$y''-\frac{e^x}{e^x+1}y'=e^{-x}+1\:$

Question

Solve the Cauchy problem $$y''-\frac{e^x}{e^x+1}y'=e^{-x}+1\:$$ $$\:y\left(0\right)=2\:$$$$y'\left(0\right)=2$$

In class we did these sort of problems but we could always write the characteristic equation for the differential equation. Like we had something along the lines of $ay''+by + c =f\left(x\right)$ and we could just write the characteristic equation as $ar^2+br+c = 0$, solve it and then get the solutions of the homogenous equation based on delta. We'd then find the solution for the particular equation and the final solution would be the sum of the two.

But how do I solve this equation? Wolfram tells me its a linear equation, so I should solve it like I described above, but how do I get the characteristic equation? Or is it done in another way?

Answer 1

Hint. Let $u=y'$ and solve the first order linear ODE $$u'-\frac{e^x}{e^x+1}u=e^{-x}+1$$ with $u(0)=y'(0)$ (note that the coefficients of the linear ODE are non constant so characteristic equation does not apply). Then by using integrating factor we get $$(e^x+1)^2D(\frac{u}{e^x+1})=(e^x+1)u'-e^xu=(e^{-x}+1)(e^x+1)$$ that is $$D(\frac{u}{e^x+1})=\frac{e^{-x}+1}{e^x+1}=e^{-x}.$$ Hence $$\frac{u(x)}{e^x+1}-\frac{u(0)}{2}=\int_0^x e^{-t}dt.$$ After finding $u$, you may obtain $y$ by integration $$y(x)=y(0)+\int_0^x u(t)dt.$$

Answer 2

Hint :.Let $y'=p $ thus $y''=p'$ so we have $\frac {dp}{dx}-f (x)p=g (x) $ which is a linear in p then you can solve for particular using $p=\frac {dy}{dx}$. If you can solve this you get a general solution with an integration constant. From given condition we have $p (0)=2$ ie at $x=0,p=2$ from here you get a particular differential equation by getting value for c. After you resubstitute $\frac {dy}{dx}=p $ . Again integrating we have another integration constant $c_1$ . As its given that $y(0)=2$ we get a particular solution after getting value of $c_1$. Hope its more clear now.