Finding the exact solution of a differential equation

Question

Let $y=f(x)$.
Is it possible to find an exact solution of the following differential equation?: \begin{equation} \ddot y+2\dot y-5xy=e^{-2x}\nonumber \end{equation}

Many thanks in advance,

--
Cesar

Answer 1

Hint:

Consider $\ddot{y_c}+2\dot{y_c}-5xy_c=0$ ,

Let $y_c=\int_Ce^{xs}K(s)~ds$ ,

Then $(\int_Ce^{xs}K(s)~ds)''+2(\int_Ce^{xs}K(s)~ds)'-5x\int_Ce^{xs}K(s)~ds=0$

$\int_Cs^2e^{xs}K(s)~ds+2\int_Cse^{xs}K(s)~ds-\int_C5e^{xs}K(s)~d(xs)=0$

$\int_C(s^2+2s)e^{xs}K(s)~ds-\int_C5K(s)~d(e^{xs})=0$

$\int_C(s^2+2s)e^{xs}K(s)~ds-[5e^{xs}K(s)]_C+\int_Ce^{xs}~d(5K(s))=0$

$\int_C(s^2+2s)e^{xs}K(s)~ds-[5e^{xs}K(s)]_C+\int_C5e^{xs}K'(s)~ds=0$

$-[5e^{xs}K(s)]_C+\int_C(5K'(s)+(s^2+2s)K(s))e^{xs}~ds=0$

$\therefore5K'(s)+(s^2+2s)K(s)=0$

$5K'(s)=-(s^2+2s)K(s)$

$\dfrac{K'(s)}{K(s)}=-\dfrac{s^2}{5}-\dfrac{2s}{5}$

$\int\dfrac{K'(s)}{K(s)}ds=\int\left(-\dfrac{s^2}{5}-\dfrac{2s}{5}\right)~ds$

$\ln K(s)=-\dfrac{s^3}{15}-\dfrac{s^2}{5}+c_1$

$K(s)=ce^{-\frac{s^3}{15}-\frac{s^2}{5}}$

$\therefore y_c=\int_Cce^{-\frac{s^3}{15}-\frac{s^2}{5}+xs}~ds$

Similar to Particular solution to a Riccati equation $y' = 1 + 2y + xy^2$, we find this general solution of $y_c$ :

$y_c=C_1\int_0^\infty e^{-\frac{t^2}{5}}\cos\left(\dfrac{t^3}{15}-xt\right)~dt+C_2\int_0^\infty\left(e^{-\frac{t^3}{15}-\frac{t^2}{5}+xt}+e^{-\frac{t^2}{5}}\sin\left(\dfrac{t^3}{15}-xt\right)\right)~dt$