<s>[INST] General solution of $(4x^2-x)y''+2(2x-1)y'-4y=12x^2-6x$ given $y_1=\frac1x$ Verify that $y_1=\frac1x$ is a solution to d.e
$\left(4x^2-x\right)y''+2\left(2x-1\right)y'-4y=0$
Find the general solution of the d.e 
$\left(4x^2-x\right)y''+2\left(2x-1\right)y'-4y=12x^2-6x$
Note: I was able to prove initial value $y_1=\frac{1}{x}$ however confused on second part. 
 [/INST] Since you already know one solution to the homogeneous equation, a straightforward method is reduction of order. You can follow the Wikipedia page for a general formula.
Let's first convert to standard form:
$$ y'' + \frac{4x-2}{x(4x-1)}y' - \frac{4}{x(4x-1)}y = \frac{12x-6}{4x-1} $$
The basic idea is to try a solution of the form $$ y(x) = y_1(x)u(x)= \frac{1}{x}u(x) $$
Plugging this in, we obtain
$$ u'' + \left(-\frac{2}{x} + \frac{4x-2}{x(4x-1)} \right)u' = \frac{12x^2-6x}{4x-1} $$
which is a standard first-order ODE in $u'$ that you can solve using the integration factor.
Here's the full working

 \begin{align} u'' - \frac{4}{4x-1}u' &= \frac{12x^2-6x}{4x-1} \\ \frac{u''}{4x-1} - \frac{4u'}{(4x-1)^2} &= \frac{12x^2-6x}{(4x-1)^2} \\ \left(\frac{u'}{4x-1}\right)' &= \frac34 - \frac{3/4}{(4x-1)^2} \\ \frac{u'}{4x-1} &= \frac{3x}{4} + \frac{3/16}{4x-1} + C \\ u' &= 3x^2 + \left(C-\frac{3}{16}\right)(4x-1) \\ u(x) &= x^3 + c_1(2x^2-x) + c_2 \\ y(x) &= x^2 + c_1(2x-1) + \frac{c_2}{x} \end{align} 

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