Solve the Second Order Nonhomogeneous ODE I'm stuck on this problem since I can't find any material in my textbook or internet to solve this. Any help?
$$2x^2 \cdot y'' - x \cdot y' + y = x$$
$$y'' - \frac{y'}{2x} + \frac{y}{2x^2} = \frac{1}{2x}$$
 A: The equation is 
$$ 2x^2y'' - xy' + y = x $$
This is an inhomogeneous Cauchy-Euler equation of second order. Make the substitution $x = e^t$ and $y(x) = u(t)$ to get
$$ 2u'' - 3u' + u = e^t $$
Which you can solve using the method of undetermined coefficients to obtain the general solution
$$ u(t) = te^t + c_1e^t + c_2e^{t/2} $$
$$ \implies y(x) = x\ln x + c_1 x + c_2 \sqrt{x} $$

An alternative approach is try to find solutions of the form $y = x^m$. Plugging this into the original equation, we obtain the characteristic polynomial
$$ 2m(m-1) - m + 1 = 0 $$
which has roots $m=1$, $m=1/2$. Therefore the homogeneous solution is
$$ y_h(x) = c_1x + c_2\sqrt{x} $$
Knowing this, we can use variation of parameters to find a particular solution of the form
$$ y_p(x) = x\ v_1(x) + \sqrt{x}\ v_2(x) $$
where the coefficient functions satisfy
$$ \left\{ \begin{aligned} x{v_1}' + \sqrt{x}{v_2}' &= 0 \\
{v_1}' + \frac{1}{2\sqrt{x}}{v_2}' &= \frac{1}{2x} \end{aligned} \right. $$
Solving the above system gives
$$ {v_1}' = \frac{1}{x}, \quad {v_2}' = -\frac{1}{\sqrt{x}} $$
Integrating and recombining gives $y_p = x\ln x - 2x$, of which the second term is absorbed into $y_h$
A: Here is another way:
$$y'' - \frac{y'}{2x} + \frac{y}{2x^2} = \frac{1}{2x}$$
$$y'' -\frac 1 2 ( \frac{y'x}{x^2} - \frac{y}{x^2})= \frac{1}{2x}$$
$$y'' -\frac 1 2 ( \frac{y'x-y}{x^2})= \frac{1}{2x}$$
$$y'' -\frac 1 2 ( \frac{y}{x})'= \frac{1}{2x}$$
Integrate:
$$y' -\frac{y}{2x}=\int \frac{dx}{2x}$$
$$y' -\frac{y}{2x}=\frac{\ln(x)}{2}+K$$
Which is easy to solve ....
