<s>[INST] Non-homogenous differential equation help 

*a) Given $x = e^u$ and $$x^2\frac{d^2y}{dx^2} - 4x \frac{dy}{dx} + 6y = 12$$ show that $$\frac{d^2y}{du^2} - 5 \frac{dy}{du} + 6y = 12$$
I was able to do this:
$x = e^u$
$$\frac{dx}{du} = e^u = x $$
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{1}{x}\frac{dy}{du}$$
$$\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x}\frac{d^2y}{du^2}\frac{du}{dx} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}$$
$$x^2\left(-\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}\right) -4x\left(\frac{1}{x}\frac{dy}{du}\right) + 6y = 12 $$
$$-\frac{dy}{du} + \frac{d^2y}{du^2} - 4\frac{dy}{du} +6y = 12 $$
$$\frac{d^2y}{du^2} - 5\frac{dy}{du} + 6y = 12 $$
part b is where I got stuck:
b) Hence solve the equation $$x^2\frac{d^2y}{dx^2} - 4x\frac{dy}{dx} + 6y = 12$$ given that $y(1) = 7$ and $y(2) = 14$
$$\frac{d^2y}{du^2} - 5\frac{dy}{du} + 6y = 12 $$
auxiliary equation: $\lambda^2 - 5λ + 6 = 0 $
Hence $\lambda = 3, 2$
Complementary function: $$y = Ae^{3u} + Be^{2u}$$
particular integral in the form $y = c$
$$\frac{dy}{dx} = 0 = \frac{d^2y}{dx^2}$$
$$6C = 12 $$
$$C = 2$$
general solution: $$y = Ae^{3u} + Be^{2u} + 2 $$
where I began to get stuck:
replacing $u = \ln x$ means you get left with
$$y = Ax^3 + Bx^2 + 2$$
$$\frac{dy}{dx} = 3Ax^2 + 2Bx $$
$$\frac{d^2y}{dx^2} = 6Ax + 2B $$
my teacher told me to "substitute this into the solution of the DE and just apply the conditions" 
(this is where I think I may me wrong, I think I might have to equate the differential equation to 0 after substituting)
through substituting these values into the differential equation you get:
$$x^2 (6Ax + 2B) - 5(3Ax^2 + 2Bx) + 6 (Ax^3 + Bx^2 + 2) = 12 $$
$$ 6Ax^3 + 2Bx^2 -15Ax^2 - 10Bx + 6Ax^3 + 6Bx^2 + 12 = 12 $$
$$ 12Ax^3 + 8Bx^2 -15Ax^2 - 10Bx = 0 $$
This may resolve very simply, but I didn't want to try without knowing I was doing the correct thing.
 [/INST] You're correct that the general solution is 
$$ y(x) = Ax^3 + Bx^2 + 2 $$
Using the given boundary conditions we have
\begin{align}
y(1) &= A + B + 2 = 7 \\
y(2) &= 8A + 4B + 2 = 14
\end{align}
or 
\begin{align}
A + B &= 5\\
8A + 4B &= 12
\end{align}
Solving the above system gives $A=-2$, $B=7$
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