8000} - 500 ]Simplify ( frac{500M}{8000} ):[ frac{500}{8000} = frac{1}{16} ], so:[ (M - 500) e^{-20k} = frac{M}{16} - 500 ]Hmm, this seems similar to what I had before. Let me write it as:[ (M - 500) e^{-20k} = frac{M - 8000}{16} ]Which is the same as before. So, I'm back to the same equation.I think I need to find ( M ) such that this equation holds. Maybe I can make an assumption or find a way to express ( M ) in terms of itself.Alternatively, perhaps I can express ( e^{-20k} ) as:[ e^{-20k} = frac{M - 8000}{16(M - 500)} ]And then, since ( e^{-20k} ) must be positive, the right-hand side must also be positive. So,[ frac{M - 8000}{16(M - 500)} > 0 ]Which implies that both numerator and denominator are positive or both negative.But ( M > 8000 ) because ( N(20) = 8000 ) and the carrying capacity is the maximum. So, ( M - 8000 > 0 ). Therefore, ( M - 500 > 0 ) as well, which is true because ( M > 8000 ). So, both numerator and denominator are positive, which is consistent.Now, let me denote ( e^{-20k} = frac{M - 8000}{16(M - 500)} ). Let me call this Equation (1).I also know that the logistic function has an inflection point at ( N = M/2 ). The growth rate is highest there. But I don't know if that helps here.Alternatively, maybe I can express ( k ) in terms of ( M ) and then use another condition, but I only have two points. Wait, perhaps I can use the fact that the solution is a logistic function and plug in ( t = 20 ) and ( N = 8000 ) into the general solution.Wait, let me write the general solution again:[ N(t) = frac{M}{1 + left(frac{M - N(0)}{N(0)}right)e^{-kt}} ]So, plugging in ( t = 20 ) and ( N(20) = 8000 ):[ 8000 = frac{M}{1 + left(frac{M - 500}{500}right)e^{-20k}} ]Which is the same equation as before.I think I need to solve for ( M ) and ( k ) simultaneously. Let me denote ( e^{-20k} = x ). Then, Equation (1) becomes:[ x = frac{M - 8000}{16(M - 500)} ]So,[ x = frac{M - 8000}{16M - 8000} ]Wait, let me compute the denominator:16(M - 500) = 16M - 8000.So,[ x = frac{M - 8000}{16M - 8000} ]But ( x = e^{-20k} ), which is a positive number less than 1 because ( k ) is positive (since it's a growth rate). So, ( x < 1 ).Let me write:[ x = frac{M - 8000}{16M - 8000} ]I can write this as:[ x = frac{M - 8000}{16M - 8000} = frac{M - 8000}{16(M - 500)} ]Wait, that's the same as before.Alternatively, let me solve for ( M ) in terms of ( x ):From ( x = frac{M - 8000}{16M - 8000} ), cross-multiplying:[ x(16M - 8000) = M - 8000 ]Expanding:[ 16xM - 8000x = M - 8000 ]Bring all terms to one side:[ 16xM - 8000x - M + 8000 = 0 ]Factor ( M ):[ M(16x - 1) - 8000x + 8000 = 0 ]So,[ M(16x - 1) = 8000x - 8000 ]Thus,[ M = frac{8000x - 8000}{16x - 1} ]Simplify numerator and denominator:Numerator: ( 8000(x - 1) )Denominator: ( 16x - 1 )So,[ M = frac{8000(x - 1)}{16x - 1} ]But ( x = e^{-20k} ), so:[ M = frac{8000(e^{-20k} - 1)}{16e^{-20k} - 1} ]Hmm, this seems a bit circular. Maybe I can find another relationship.Wait, I also know that the initial condition is ( N(0) = 500 ), which we already used to get the general solution. Maybe I can think about the derivative at ( t = 0 ). Let me compute ( dN/dt ) at ( t = 0 ):[ frac{dN}{dt}bigg|_{t=0} = k cdot 500 left(1 - frac{500}{M}right) ]But I don't know the value of this derivative, so that might not help directly.Alternatively, perhaps I can assume that the growth is still in the exponential phase, but I don't think that's necessarily true because 20 years is a long time, and the membership has grown significantly.Wait, maybe I can make an assumption about the value of ( M ). Let me think: If ( N(20) = 8000 ), which is much larger than ( N(0) = 500 ), but still, ( M ) must be larger than 8000. Maybe ( M ) is not too much larger, but I don't know.Alternatively, perhaps I can use the fact that the logistic function can be rewritten in terms of the initial slope.Wait, let me think differently. The logistic equation can be rewritten as:[ frac{dN}{dt} = rN left(1 - frac{N}{K}right) ]Where ( r ) is the intrinsic growth rate and ( K ) is the carrying capacity. In our case, ( k = r ) and ( M = K ).The solution is:[ N(t) = frac{K}{1 + left(frac{K - N_0}{N_0}right)e^{-rt}} ]Which is what we have.Given that, perhaps I can express the ratio ( frac{N(t)}{K - N(t)} ) as:[ frac{N(t)}{K - N(t)} = frac{N_0}{K - N_0} e^{rt} ]Let me verify that.Starting from the solution:[ N(t) = frac{K}{1 + left(frac{K - N_0}{N_0}right)e^{-rt}} ]Let me solve for ( frac{N(t)}{K - N(t)} ):[ frac{N(t)}{K - N(t)} = frac{frac{K}{1 + C e^{-rt}}}{K - frac{K}{1 + C e^{-rt}}} ]Where ( C = frac{K - N_0}{N_0} ).Simplify denominator:[ K - frac{K}{1 + C e^{-rt}} = K left(1 - frac{1}{1 + C e^{-rt}}right) = K left(frac{C e^{-rt}}{1 + C e^{-rt}}right) ]So,[ frac{N(t)}{K - N(t)} = frac{frac{K}{1 + C e^{-rt}}}{K cdot frac{C e^{-rt}}{1 + C e^{-rt}}} = frac{1}{C e^{-rt}} = frac{e^{rt}}{C} ]But ( C = frac{K - N_0}{N_0} ), so:[ frac{N(t)}{K - N(t)} = frac{e^{rt} N_0}{K - N_0} ]Which can be written as:[ frac{N(t)}{K - N(t)} = frac{N_0}{K - N_0} e^{rt} ]Yes, that's correct. So, taking the natural logarithm of both sides:[ lnleft(frac{N(t)}{K - N(t)}right) = lnleft(frac{N_0}{K - N_0}right) + rt ]This is a linear equation in terms of ( t ), which might be easier to handle.Given that, let's plug in our values.At ( t = 0 ):[ lnleft(frac{500}{M - 500}right) = lnleft(frac{500}{M - 500}right) + 0 ]Which is consistent.At ( t = 20 ):[ lnleft(frac{8000}{M - 8000}right) = lnleft(frac{500}{M - 500}right) + 20k ]So, rearranging:[ 20k = lnleft(frac{8000}{M - 8000}right) - lnleft(frac{500}{M - 500}right) ]Using logarithm properties:[ 20k = lnleft(frac{8000}{M - 8000} cdot frac{M - 500}{500}right) ]Simplify the fraction inside the log:[ frac{8000}{M - 8000} cdot frac{M - 500}{500} = frac{8000 cdot (M - 500)}{500 cdot (M - 8000)} ]Simplify 8000/500 = 16:[ = frac{16(M - 500)}{M - 8000} ]So,[ 20k = lnleft(frac{16(M - 500)}{M - 8000}right) ]But from earlier, we had:[ e^{-20k} = frac{M - 8000}{16(M - 500)} ]Which can be written as:[ e^{-20k} = frac{1}{frac{16(M - 500)}{M - 8000}} ]So,[ e^{-20k} = frac{1}{frac{16(M - 500)}{M - 8000}} = frac{M - 8000}{16(M - 500)} ]Which is consistent with what we had before.But now, from the linear equation approach, we have:[ 20k = lnleft(frac{16(M - 500)}{M - 8000}right) ]So,[ k = frac{1}{20} lnleft(frac{16(M - 500)}{M - 8000}right) ]But earlier, from the other approach, we had:[ k = -frac{1}{20} lnleft(frac{M - 8000}{16(M - 500)}right) ]Which is the same because:[ lnleft(frac{16(M - 500)}{M - 8000}right) = -lnleft(frac{M - 8000}{16(M - 500)}right) ]So, both expressions for ( k ) are consistent.Now, I need to find ( M ) such that this equation holds. Let me denote ( y = M ). Then, the equation becomes:[ k = frac{1}{20} lnleft(frac{16(y - 500)}{y - 8000}right) ]But I also have the expression for ( k ) in terms of ( y ):From Equation (1):[ e^{-20k} = frac{y - 8000}{16(y - 500)} ]But from the linear equation approach, we have:[ 20k = lnleft(frac{16(y - 500)}{y - 8000}right) ]Which implies:[ e^{20k} = frac{16(y - 500)}{y - 8000} ]So,[ e^{-20k} = frac{y - 8000}{16(y - 500)} ]Which is consistent.So, essentially, I have:[ e^{-20k} = frac{y - 8000}{16(y - 500)} ]But I also have:[ e^{-20k} = frac{y - 8000}{16(y - 500)} ]Wait, that's the same equation. So, I'm going in circles.Perhaps I need to make an assumption or find a way to solve for ( y ).Let me denote ( z = y - 8000 ). Then, ( y = z + 8000 ). Let me substitute this into the equation.So,[ e^{-20k} = frac{z}{16(z + 8000 - 500)} = frac{z}{16(z + 7500)} ]But from the linear equation approach, we have:[ 20k = lnleft(frac{16(y - 500)}{y - 8000}right) = lnleft(frac{16(z + 7500)}{z}right) ]So,[ 20k = lnleft(frac{16(z + 7500)}{z}right) ]But ( e^{-20k} = frac{z}{16(z + 7500)} ), so:[ e^{-20k} = frac{z}{16(z + 7500)} ]But ( e^{-20k} = frac{1}{e^{20k}} ), and from the linear equation:[ e^{20k} = frac{16(z + 7500)}{z} ]So,[ e^{-20k} = frac{z}{16(z + 7500)} ]Which is consistent.Wait, maybe I can set ( e^{-20k} = frac{z}{16(z + 7500)} ) and ( e^{-20k} = frac{z}{16(z + 7500)} ). Hmm, that's the same thing.I think I need to find ( z ) such that:[ frac{z}{16(z + 7500)} = e^{-20k} ]But I don't know ( k ). Alternatively, maybe I can express ( k ) in terms of ( z ) and substitute back.Wait, from the linear equation:[ 20k = lnleft(frac{16(z + 7500)}{z}right) ]So,[ k = frac{1}{20} lnleft(frac{16(z + 7500)}{z}right) ]So, substituting this into the equation for ( e^{-20k} ):[ e^{-20k} = e^{- lnleft(frac{16(z + 7500)}{z}right)} = frac{z}{16(z + 7500)} ]Which is consistent.So, I'm not making progress here. Maybe I need to make an assumption or use numerical methods.Wait, perhaps I can assume that ( M ) is not too much larger than 8000, so that ( M - 8000 ) is small compared to ( M ). But I don't know if that's valid.Alternatively, maybe I can try plugging in some values for ( M ) and see if the equation holds.Let me try ( M = 10000 ). Then,From Equation (1):[ e^{-20k} = frac{10000 - 8000}{16(10000 - 500)} = frac{2000}{16 times 9500} = frac{2000}{152000} = frac{1}{76} approx 0.01316 ]So,[ -20k = ln(1/76) approx ln(0.01316) approx -4.32 ]Thus,[ k approx frac{4.32}{20} approx 0.216 ]Now, let's check if with ( M = 10000 ) and ( k = 0.216 ), does ( N(20) = 8000 )?Using the logistic equation:[ N(20) = frac{10000}{1 + left(frac{10000 - 500}{500}right)e^{-0.216 times 20}} ]Calculate ( frac{10000 - 500}{500} = frac{9500}{500} = 19 ).Calculate ( e^{-0.216 times 20} = e^{-4.32} approx 0.01316 ).So,[ N(20) = frac{10000}{1 + 19 times 0.01316} = frac{10000}{1 + 0.25} = frac{10000}{1.25} = 8000 ]Perfect! So, ( M = 10000 ) and ( k approx 0.216 ) satisfy the equation.Wait, so that works. So, ( M = 10000 ) and ( k approx 0.216 ).But let me compute ( k ) more accurately.From earlier:[ e^{-20k} = frac{2000}{16 times 9500} = frac{2000}{152000} = frac{1}{76} ]So,[ -20k = ln(1/76) = -ln(76) ]Thus,[ k = frac{ln(76)}{20} ]Compute ( ln(76) ):( ln(70) approx 4.248 ), ( ln(80) approx 4.382 ). Since 76 is closer to 80, maybe around 4.33.But let me compute it accurately.( ln(76) approx 4.3307 )So,[ k = frac{4.3307}{20} approx 0.2165 ]So, ( k approx 0.2165 ) per year.Therefore, the values are ( M = 10000 ) and ( k approx 0.2165 ).Let me double-check with ( M = 10000 ) and ( k = ln(76)/20 ).Compute ( N(20) ):[ N(20) = frac{10000}{1 + 19 e^{-20k}} ]Since ( 20k = ln(76) ), so ( e^{-20k} = 1/76 ).Thus,[ N(20) = frac{10000}{1 + 19 times (1/76)} = frac{10000}{1 + 19/76} ]Simplify ( 19/76 = 1/4 ), so:[ N(20) = frac{10000}{1 + 0.25} = frac{10000}{1.25} = 8000 ]Yes, that checks out.So, Sub-problem 1 answer is ( M = 10000 ) and ( k = frac{ln(76)}{20} approx 0.2165 ).Now, moving on to Sub-problem 2: Find the time ( t ) when the number of members reaches 90% of the carrying capacity ( M ).So, 90% of ( M ) is ( 0.9 times 10000 = 9000 ).We need to find ( t ) such that ( N(t) = 9000 ).Using the logistic equation:[ 9000 = frac{10000}{1 + 19 e^{-kt}} ]Solve for ( t ).First, rearrange:[ 1 + 19 e^{-kt} = frac{10000}{9000} = frac{10}{9} ]So,[ 19 e^{-kt} = frac{10}{9} - 1 = frac{1}{9} ]Thus,[ e^{-kt} = frac{1}{9 times 19} = frac{1}{171} ]Take natural logarithm:[ -kt = lnleft(frac{1}{171}right) = -ln(171) ]So,[ t = frac{ln(171)}{k} ]We know ( k = frac{ln(76)}{20} ), so:[ t = frac{ln(171)}{frac{ln(76)}{20}} = frac{20 ln(171)}{ln(76)} ]Compute ( ln(171) ) and ( ln(76) ):( ln(171) approx 5.141 )( ln(76) approx 4.3307 )So,[ t approx frac{20 times 5.141}{4.3307} approx frac{102.82}{4.3307} approx 23.75 ]So, approximately 23.75 years.But let me compute it more accurately.First, compute ( ln(171) ):171 is between ( e^5 approx 148.41 ) and ( e^{5.1} approx 170.0 ). Wait, ( e^{5.1} approx 170.0 ), so ( ln(171) approx 5.1 + ) a little bit.Compute ( e^{5.1} = e^{5} times e^{0.1} approx 148.41 times 1.10517 approx 148.41 times 1.105 ≈ 164.0 ). Hmm, that's not right. Wait, no, ( e^{5.1} ) is actually approximately 170.0.Wait, let me use a calculator-like approach.We know that ( e^{5} = 148.413 )( e^{5.1} = e^{5} times e^{0.1} ≈ 148.413 times 1.10517 ≈ 148.413 times 1.105 ≈ 164.0 ). Wait, that can't be right because 5.1 is 5 + 0.1, so e^{5.1} ≈ 170.0.Wait, maybe I should use a better approximation.Alternatively, use the fact that ( ln(171) = ln(170) + ln(1.00588) approx ln(170) + 0.00586 ).We know ( ln(170) approx 5.135 ), so ( ln(171) ≈ 5.135 + 0.00586 ≈ 5.1409 ).Similarly, ( ln(76) approx 4.3307 ).So,[ t ≈ frac{20 times 5.1409}{4.3307} ≈ frac{102.818}{4.3307} ≈ 23.75 ]So, approximately 23.75 years.But let me compute it more precisely.Compute 102.818 / 4.3307:4.3307 * 23 = 99.60614.3307 * 23.75 = 4.3307*(23 + 0.75) = 99.6061 + 3.2479 ≈ 102.854Which is very close to 102.818, so t ≈ 23.75 - a tiny bit.Compute 4.3307 * 23.75 = 102.854But we have 102.818, which is 102.854 - 0.036.So, 0.036 / 4.3307 ≈ 0.0083So, t ≈ 23.75 - 0.0083 ≈ 23.7417So, approximately 23.74 years.So, about 23.74 years.But let me see if I can express this in terms of exact logarithms.We have:[ t = frac{20 ln(171)}{ln(76)} ]But 171 = 9*19, and 76 = 4*19. So,[ ln(171) = ln(9) + ln(19) = 2ln(3) + ln(19) ][ ln(76) = ln(4) + ln(19) = 2ln(2) + ln(19) ]So,[ t = frac{20(2ln(3) + ln(19))}{2ln(2) + ln(19)} ]But I don't think this simplifies further, so it's better to leave it as is or compute the numerical value.So, approximately 23.74 years.But let me check if this makes sense. Since at t=20, N=8000, and M=10000, so 9000 is 1000 more. Given the logistic curve, the growth slows down as it approaches M, so it should take a bit more than 20 years to reach 9000. 23.74 years seems reasonable.Alternatively, let me compute the exact value using more precise logarithms.Using calculator-like precision:Compute ( ln(171) ):We know that ( ln(170) ≈ 5.1357 ), and ( ln(171) ≈ 5.1357 + (1/170) ≈ 5.1357 + 0.00588 ≈ 5.1416 )Similarly, ( ln(76) ≈ 4.3307 )So,[ t = frac{20 times 5.1416}{4.3307} ≈ frac{102.832}{4.3307} ≈ 23.74 ]Yes, so 23.74 years.Therefore, the time when the number of members reaches 90% of the carrying capacity is approximately 23.74 years.But let me express this as an exact expression:[ t = frac{20 ln(171)}{ln(76)} ]Alternatively, since 171 = 9*19 and 76 = 4*19, we can write:[ t = frac{20 ln(9 times 19)}{ln(4 times 19)} = frac{20 (ln 9 + ln 19)}{ln 4 + ln 19} ]But I don't think this is necessary unless specified.So, summarizing:Sub-problem 1: ( M = 10000 ), ( k = frac{ln(76)}{20} approx 0.2165 )Sub-problem 2: ( t approx 23.74 ) years.But let me check if I can express ( t ) in terms of ( ln(171)/ln(76) times 20 ).Yes, that's the exact form.Alternatively, since ( ln(171) = ln(9 times 19) = ln(9) + ln(19) ), and ( ln(76) = ln(4 times 19) = ln(4) + ln(19) ), we can write:[ t = frac{20 (ln 9 + ln 19)}{ln 4 + ln 19} ]But again, unless required, the numerical value is sufficient.So, final answers:Sub-problem 1: ( M = 10000 ), ( k = frac{ln(76)}{20} ) or approximately 0.2165.Sub-problem 2: ( t approx 23.74 ) years.But let me express ( k ) exactly as ( frac{ln(76)}{20} ) and ( t ) as ( frac{20 ln(171)}{ln(76)} ).Alternatively, since 171 = 9*19 and 76 = 4*19, we can write:[ t = frac{20 ln(9 times 19)}{ln(4 times 19)} = frac{20 (ln 9 + ln 19)}{ln 4 + ln 19} ]But I think the exact form is fine as ( frac{20 ln(171)}{ln(76)} ).So, to wrap up:Sub-problem 1:- ( M = 10000 )- ( k = frac{ln(76)}{20} )Sub-problem 2:- ( t = frac{20 ln(171)}{ln(76)} approx 23.74 ) years.I think that's it.</think>

🧠:<think>Alright, so I have this problem where a fiction writer is analyzing word frequencies from classic novels. They're using some differential equations to model the frequencies, both deterministic and stochastic. I need to solve two parts here. Let me take them one by one.Starting with part 1: The differential equation given is [frac{df(t)}{dt} = alpha f(t) - beta f(t)^2 + gamma sin(omega t)]Hmm, okay. So this is a first-order ordinary differential equation (ODE). It looks like a logistic growth model with an additional sinusoidal term. The logistic term is (alpha f(t) - beta f(t)^2), which typically models growth with carrying capacity, and then there's this oscillating term (gamma sin(omega t)) which probably introduces some periodic influence on the word frequency.I need to find the general solution for (f(t)). Let me think about how to approach this. The equation is nonlinear because of the (f(t)^2) term, which complicates things. Nonlinear ODEs can be tricky because they don't always have closed-form solutions. However, maybe I can find an integrating factor or use some substitution to linearize it.Wait, let's write the equation again:[frac{df}{dt} + (beta f - alpha) f = gamma sin(omega t)]Hmm, actually, it's written as:[frac{df}{dt} = alpha f - beta f^2 + gamma sin(omega t)]Which can be rearranged as:[frac{df}{dt} + (beta f - alpha) f = gamma sin(omega t)]But that doesn't seem to help much. Maybe I can consider this as a Bernoulli equation? Bernoulli equations have the form:[frac{df}{dt} + P(t) f = Q(t) f^n]In our case, comparing:[frac{df}{dt} - (alpha - beta f) f = gamma sin(omega t)]Wait, no, that's not quite the standard Bernoulli form. Let me try to rewrite the original equation:[frac{df}{dt} + (beta f - alpha) f = gamma sin(omega t)]Hmm, actually, it's:[frac{df}{dt} = (alpha - beta f) f + gamma sin(omega t)]So, this is a Riccati equation because it's quadratic in (f). Riccati equations are generally difficult to solve unless we have a particular solution. The standard Riccati equation is:[frac{df}{dt} = Q(t) + R(t) f + S(t) f^2]Comparing, we have (Q(t) = gamma sin(omega t)), (R(t) = -beta f + alpha), and (S(t) = -beta). Wait, no, actually, let me see:Wait, the standard Riccati is:[frac{df}{dt} = Q(t) + R(t) f + S(t) f^2]In our case:[frac{df}{dt} = alpha f - beta f^2 + gamma sin(omega t)]So, that would mean (Q(t) = gamma sin(omega t)), (R(t) = alpha), and (S(t) = -beta). So yes, it's a Riccati equation with constant coefficients except for the (Q(t)) term which is time-dependent.Riccati equations are tough because they don't generally have solutions in terms of elementary functions unless we can find a particular solution. Since the equation has a sinusoidal forcing term, maybe we can look for a particular solution of the form (f_p(t) = A sin(omega t) + B cos(omega t)). Let me try that.Assume (f_p(t) = A sin(omega t) + B cos(omega t)). Then, its derivative is:[f_p'(t) = A omega cos(omega t) - B omega sin(omega t)]Plugging into the ODE:[A omega cos(omega t) - B omega sin(omega t) = alpha (A sin(omega t) + B cos(omega t)) - beta (A sin(omega t) + B cos(omega t))^2 + gamma sin(omega t)]Let me expand the right-hand side:First, compute (alpha f_p):[alpha A sin(omega t) + alpha B cos(omega t)]Then, compute (-beta f_p^2):[-beta (A^2 sin^2(omega t) + 2AB sin(omega t)cos(omega t) + B^2 cos^2(omega t))]Using the identity (sin^2 x = frac{1 - cos(2x)}{2}) and (cos^2 x = frac{1 + cos(2x)}{2}), and (sin x cos x = frac{sin(2x)}{2}), we can rewrite the squared term:[-beta left( frac{A^2 (1 - cos(2omega t))}{2} + AB sin(2omega t) + frac{B^2 (1 + cos(2omega t))}{2} right )]Simplify:[-beta left( frac{A^2 + B^2}{2} - frac{A^2 - B^2}{2} cos(2omega t) + AB sin(2omega t) right )]So, putting it all together, the right-hand side becomes:[alpha A sin(omega t) + alpha B cos(omega t) - beta left( frac{A^2 + B^2}{2} - frac{A^2 - B^2}{2} cos(2omega t) + AB sin(2omega t) right ) + gamma sin(omega t)]Now, let's collect like terms:1. Terms with (sin(omega t)):[alpha A sin(omega t) + gamma sin(omega t)]2. Terms with (cos(omega t)):[alpha B cos(omega t)]3. Terms with (sin(2omega t)):[- beta AB sin(2omega t)]4. Terms with (cos(2omega t)):[beta frac{A^2 - B^2}{2} cos(2omega t)]5. Constant terms:[- beta frac{A^2 + B^2}{2}]So, the right-hand side is:[(alpha A + gamma) sin(omega t) + alpha B cos(omega t) - beta AB sin(2omega t) + beta frac{A^2 - B^2}{2} cos(2omega t) - beta frac{A^2 + B^2}{2}]Now, equate this to the left-hand side:Left-hand side is:[A omega cos(omega t) - B omega sin(omega t)]So, equating coefficients of like terms:1. Coefficient of (sin(omega t)):Left: (-B omega)Right: (alpha A + gamma)So,[- B omega = alpha A + gamma quad (1)]2. Coefficient of (cos(omega t)):Left: (A omega)Right: (alpha B)So,[A omega = alpha B quad (2)]3. Coefficient of (sin(2omega t)):Left: 0Right: (- beta AB)So,[0 = - beta AB quad (3)]4. Coefficient of (cos(2omega t)):Left: 0Right: (beta frac{A^2 - B^2}{2})So,[0 = beta frac{A^2 - B^2}{2} quad (4)]5. Constant terms:Left: 0Right: (- beta frac{A^2 + B^2}{2})So,[0 = - beta frac{A^2 + B^2}{2} quad (5)]Now, let's analyze these equations.From equation (3): (0 = - beta AB). Assuming (beta neq 0), then either (A = 0) or (B = 0).Similarly, from equation (4): (0 = beta frac{A^2 - B^2}{2}). Again, assuming (beta neq 0), then (A^2 = B^2), so (A = pm B).From equation (5): (0 = - beta frac{A^2 + B^2}{2}). Again, assuming (beta neq 0), then (A^2 + B^2 = 0), which implies (A = 0) and (B = 0).But if (A = 0) and (B = 0), then from equation (1): (-0 = alpha * 0 + gamma), which implies (gamma = 0). But (gamma) is a given constant, not necessarily zero.This suggests that our initial assumption of a particular solution of the form (A sin(omega t) + B cos(omega t)) may not work unless (gamma = 0), which isn't necessarily the case. Therefore, perhaps we need a different approach.Alternatively, maybe we can use the method of variation of parameters or look for an integrating factor. But since the equation is Riccati, another approach is to use substitution to linearize it.Let me recall that for a Riccati equation:[frac{df}{dt} = Q(t) + R(t) f + S(t) f^2]If we have one particular solution (f_p(t)), we can use the substitution (f(t) = frac{1}{S(t)} frac{d}{dt} ln phi(t)), which transforms the Riccati equation into a linear ODE for (phi(t)).But in our case, (S(t) = -beta), which is constant. So, the substitution would be:[f(t) = frac{1}{-beta} frac{d}{dt} ln phi(t)]Let me compute this:Let (f(t) = -frac{1}{beta} frac{phi'(t)}{phi(t)}). Then,[f'(t) = -frac{1}{beta} left( frac{phi''(t)}{phi(t)} - frac{(phi'(t))^2}{phi(t)^2} right )]Plugging into the Riccati equation:[-frac{1}{beta} left( frac{phi''}{phi} - left( frac{phi'}{phi} right )^2 right ) = gamma sin(omega t) + alpha left( -frac{1}{beta} frac{phi'}{phi} right ) + (-beta) left( -frac{1}{beta} frac{phi'}{phi} right )^2]Simplify term by term.Left-hand side:[-frac{1}{beta} frac{phi''}{phi} + frac{1}{beta} left( frac{phi'}{phi} right )^2]Right-hand side:[gamma sin(omega t) - frac{alpha}{beta} frac{phi'}{phi} + frac{1}{beta} left( frac{phi'}{phi} right )^2]Now, let's bring all terms to one side:Left-hand side - Right-hand side:[-frac{1}{beta} frac{phi''}{phi} + frac{1}{beta} left( frac{phi'}{phi} right )^2 - gamma sin(omega t) + frac{alpha}{beta} frac{phi'}{phi} - frac{1}{beta} left( frac{phi'}{phi} right )^2 = 0]Simplify:The (frac{1}{beta} left( frac{phi'}{phi} right )^2) terms cancel out.So,[-frac{1}{beta} frac{phi''}{phi} - gamma sin(omega t) + frac{alpha}{beta} frac{phi'}{phi} = 0]Multiply both sides by (-beta phi):[phi'' - alpha phi' + beta gamma sin(omega t) phi = 0]So, we have a linear second-order ODE for (phi(t)):[phi''(t) - alpha phi'(t) + beta gamma sin(omega t) phi(t) = 0]Hmm, this seems complicated. It's a linear ODE with time-dependent coefficients because of the (sin(omega t)) term. Solving this might not be straightforward. Maybe we can look for solutions using methods for linear ODEs with periodic coefficients, like Floquet theory, but that might be beyond the scope here.Alternatively, perhaps we can assume a particular solution for (phi(t)) as well, but this seems recursive because we were trying to find (f(t)) in terms of (phi(t)), and now we have another ODE for (phi(t)).Wait, maybe instead of assuming a particular solution for (f(t)), we can use another substitution. Let me think.Alternatively, perhaps we can use the method of averaging or perturbation if (gamma) is small, but the problem doesn't specify that.Alternatively, maybe we can write the equation in terms of (g(t) = 1/f(t)), but let's try that.Let (g(t) = 1/f(t)). Then,[f(t) = 1/g(t)][f'(t) = -frac{g'(t)}{g(t)^2}]Plugging into the ODE:[-frac{g'}{g^2} = alpha frac{1}{g} - beta frac{1}{g^2} + gamma sin(omega t)]Multiply both sides by (-g^2):[g' = -alpha g + beta - gamma g^2 sin(omega t)]Hmm, so we get:[g'(t) + gamma sin(omega t) g(t)^2 = beta - alpha g(t)]This is a Bernoulli equation in terms of (g(t)). Bernoulli equations can be linearized by substituting (h(t) = g(t)^{1 - n}), where (n) is the exponent on (g(t)). In this case, the equation is:[g' + gamma sin(omega t) g^2 = beta - alpha g]So, it's a Bernoulli equation with (n = 2). Let me set (h(t) = 1/g(t)), so (n = 2), then (h = g^{-1}), so (g = 1/h), and (g' = -h'/h^2).Substituting into the equation:[- frac{h'}{h^2} + gamma sin(omega t) left( frac{1}{h^2} right ) = beta - alpha left( frac{1}{h} right )]Multiply both sides by (-h^2):[h' - gamma sin(omega t) = -beta h^2 + alpha h]Rearranged:[h' = -beta h^2 + alpha h + gamma sin(omega t)]Wait, that's similar to the original equation but with a sign change on the quadratic term. Hmm, not sure if that helps. Maybe not the best substitution.Alternatively, perhaps I should consider that the equation is a forced logistic equation, and look for solutions in terms of known functions or use numerical methods. But since the problem asks for the general solution, I might need to express it in terms of integrals or special functions.Alternatively, perhaps we can write the equation as:[frac{df}{dt} + (beta f - alpha) f = gamma sin(omega t)]Which is a Bernoulli equation with (n = 2). Let me use the standard substitution for Bernoulli equations: (v = f^{1 - n} = f^{-1}). So, (v = 1/f), then (dv/dt = -f^{-2} df/dt).So, substituting into the equation:[- frac{dv}{dt} = (alpha f - beta f^2) + gamma sin(omega t)]But (f = 1/v), so:[- frac{dv}{dt} = alpha frac{1}{v} - beta frac{1}{v^2} + gamma sin(omega t)]Multiply both sides by (-1):[frac{dv}{dt} = -alpha frac{1}{v} + beta frac{1}{v^2} - gamma sin(omega t)]Hmm, this doesn't seem to linearize the equation. Maybe I made a mistake in substitution.Wait, let's go back. The standard Bernoulli substitution is (v = f^{1 - n}), so for (n = 2), (v = f^{-1}). Then, (dv/dt = -f^{-2} df/dt). So, the original equation:[frac{df}{dt} = alpha f - beta f^2 + gamma sin(omega t)]Multiply both sides by (-f^{-2}):[- f^{-2} frac{df}{dt} = - alpha f^{-1} + beta - gamma f^{-2} sin(omega t)]But ( - f^{-2} df/dt = dv/dt ), so:[frac{dv}{dt} = - alpha v + beta - gamma v^2 sin(omega t)]Ah, that's better. So now we have:[frac{dv}{dt} + gamma sin(omega t) v^2 = beta - alpha v]Which is a Bernoulli equation in terms of (v(t)). But wait, this is similar to what I had before. It's still nonlinear because of the (v^2) term. So, perhaps this substitution didn't help.Alternatively, maybe I can write it as:[frac{dv}{dt} + alpha v = beta - gamma sin(omega t) v^2]Which is still a Riccati equation. Hmm, seems like we're going in circles.Maybe another approach: since the equation is nonlinear, perhaps we can look for an integrating factor or use some other technique. Alternatively, if we can't find an exact solution, we might have to express the solution in terms of integrals or use perturbation methods if (gamma) is small.Alternatively, perhaps we can write the equation in terms of (u(t) = f(t) - f_s(t)), where (f_s(t)) is a particular solution. But without knowing a particular solution, that might not help.Wait, maybe I can consider the homogeneous equation first:[frac{df}{dt} = alpha f - beta f^2]This is a logistic equation, which has the solution:[f(t) = frac{alpha}{beta} cdot frac{1}{1 + C e^{-alpha t}}]Where (C) is a constant determined by initial conditions. So, the homogeneous solution is known.Now, the nonhomogeneous term is (gamma sin(omega t)). Maybe we can use the method of variation of parameters. For linear equations, variation of parameters works, but our equation is nonlinear. However, perhaps we can use a similar idea.Alternatively, since the homogeneous solution is known, maybe we can write the general solution as the homogeneous solution plus a particular solution. But since the equation is nonlinear, the superposition principle doesn't apply, so that might not work.Alternatively, perhaps we can use the Green's function approach, but again, for nonlinear equations, that's complicated.Wait, another thought: since the equation is a Riccati equation, and we know that Riccati equations can sometimes be transformed into linear equations if we have a particular solution. But as we saw earlier, finding a particular solution is non-trivial because of the sinusoidal term.Alternatively, maybe we can assume that the particular solution is small and use perturbation methods. Let me consider that.Assume that (gamma) is small, so the particular solution due to the sinusoidal term is small compared to the homogeneous solution. Then, we can write (f(t) = f_h(t) + f_p(t)), where (f_h(t)) is the homogeneous solution and (f_p(t)) is a small perturbation.But this might not be valid if (gamma) is not small. Since the problem doesn't specify, I can't assume that.Alternatively, perhaps we can use the method of averaging, which is used for nonlinear oscillations. But I'm not sure.Wait, maybe I can use the substitution (f(t) = frac{alpha}{beta} + g(t)), shifting the solution to eliminate the linear term. Let me try that.Let (f(t) = frac{alpha}{beta} + g(t)). Then, plugging into the ODE:[frac{d}{dt} left( frac{alpha}{beta} + g right ) = alpha left( frac{alpha}{beta} + g right ) - beta left( frac{alpha}{beta} + g right )^2 + gamma sin(omega t)]Simplify:Left-hand side:[g'(t)]Right-hand side:[alpha frac{alpha}{beta} + alpha g - beta left( frac{alpha^2}{beta^2} + frac{2 alpha}{beta} g + g^2 right ) + gamma sin(omega t)]Simplify term by term:1. (alpha frac{alpha}{beta} = frac{alpha^2}{beta})2. (alpha g)3. (- beta frac{alpha^2}{beta^2} = - frac{alpha^2}{beta})4. (- beta frac{2 alpha}{beta} g = -2 alpha g)5. (- beta g^2)6. (+ gamma sin(omega t))So, combining terms:[frac{alpha^2}{beta} + alpha g - frac{alpha^2}{beta} - 2 alpha g - beta g^2 + gamma sin(omega t)]Simplify:The (frac{alpha^2}{beta}) terms cancel out.[alpha g - 2 alpha g = - alpha g]So, right-hand side becomes:[- alpha g - beta g^2 + gamma sin(omega t)]Thus, the equation for (g(t)) is:[g'(t) = - alpha g - beta g^2 + gamma sin(omega t)]Hmm, that's still a nonlinear equation, but perhaps it's easier to handle. It's a Riccati equation with constant coefficients except for the sinusoidal term.Alternatively, maybe we can linearize it around the equilibrium point. The equilibrium points are found by setting (g' = 0):[0 = - alpha g - beta g^2 + gamma sin(omega t)]But since (gamma sin(omega t)) is time-dependent, the equilibrium is also time-dependent, which complicates things.Alternatively, perhaps we can consider the equation in the form:[g'(t) + alpha g(t) + beta g(t)^2 = gamma sin(omega t)]This is still a Riccati equation. Maybe we can use the substitution (g(t) = frac{u'(t)}{beta u(t)}), which sometimes works for Riccati equations.Let me try that substitution. Let (g(t) = frac{u'(t)}{beta u(t)}). Then,[g'(t) = frac{u''(t)}{beta u(t)} - frac{(u'(t))^2}{beta u(t)^2}]Plugging into the equation:[frac{u''}{beta u} - frac{(u')^2}{beta u^2} + alpha frac{u'}{beta u} + beta left( frac{u'}{beta u} right )^2 = gamma sin(omega t)]Simplify term by term:1. (frac{u''}{beta u})2. (- frac{(u')^2}{beta u^2})3. (+ frac{alpha u'}{beta u})4. (+ beta frac{(u')^2}{beta^2 u^2} = frac{(u')^2}{beta u^2})So, combining terms:The (- frac{(u')^2}{beta u^2}) and (+ frac{(u')^2}{beta u^2}) cancel out.So, we're left with:[frac{u''}{beta u} + frac{alpha u'}{beta u} = gamma sin(omega t)]Multiply both sides by (beta u):[u'' + alpha u' = beta gamma u sin(omega t)]So, we have a linear second-order ODE:[u''(t) + alpha u'(t) - beta gamma u(t) sin(omega t) = 0]This is still a difficult equation because of the (sin(omega t)) term. It's a linear ODE with time-dependent coefficients. Solving this exactly might not be possible without special functions or using methods like Green's functions or perturbation.Alternatively, perhaps we can use the method of undetermined coefficients, but since the nonhomogeneous term is (sin(omega t)), and the equation has variable coefficients, it's not straightforward.Alternatively, maybe we can use the method of variation of parameters. Let me consider the homogeneous equation:[u'' + alpha u' = 0]The characteristic equation is (r^2 + alpha r = 0), so roots (r = 0) and (r = -alpha). So, the general solution to the homogeneous equation is:[u_h(t) = C_1 + C_2 e^{-alpha t}]Now, for the nonhomogeneous equation:[u'' + alpha u' = beta gamma u sin(omega t)]Wait, no, actually, the nonhomogeneous term is (- beta gamma u sin(omega t)), but it's multiplied by (u(t)), making it nonlinear. So, actually, the equation is:[u'' + alpha u' - beta gamma u sin(omega t) = 0]Which is a linear ODE with variable coefficients because of the (sin(omega t)) term. This is a type of Hill's equation, which generally doesn't have solutions in terms of elementary functions. So, perhaps we can't find an exact solution here.Given that, maybe the best approach is to accept that the general solution can't be expressed in closed form and instead express it in terms of integrals or use numerical methods. However, the problem asks for the general solution, so perhaps we can write it using the integrating factor method for the Riccati equation.Wait, another thought: since the equation is a Riccati equation, and we have transformed it into a linear second-order ODE, perhaps we can express the solution in terms of the homogeneous solutions and particular solutions. But without knowing a particular solution, it's difficult.Alternatively, perhaps we can use the method of Green's functions for the linear ODE. The equation is:[u'' + alpha u' - beta gamma sin(omega t) u = 0]This is a linear ODE with variable coefficients. The Green's function approach involves finding two linearly independent solutions to the homogeneous equation and then constructing the particular solution using those.But as I mentioned earlier, the homogeneous equation is (u'' + alpha u' - beta gamma sin(omega t) u = 0), which is difficult to solve exactly. So, perhaps we can't proceed further analytically.Given that, maybe the best we can do is express the solution in terms of integrals or special functions, but I'm not sure. Alternatively, perhaps the problem expects a qualitative analysis rather than an explicit solution.Wait, the problem says "determine the general solution for (f(t)), discussing the conditions under which this solution is valid." Maybe it's expecting a solution in terms of integrals or recognizing it as a Riccati equation and stating that it can be transformed into a linear second-order ODE, but without solving it explicitly.Alternatively, perhaps the problem is expecting the use of an integrating factor for the Bernoulli equation, but as we saw, that leads to a complicated equation.Wait, going back to the substitution (v = 1/f), which led us to:[v' = -alpha v + beta - gamma v^2 sin(omega t)]This is a Bernoulli equation with (n = 2). The standard method for Bernoulli equations is to use the substitution (w = v^{1 - n} = v^{-1}), but that didn't help earlier. Alternatively, we can use the integrating factor method for Bernoulli equations.Wait, actually, for a Bernoulli equation of the form:[v' + P(t) v = Q(t) v^n]The substitution is (w = v^{1 - n}), which linearizes the equation. In our case, the equation is:[v' + alpha v = beta - gamma sin(omega t) v^2]So, it's:[v' - gamma sin(omega t) v^2 + alpha v = beta]Which is a Bernoulli equation with (n = 2). So, let me set (w = v^{1 - 2} = v^{-1}). Then, (v = w^{-1}), and (v' = -w^{-2} w').Substituting into the equation:[- w^{-2} w' - gamma sin(omega t) w^{-2} + alpha w^{-1} = beta]Multiply both sides by (-w^2):[w' + gamma sin(omega t) - alpha w = -beta w^2]Hmm, that doesn't seem to help. It still has a quadratic term in (w). Maybe I made a mistake in substitution.Wait, let's do it step by step. The standard Bernoulli substitution is:Given (v' + P(t) v = Q(t) v^n), set (w = v^{1 - n}). Then, (v = w^{1/(1 - n)}), and (v' = frac{1}{1 - n} w^{n/(1 - n)} w').In our case, (n = 2), so (w = v^{-1}), (v = w^{-1}), (v' = -w^{-2} w').Plugging into the equation:[- w^{-2} w' + alpha w^{-1} = beta - gamma sin(omega t) w^{-2}]Multiply both sides by (-w^2):[w' - alpha w = -beta w^2 + gamma sin(omega t)]Which is:[w' + gamma sin(omega t) = alpha w + beta w^2]This is still a Riccati equation in terms of (w(t)). So, it seems we're stuck in a loop.Given that, perhaps the conclusion is that the general solution cannot be expressed in terms of elementary functions and requires special functions or numerical methods. Therefore, the solution is valid under the condition that we can solve the resulting linear second-order ODE, which may not be possible analytically for arbitrary (omega) and (gamma).Alternatively, if we consider small (gamma), we might use perturbation methods, expanding (f(t)) as a series in (gamma). But since the problem doesn't specify, I can't assume that.So, summarizing, the differential equation is a Riccati equation, which can be transformed into a linear second-order ODE, but solving that ODE analytically is not straightforward unless specific conditions are met (like particular values of (omega) and (gamma)). Therefore, the general solution may not be expressible in closed form and would require numerical methods or special functions.Moving on to part 2: The writer introduces a stochastic term (xi(t)) as white noise, leading to the SDE:[df(t) = left( alpha f(t) - beta f(t)^2 + gamma sin(omega t) right ) dt + sigma dW(t)]We need to derive the Fokker-Planck equation corresponding to this SDE and discuss the steady-state distribution of (f(t)).First, recalling that the Fokker-Planck equation (also known as the Kolmogorov forward equation) describes the time evolution of the probability density function (P(f, t)) of the stochastic process (f(t)). For an SDE of the form:[df(t) = mu(f, t) dt + sigma(f, t) dW(t)]The corresponding Fokker-Planck equation is:[frac{partial P}{partial t} = - frac{partial}{partial f} left[ mu(f, t) P(f, t) right ] + frac{1}{2} frac{partial^2}{partial f^2} left[ sigma^2(f, t) P(f, t) right ]]In our case, the drift coefficient (mu(f, t)) is:[mu(f, t) = alpha f - beta f^2 + gamma sin(omega t)]And the diffusion coefficient (sigma(f, t)) is constant, (sigma). So, (sigma^2(f, t) = sigma^2).Therefore, the Fokker-Planck equation becomes:[frac{partial P}{partial t} = - frac{partial}{partial f} left[ (alpha f - beta f^2 + gamma sin(omega t)) P(f, t) right ] + frac{sigma^2}{2} frac{partial^2 P}{partial f^2}]That's the Fokker-Planck equation for this system.Now, discussing the steady-state distribution. The steady-state distribution is the time-independent solution (P_{ss}(f)) which satisfies:[0 = - frac{partial}{partial f} left[ (alpha f - beta f^2 + gamma sin(omega t)) P_{ss}(f) right ] + frac{sigma^2}{2} frac{partial^2 P_{ss}}{partial f^2}]Wait, but in the steady state, the distribution doesn't depend on time, so the term (gamma sin(omega t)) is still time-dependent. This complicates things because the steady state would have to account for the periodic driving.Alternatively, if we consider the system under the influence of the periodic forcing, the steady-state distribution might be a periodic function in time. However, finding an explicit form for (P_{ss}(f)) is non-trivial due to the nonlinear drift term and the periodic forcing.Alternatively, if we average over the periodic forcing, perhaps we can find an effective steady-state distribution. But that would require knowing the statistics over one period, which might not be straightforward.Alternatively, if the forcing is weak or if we can separate the time scales, maybe we can use perturbation methods or adiabatic approximation. But without more information, it's hard to say.Alternatively, perhaps in the absence of the periodic forcing ((gamma = 0)), the steady-state distribution can be found. Let me consider that case.If (gamma = 0), the SDE becomes:[df = (alpha f - beta f^2) dt + sigma dW]The Fokker-Planck equation is:[frac{partial P}{partial t} = - frac{partial}{partial f} [ (alpha f - beta f^2) P ] + frac{sigma^2}{2} frac{partial^2 P}{partial f^2}]To find the steady-state, set (frac{partial P}{partial t} = 0):[- frac{partial}{partial f} [ (alpha f - beta f^2) P ] + frac{sigma^2}{2} frac{partial^2 P}{partial f^2} = 0]This is a second-order ODE for (P(f)). Let me write it as:[frac{sigma^2}{2} frac{d^2 P}{df^2} - (alpha f - beta f^2) frac{dP}{df} = 0]Let me set (Q = frac{dP}{df}), then:[frac{sigma^2}{2} frac{dQ}{df} - (alpha f - beta f^2) Q = 0]This is a first-order linear ODE for (Q):[frac{dQ}{df} - frac{2}{sigma^2} (alpha f - beta f^2) Q = 0]The integrating factor is:[mu(f) = expleft( - frac{2}{sigma^2} int (alpha f - beta f^2) df right )]Compute the integral:[int (alpha f - beta f^2) df = frac{alpha}{2} f^2 - frac{beta}{3} f^3 + C]So,[mu(f) = expleft( - frac{2}{sigma^2} left( frac{alpha}{2} f^2 - frac{beta}{3} f^3 right ) right ) = expleft( - frac{alpha}{sigma^2} f^2 + frac{2 beta}{3 sigma^2} f^3 right )]Thus, the solution for (Q) is:[Q(f) = C expleft( frac{alpha}{sigma^2} f^2 - frac{2 beta}{3 sigma^2} f^3 right )]Therefore, (P(f)) is the integral of (Q(f)):[P(f) = int C expleft( frac{alpha}{sigma^2} f^2 - frac{2 beta}{3 sigma^2} f^3 right ) df + D]However, this integral doesn't have a closed-form solution in terms of elementary functions. It might be expressible in terms of special functions, but generally, the steady-state distribution for such a nonlinear SDE is not analytically solvable.Therefore, in the case of (gamma = 0), the steady-state distribution is given implicitly by the integral above, but it's not expressible in a simple form. When (gamma neq 0), the periodic forcing complicates things further, and the steady-state distribution would likely be time-dependent and periodic as well.In summary, the Fokker-Planck equation is derived as above, and the steady-state distribution exists but cannot be expressed in a simple closed form due to the nonlinearity and periodic forcing. It would require numerical methods or further approximations to characterize it.Final Answer1. The general solution for ( f(t) ) is expressed in terms of a Riccati equation and cannot be simplified into elementary functions without specific conditions. The solution is valid under the condition that the resulting linear second-order ODE can be solved, typically requiring numerical methods or special functions.2. The Fokker-Planck equation is: [ frac{partial P}{partial t} = - frac{partial}{partial f} left[ (alpha f - beta f^2 + gamma sin(omega t)) P right ] + frac{sigma^2}{2} frac{partial^2 P}{partial f^2} ] The steady-state distribution exists but is not expressible in a simple closed form due to the nonlinear drift and periodic forcing, necessitating numerical methods for characterization.boxed{f(t) text{ is governed by a Riccati equation with no elementary solution, and the Fokker-Planck equation describes the probability density with a non-trivial steady-state.}}</think>