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. 
 A: If you have a solution of the homogeneous equation associated to the linear equation, you can make a reduction of order on the equation. It is particularly useful here, because we know how to solve a first order linear equation. 
If we write the equation in the form 
$$ y''+p(x)y+q(x)y = g(x), $$
since we know that a solution of the associated homogeneous linear equation $ y''+p(x)y+q(x)y = 0\ $ is  $\ y_h = 1/x$, we can look for a solution to the non-homogeneous equation of the form $y_s=y_hu$, where $u$ is a function we must determine. Therefore, we have 
$$ y_s ' = y_h 'u + y_h,u' \ , \ y_s '' = y_h '' u +2y_h ' u' + y_h u'' $$
putting it in the equation leads to
$$ y_h '' u + 2 y_h ' u + y_h u'' + p(y_h ' u + y_h u') + q (y_h u) = g $$
$$ \implies u(y_h '' + p y_h ' + qy_h) + 2y_h ' u' + y_h u'' + py_h u' =g $$
$$ \implies 2y_h ' u' + y_h u'' + py_h u' = g $$
So we just have to solve this first order linear equation for $u$ to get a solution of the equation $y''+p(x)y+q(x)y = g(x)$. The same can be done to find another linear independent solution to the homogeneous equation.
A: 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} 

A: Remark. I need to write a separate answer for the other part of the question because MathJax is disturbingly slow.  However, I shall use notations from my first post.  Hence, please read that answer first.

You can use the Wronskian technique as mrtaurho suggests.   A Wronskian $W(x)$ of the differential equation is
$$W(x)=\det\left(\begin{bmatrix} \upsilon_1(x)&\upsilon_2(x)\\ \upsilon_1'(x)&\upsilon_2'(x)\end{bmatrix}\right)=-\frac{8(4x-1)}{x^2}\,,$$
where $\upsilon_1(x):=\dfrac{(4x-1)^2}{x}$ and $\upsilon_2(x):=\dfrac{1}{x}$ are linearly independent solutions I found in my first answer.  Then, a particular solution $y:=y_p$ to
$$y''(x)-u(x)\,y'(x)+v(x)\,y(x)=\frac{6(2x-1)}{4x-1}=:t(x)\tag{$\Box$}$$
is given by
$$y_p(x)=-\upsilon_1(x)\,\int\,\frac{\upsilon_2(x)\,t(x)}{W(x)}\,\text{d}x+\upsilon_2(x)\,\int\,\frac{\upsilon_1(x)\,t(x)}{W(x)}\,\text{d}x\,.$$

That is, a solution is $$\begin{align}y_p(x) &=\frac{3(4x-1)^2}{4x}\,\int\,\frac{x(2x-1)}{(4x-1)^2}\,\text{d}x-\frac{3}{4x}\,\int\,x(2x-1)\,\text{d}x \\ &=\frac{3(4x-1)^2}{4x}\,\left(\frac{8x^2-4x+1}{16(4x-1)}+\frac{1}{16}\right)-\frac{3}{4x}\,\left(\frac{2x^3}{3}-\frac{x^2}{2}\right) \\ &=\frac{3x(4x-1)}{8}-\frac{x(4x-3)}{8}=x^2 \,.\end{align}$$ Thus, all solutions to ($\Box$) take the form $$\begin{align} y(x)&=y_p(x)+A\,\upsilon_1(x)+B\,\upsilon_2(x) \\ &=x^2+A\,\left(\frac{(4x-1)^2}{x}\right)+B\,\left(\frac{1}{x}\right)=x^2+A'\,(2x-1)+\frac{B'}{x}\,,\end{align}$$ for some constants $A,A',B,B'$.



Alternatively, you can solve for a particular solution $y:=y_p$ to
$$\big(D-f(X)\big)\,\big(D-g(X)\big)\,y=t\,.$$
By setting $z_p:=\big(D-g(X)\big)\,y_p$, we get a solution
$$z_p(x)=\frac{1}{\nu(x)}\,\int\,\nu(x)\,t(x)\,\text{d}x\,,\text{ with }\nu(x):=\exp\left(-\int\,f(x)\,\text{d}x\right)=x(4x-1)\,.$$
By picking an appropriate integral constant, we have
$$z_p(x)=\frac{1}{x(4x-1)}\,\int\,6x(2x-1)\,\text{d}x=\frac{x^2(4x-3)}{x(4x-1)}=\frac{x(4x-3)}{4x-1}\,.$$

Then, we solve for a solution $y_p$ to $\big(D-g(X)\big)\,y_p=z_p$.  We get that $$y_p(x)=\frac{1}{\mu(x)}\,\int\,\mu(x)\,z(x)\,\text{d}x=\frac{(4x-1)^2}{x}\,\int\,\frac{x(4x-3)}{(4x-1)^3}\,\text{d}x\,,$$ where $\mu(x)=\dfrac{x}{(4x-1)^2}$ is as before.  Ergo, picking one integral constant, we obtain a particular solution $y_p$ to the nonhomogeneous differential equation ($\Box$): $$y_p(x)=\frac{(4x-1)^2}{x}\,\left(\frac{x^3}{(4x-1)^2}\right)=x^2\,.$$

A: HINT
First of all bring the equation in the form 
$$y''+p(x)y'+q(x)y~=~g(x)$$
Use the fact that there are two different ways to compute the Wronskian of a fundamental set of solutions:
$$\begin{align}
&(1)W(y_1,y_2)~=~\begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix}~=~y_1y_2'-y_2y_1'\\
&(2)W(y_1,y_2)~=~e^{-\int p(x)dx}
\end{align}$$
From thereone you can compute the general solution to the homogenous equation by solving the ODE for $y_2$. After this you can do variation of constants to determine the solution to your given equation.
A: Not an answer just a hint
$$\left(4x^2-x\right)y''+2\left(2x-1\right)y'-4y=0$$
Assume that $y_2=P(x)$ is a solution. Where
$$P(x)=\sum_{n=0}^ma_nx^n$$
Whats the degree of $P(x)$ ? plug the solution you get that
$$a_nn(n-1)x^n=0 \,\, \forall x \implies n=0, n=1 $$
For $n=0$ you get the trivial solution $y=0$
Try $n=1$
$$ y_2=ax+b$$
$$2\left(2x-1\right)a-4(ax+b)=0$$
$$\implies a=-2b$$
Therefore
$$y_2=-2bx+b=b(1-2x)$$
Therefore the solution for the homogeneous equation is
$$y_h=\frac {c_1}x+c_2(1-2x)$$
Use variation of constant method to find $y_p$
A: Warning. This is a very wild solution for the first part of the problem (i.e., solving the homogeneous differential equation), intended only to illustrate its beauty (if you would agree), and not intended to be practical.  This solution hinges on some heavy guess work, but it does not assume to know any solution to the required differential equaiton. That is, I was looking for a solution $g$ to the differential equation
$$g'(x)+\big(g(x)\big)^2-u(x)\,g(x)+v(x)=0\,,\tag{$\star$}$$
where $u(x):=-\dfrac{2(2x-1)}{x(4x-1)}$ and $v(x):=-\dfrac{4}{x(4x-1)}$.  I guessed that $g(x)=\dfrac{ax+b}{x(4x-1)}$ for some $a$ and $b$, and then I solved for $a$ and $b$.  I shall omit the gruesome detail of how I obtained $a$ and $b$, but it is really nothing but expanding terms in ($\star$).

Let
$$f(x):=-\frac{8x-1}{x(4x-1)}\text{ and }g(x):=\frac{4x+1}{x(4x-1)}\,.$$ Observe that
$$f(x)+g(x)=u(x)\text{ and }f(x)\,g(x)-g'(x)=v(x)\,.$$
Define the operators $D$ and $X$ by $(D\,h)(x):=h'(x)$ and $(X\,h)(x):=x\,h(x)$.  For any function $\phi$, we also define the operator $\phi(X)$ to be $\big(\phi(X)\,h\big)(x):=\phi(x)\,h(x)$.  Note that
$$D^2-u(X)\,D+v(X)=\big(D-f(X)\big)\,\big(D-g(X)\big)\,.\tag{*}$$
Suppose that a function $y$ satisfies $$\big(D^2-u(X)\,D+v(X)\big)\,y=0\,.\tag{#}$$  Let $z:=\big(D-g(X)\big)\,y$.  By (*) and (#),
$$\big(D-f(X)\big)\,z=0\,.$$
Consequently,
$$z'(x)-f(x)\,z(x)=0\text{ or }z(x)=\exp\left(\int\,f(x)\,\text{d}x\right)\,.$$
This is a not-so-difficult work, and the result is
$$z(x)=\frac{k}{x(4x-1)}\text{ for some constant }k\,.$$

Now, we want to solve $\big(D-g(X)\big)\,y=z$, or $$y'(x)-g(x)\,y(x)=z(x)\,.$$ That is, $$y(x)=\frac{1}{\mu(x)}\,\int\,\mu(x)\,z(x)\,\text{d}x\,,\text{ with }\mu(x):=\exp\left(-\int\,g(x)\,\text{d}x\right)\,.$$ We see that $\mu(x)=\dfrac{x}{(4x-1)^2}$.  Therefore, $$y(x)=\frac{(4x-1)^2}{x}\,\int\,\frac{x}{(4x-1)^2}\,\left(\frac{k}{x(4x-1)}\right)\,\text{d}x\,.$$ Ergo, $$y(x)=\frac{(4x-1)^2}{x}\,\Biggl(A+B\,\left(\frac{1}{(4x-1)^2}\right)\Biggr)\,,$$ where $A$ is a constant and $B:=-\dfrac{k}{8}$. This means $$y(x)=A\,\left(\frac{(4x-1)^2}{x}\right)+B\left(\frac{1}{x}\right)\,.$$

Thus, the given solution $y_1(x)=\dfrac{1}{x}$ corresponds to $(A,B):=(0,1)$.  The linear solution $y_2(x)=2x-1$ found by Isham comes from $(A,B):=\left(\dfrac{1}{8},-\dfrac{1}{8}\right)$.  

P.S.  This solution is inspired by a method of solving the quantum harmonic oscillator problem.  One such method writes $$D^2-X^2+I=(D-X)\,(D+X)\,,$$
where $I$ is the identity operator.
