Cauchy Euler Equation How do I find the particular integral of 
$$(x^2D^2+xD-1)y=x^2 e^{2x}$$
where $D=\frac{d}{dx}$? I have tried using the substitution $x=e^z$, but I got stuck. 
 A: You should have first tried the direct Euler-Cauchy approach by computing the characteristic polynomial $0=m(m-1)+m-1=m^2-1$ giving $x,x^{-1}$ as basis solutions. The right side is not of the form $x^r\ln(x)^p$ or a sum of such terms, so that the method of undetermined coefficients does not work. 
Use the variation-of-constant method
\begin{align}
y_p(z)&=xv(x)+x^{-1}w(x),\\
0&=xv'(x)+x^{-1}w'(x)\\
y_p'(x)=v(x)-x^{-2}w(x)
x^2e^{2x}=x(xy_p'(x))'-y_p(x)&=x^2v'(x)-w'(x)
\end{align}
so that one coefficient integration is readily solvable
$$
v'(x)=\frac12e^{2x}\implies v(x)=\frac14e^{2x}+C
$$
while the more complicated reads as
\begin{align}
w'(x)=-\frac12x^2e^{2x}\implies 
w(x)=-\frac12\int x^2e^{2x}dx
&=-\frac14x^2e^{2x}+\frac12\int xe^{2x}dx
\\
&=-\frac14x^2e^{2x}+\frac14xe^{2x}-\frac18e^{2x}+D
\end{align}
Combined this results in
\begin{align}
y_p(x)&=\frac14e^{2x}-\frac18x^{-1}e^{2x}\\
&\color{blue}{\text{test: }\begin{aligned}[t]
xy_p'(x)&=\frac12xe^{2x}+\frac18x^{-1}e^{2x}-\frac14e^{2x}\\
x(xy_p'(x))'&=\frac12xe^{2x}+x^2e^{2x}-\frac18x^{-1}e^{2x}+\frac14e^{2x}-\frac12xe^{2x}
\\
&=x^2e^{2x}+y_p(x)
\end{aligned}}
\end{align}
A: Hint: try $y=\sum\limits_{k=0}^\infty a_nx^{n}$. The coefficient of $x^{n}$ on the left side becomes $(n^{2}-1)a_n$. Equate this to the coefficient of $x^{n}$ on the right side. 
A: Render $y=u/x$ where $1/x$ is the homogeneous solution with the lower power of $x$ (using the other solution could complicate the integral you eventually get by forcing unneeded fractions in the integrand).  Terms containing $u$ with no derivative cancel out and then
$x\dfrac{d^2u}{dx^2}-\dfrac{du}{dx}=x^2e^{2x}$
$\color{blue}{\text{When you "vary the coefficient" on a homogeneous solution,}}$
$\color{blue}{\text{adding a constant just adds multiples of the homogeneous solution.}}$
$\color{blue}{\text{So only derivatives of your variable coefficient will have any effect.}}$
Introduce an integrating factor $1/x^2$ which matches the left side with the product rule for derivatives:
$\dfrac{1}{x}\dfrac{d^2u}{dx^2}-\dfrac{1}{x^2}\dfrac{du}{dx}=e^{2x}$
$\dfrac{d}{dx}(\dfrac{1}{x}\dfrac{du}{dx})=e^{2x}$
$\color{blue}{\text{The ratio of homogeneous solutions is }x^{-1}/x=x^{-2}.}$
$\color{blue}{\text{That will become the coefficient of the first derivative in the exact differential.}}$
$\color{blue}{\text{We choose our integrating factor accordingly.}}$
Integrating (and leaving out constants since only a particular solution is needed):
$\dfrac{du}{dx}=\dfrac{xe^{2x}}{2}$
$u=\dfrac{(x-2)e^{2x}}{4}$ with integration by parts
Back substituting:
$\color{blue}{y=\dfrac{(x-2)e^{2x}}{4x}}$
