How to solve this ODE with powerseries approach I want to solve this second order inhomogeneous ODE:
$$ x^2y''-3xy'+4y= \log x $$
I know this is a Cauchy-Euler ODE but i s there a way to solve it with powerseries approach?  with
$$ y= \sum_{n=0}^{ \infty} a_n x^n , y'= \sum_{n=1}^{ \infty}na_n x^{n-1}, y''=\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}$$
thanks for any help!
 A: By "chance", plugging $y=\log x$ in the LHS we get $$-\frac{x^2}{x^2}-3\frac xx+4\log x$$ so that $$\frac{\log x + 1}4$$ is a particular solution.
Now in the power series for the homogeneous equation, the coefficient of the powers of $x$ are
$$0\to 4a_0,
\\1\to-3a_1+4a_1,
\\2\to2a_2-6a_2+4a_2=0,
\\n>1\to n(n-1)a_n-3na_n+4a_n=(n-1)^2a_n.$$
Finally, only the term
$$a_2x^2$$ is possible.
A: $$x^2y''-3xy'+4y=0$$
Solve the homogeneous differential equation first:
$$4\sum_{n=0}^{ \infty} a_n x^n -3x \sum_{n=1}^{ \infty}na_n x^{n-1}+x^2\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}=0$$
For $n=0 \implies a_0=0$
For $n=1 \implies -3aa_1+4a_1=0 \implies a_1=0$
For $ \ge 2$ we have:
$$\sum_{n=2}^{ \infty} (4a_n  -3 na_n + n(n-1)a_n )x^{n}=0$$
$$\sum_{n=2}^{ \infty} (n-2)^2a_nx^{n}=0$$
This must be true $\forall \; x$ so that the coefficents must be equal zero:
$$(n-2)^2a_n=0$$
$$a_n =0 \;\; \forall \;\; n \ne 2$$
So that the solution is:
$$y_1=\sum_{n=0}^\infty a_nx^n=a_2x^2$$
You can use $y_2=y_1v(x)$ to find the second solution to the homogeneous differential equation.
For the particular solution Yves has already answered it.
