Solve equation $L(t) = \int_0^t L(t)\, P\frac{(1+e^{-L(t)})^2}{4e^{-L(t)}}\, \mathrm{d}t$

Question

I need help for solving the following equation: $$L(t) = \int_0^t L(t) \, P\cdot\frac{(1+e^{-L(t)})^2}{4e^{-L(t)}} \, \mathrm{d}t, \quad $L(0)=C $$

By splitting it up in the form $$\frac{dL(t)}{\frac{4e^{-L(t)}}{(1+e^{-L(t)})^2}} = P \, \mathrm{d}t$$

I obtained: $$ P \, t=1/2 \left(L(t) + \sinh(L(t)) \,\cosh(L(t)\right)+ \operatorname{constant} $$

but I'm not sure if this is OK and I need it as a function of just $L(t)=\ldots$, is this possible at all?

Answer 1

I think you have a wrong expression for $L'(t)$. Your ODE for $L$ should be

$$ L'(t) = \frac{P}{4} \frac{\left(1+e^{-L(t)}\right)^2}{e^{-L(t)}}, \quad L(0) = C$$

Mathematica gives the solution to this problem as

$$ L(t) = \log \left(\frac{1}{\frac{1}{e^C+1}-\frac{P t}{4}}-1\right) $$

Answer 2

$$L(t) = \int_0^t L(t) \, P\cdot\frac{(1+e^{-L(t)})^2}{4e^{-L(t)}} \, \mathrm{d}t, \qquad L(0)=C \tag 1$$ $$\frac{\mathrm{d}L}{\mathrm{d}t}=\frac{P}{4}\cdot L(t)(1+e^{-L(t)})^2e^{L(t)} \quad\to\quad dt=\frac{4}{P}\frac{\mathrm{d}L}{L(1+e^{-L})^2e^{L} }$$ $$t=\frac{4}{P}\int \frac{\mathrm{d}L}{L(1+e^{-L})^2e^{L} }+\text{constant}$$ $$ L(t=0)=C \quad\to\quad t(L=C)=0 \quad\to\quad t(L)=\frac{4}{P}\int_C^L \frac{\mathrm{d}x}{x(1+e^{-x})^2e^{x} }$$ $$L(t)\text{ is the inverse function.}$$ There is no closed form for the integral and even more for the inverse function. So, you have to use numerical method of solving to go further.

Comment : If it is a textbook problem, a solution is expected with usual functions. So, equation $(1)$ is probably false. Since you didn't edit the original wording of the problem and the steps to go to eq.$(1)$, it is impossible to show you where you made the mistake.