Particular integral of a differential equation having a complicated exponential function So I have to solve the differential equation
$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=x^2e^{2x}$
I identified this as a differential in Cauchy-Euler form, and used the substitution x=e^t to obtain $\frac{d^2y}{dx^2}-y=e^{2(t+e^t)}$
The Complementary function is pretty easy. Now I cannot understand how to find the Particular Integral.
I rearranged the equation as $y=\frac{e^{2t}.e^{2e^t}}{D^2-1}$ where $D=\frac{d}{dt}$.
Then I substituted $D+2$ in place of $D$ according to the known rule. Now what should I do? Also, is what I've done till now correct? I've worked with differential equations where exponential functions were paired with trigonometric or linear functions, but not such functions before. Any help is appreciated. Thanks in advance.
 A: Another way: The ODE is linear and exact so it can be written as $$\frac{\mathrm d}{\mathrm dx} \left(P(x) y'(x)+Q(x) y(x)\right)=x^2\exp(2x)$$ for some functions $P$ and $Q$. By collecting the derivatives of $y$ you get a (very!) easy equation system in $P,Q,P',Q'$ with solution $P(x)=x^2$ and $Q(x)=-x$. So the equation can be written as $$\frac{\mathrm d}{\mathrm dx} \left(x^2 y'(x)-x y(x)\right)=x^2\exp(2x).$$ Now integrate both sides (the inhomogenous term can be integrated by parts) $$x^2 y'(x)-x y(x)=\frac14 \exp(2x)(2x^2-2x+1)+c_1.$$ By dividing by $x^2$ we get the ODE $$y'(x)=a(x) y(x)+b(x)$$ where $a(x)=\frac1x$ and $b(x)=\frac{1}{4x^2}\big(\exp(2x)(2x^2-2x+1)+c_1\big)$.
By variation of parameters, the general solution is $$y(x) = e^{A(x)}\cdot\left[\int_{x_0}^x b(t)e^{-A(t)}{\rm d}t + c_2 \right]$$ where $A(x)=\int_{x_0}^x a(t)\,\mathrm dt$ and $x_0$ is arbitrary. In our case this is equal to $$y(x) =\frac{c_1}{x}+c_2 x+\frac{e^{2 x} (2 x-1)}{8 x}$$
A: Why not to try instead
$$y=\frac z x \implies x z''-z'=x^2 e^{2x}$$ Now, reduction of order $p=z'$ to get
$$x p'-p=x^2 e^{2x}\implies p=\frac{1}{2}x e^{2 x} +c_1 x$$ It looks to be simple.
A: $$\frac{d^2y}{dx^2}-y=e^{2(t+e^t)}$$
$$(\frac{d^2y}{dx^2}\color {red}{-y')+(y'}-y)=e^{2(t+e^t)}$$
Integrating factor is $ \mu(t)=e^{-t}$ :
$$(y'e^{-t})'+(ye^{-t})'=e^{(t+2e^t)}$$
Integrate:
$$y'e^{-t}+ye^{-t}=\int e^{(t+2e^t)}dt$$
Substitute $$u=e^t \implies du =udt$$
$$y'e^{-t}+ye^{-t}=\int e^{2u}du$$
$$y'e^{-t}+ye^{-t}=\frac 12 e^{2e^t}+C_1$$
$$(ye^{t})'=\frac 12 e^{2t+2e^t}+C_1e^{2t}$$
Integrate again to get the final answer.
$$ye^{t}= \frac 12 \int e^{2t+2e^t}dt+C_1e^{2t}+C_2$$
$$ye^{t}=  \frac 12 \int ue^{2u}du+C_1e^{2t}+C_2$$
$$ye^{t}=  \frac 14  e^te^{2e^t}-\frac 18  e^{2e^t}+C_1e^{2t}+C_2$$
Finally:
$$\boxed {y(t)=  \frac 14  e^{2e^t}-\frac 18  e^{2e^t-t}+C_1e^{t}+C_2e^{-t}}$$
Note that the easiest way is to write it as
$$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=x^2e^{2x}$$
$$\frac{d^2y}{dx^2}+\frac{dy}{xdx}-\frac 1 {x^2}y=e^{2x}$$
$$\frac{d^2y}{dx^2}+(\frac y{x})'=e^{2x}$$
And integrate:
$$y'+\frac y{x}=\frac 12e^{2x}+C_1$$
$$xy'+ y=\frac 12xe^{2x}+C_1x$$
$$(xy)'=\frac 12xe^{2x}+C_1x$$
$$xy=\frac 14xe^{2x}-\frac 18e^{2x}+C_1x^2+C_2$$
$$\boxed {y=\frac 14e^{2x}-\frac 1 {8x}e^{2x}+C_1x+\dfrac {C_2}x}$$
