Non-homogeneous 2nd order Euler-Cauchy differential equation
Sophia Terry 
I have the following differential equation to solve:
$$x^2y''-2y=x^3e^x$$
I've identified the homogeneous equation as an Euler-Cauchy equation, and I've managed to obtain the complementary solution, being:
$$y_c=c_1\frac{1}{x}+c_2x^2$$
But I have no idea how to obtain the particular solution. How can I do it?
$\endgroup$4 Answers
$\begingroup$Let me change your notation and put $y_h(x)$ as the solution you have obtained. Then:
$$y(x) = c_1 y_1(x) + c_2 y_2(x) + y_p(x),$$
where $y_p(x)$ is the particular solution of your ode. Then make use of variation of parameters and try solutions $y(x) = C(x) y_1(x)$ (for example; you could have chosen $y = C y_2$). Substitute back in the ode and you will have:
$$x^2 (C''y_1 + 2 C' y_1' + C y_1'') - 2 C y_1 = C'' x^2 + C'(2y_1' x^2/y_1) + C(x^2 y''_1 - 2 y_1) =x^3 e^x,$$
the coefficient in $C$ must vanish since $y_1$ is a solution of the homogenous part of the equation. Simplify a bit and obtain a "false" second order differential equation for $C(x)$:
$$ C'' + C' 2y'_1/y_1 = x e^x/y_1,$$
which can be solved in terms of an integrating factor, $u = e^{\int 2y'_1/y_1 \,dx} = y_1^2$, as follows:
$$\frac{d}{dx}\left( C' u\right) = u x e^x/y_1 = y_1 x e^x \Leftrightarrow C' = \frac{1}{y_1^2} \int y_1 x e^x \, dx + \frac{A}{y_1^2},$$
and hence the solution for $C$:
$$C = B + A \int\frac{dx}{y_1^2} + \int\left(\frac{1}{y_1^2} \int y_1 x e^x \, dx\right) \, dx,$$
where $A$ and $B$ are constants of integration. This readily leads you to the solution, $y(x) = C(x) y_1$:
$$y(x) =\frac{A x^2}{3}+\frac{B}{x}+\frac{2 e^x}{x}+e^x( x -2 ),$$
where you can make $c_1 = \frac{A}{3}$ (and $c_2 = B$).
I hope this may be useful to you.
Cheers!
$\endgroup$ 1 $\begingroup$Further to the other posts above, we can manipulate to yield a tractable equation.
Try using the fact: $$ x^{2}\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}\right)-2x\frac{\mathrm{d}y}{\mathrm{d}x}. $$ Inserting into the original equation, yields: $$ \frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}\right)-2x\frac{\mathrm{d}y}{\mathrm{d}x}-2y = x^{3}\mathrm{e}^{x} $$ Hopefully you can see that the second and third term in is just simply $$ -2\frac{\mathrm{d}}{\mathrm{d}x}\left(xy\right). $$
Combining them all into a single equation $$ \frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}\right)-2\frac{\mathrm{d}}{\mathrm{d}x}\left(xy\right) = x^{3}\mathrm{e}^{x} $$ Which you should be able to solve for there. If you need any further help please ask :).
$\endgroup$ $\begingroup$If $$ z(\log x)=y(x), $$ then $$ z'(\log x)/x=y'(x),\quad \big(z''(\log x)-z'(\log x)\big)/x^2=y''(x). $$ Thus $$ x^2y''-2y=z''(\log x)-z'(\log x)-2z(\log x)=x^3\mathrm{e}^{x} $$ or $$ z''(x)-z'(x)-2z(x)=\mathrm{e}^{3x}\mathrm{e}^{\mathrm{e}^x}. $$ I can't see any simple special solution for this equation. You need to use the variation of parameters to obtain an expression of the solution.
$\endgroup$ $\begingroup$For the particular solution the RHS divided by the term besides $y''$ so
$$x^2y''-2y=x^3e^x \qquad \rightarrow \qquad y''-2\frac{y}{x^2}=xe^x$$
$$f(x)=xe^x$$ From here, depends of you if you want Undeterminated Coefficientes method of Variation of Parameters.
I did it by Variation of Parameters $\ y_1=x^{-1}\ $,$\ y_2=x^2\ $, $\ W=3\ $, $\ W_1=-x^3e^x\ $, $\ W_2=e^x\ $
$$y_p(x)=e^x(x-2+2x^{-1})$$
$$y(x)=(2e^x+C_1)x^{-1}+C_2x^2+(x-2)e^x$$
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