I need to solve the following ODE:
$$y''(x)-y=\textrm{sech}(x)$$
where $\textrm{sech}(x)$ is the hyperbolic secant function.
Now, finding the complementary function is trivial, since the auxillary equation has roots $1,-1$, the C.F. is $C_1e^x+C_2e^{-x}$ where $C_1,C_2$ are arbitrary constants.
Now, for the particular integral, I'm having trouble because of $\textrm{sech}(x)$. It would be easy if it was $\sinh(x)$ or $\cosh(x)$ because I can simply use the definition and split them into exponential components and then use the formula for finding the PI of $e^{ax}$ where $a$ is a constant.
One way I thought to do this was to use the fact that $\displaystyle\frac 1{D-a}V=e^{ax}\int e^{-ax}V\,\mathrm dx$ but it seems like this will be very tedious.
Another way I thought to do this was to perform a change of variables $(z=\log(x))$ which would give me the following ODE to solve:
$$z^2\frac{\textrm{d}^2y}{\textrm{d}z^2}-y=\frac{2z}{z^2+1}$$
but I'm not able to solve this one either. I see that this is an equation of the Euler-Cauchy type (homogeneous) but since the method for that uses the reverse change of variables of what I did, I can't follow that method.
Can someone help? Thanks in advance!
Best Answer
Variation of Parameters seems easy in this case. It is a second-order linear nonhomogeneous ODE: $$y''+p(x)y'+q(x)y=r(x)$$
The basis solutions are $e^{-x}$ and $e^x$. Therefore, the Wronskian is: $$W(x)=\begin{vmatrix} e^{-x} & e^x \\ \frac{d}{dx}(e^{-x}) & \frac{d}{dx}(e^x) \end{vmatrix}=\begin{vmatrix} e^{-x} & e^x \\ -e^{-x} & e^x \end{vmatrix}=2$$ Therefore, our particular solution will be given by: $$y_p(x)=v_1(x)\cdot e^{-x}+v_2(x)\cdot e^x \tag{1}$$ Where: $$v_1(x)=-\int \frac{r(x)\cdot e^{x}}{W(x)}~dx=-\frac{1}{2}\int e^x \operatorname*{sech}(x)~dx$$ And: $$v_2(x)=\int \frac{r(x)\cdot e^{-x}}{W(x)}~dx=\frac{1}{2}\int e^{-x} \operatorname*{sech}(x)~dx$$
You can use the fact that: $$\operatorname*{sech}(x)\equiv \frac{1}{\cosh(x)}\equiv \frac{2}{e^x+e^{-x}}$$
All that remains to do is to evaluate these integrals, substitute them into $(1)$, and then use the fact that $y(x)=y_c(x)+y_p(x)$.