How to find a solution to a differential equation based on another, given solution Let's say I have the DE:
$$
(x^2 - 2x)y'' - (x^2 - 2)y' + (2x - 2)y = 0
$$
And I have one possible solution to the DE:
$$
y_1(x) = e^x
$$
How would I go about solving this? I could solve the actual DE, but then what is the point of supplying a possible solution? Where does the solution $y_1$ come into play?
 A: You can proceed using Abel's integration identity. 
In general for differential equations of the form
$$
\sum\limits_{k=0}^n a_k(x)y^{(n-k)}(x)=0
$$
we can consider its solutions $y_1(x),\ldots,y_n(x)$ and define so called Wronskian
$$
W(y_1,\ldots,y_n)(x)=
\begin{pmatrix}
y_1(x)&&y_2(x)&&\ldots&&y_n(x)\\
y'_1(x)&&y'_2(x)&&\ldots&&y'_n(x)\\
\ldots&&\ldots&&\ldots&&\ldots\\
y'_1(x)&&y'_2(x)&&\ldots&&y'_n(x)\\
\end{pmatrix}
$$
Then we have the following identity
$$
\det W(x)=\det W(x_0) e^{-\int\limits_{x_0}^x \frac{a_1(t)}{a_0(t)}dt}
$$
In particular for your problem we have the following differential equation
$$
\begin{vmatrix}
y_1(x)&&y_2(x)\\
y'_1(x)&&y'_2(x)
\end{vmatrix}=C e^{-\int\frac{-(x^2-2)}{x^2-2x}dx}
$$
with $y_1(x)=e^x$. Which reduces to
$$
y'_2(x)e^x-y_2(x)e^x=C e^{\int\frac{x^2-2}{x^2-2x}dx}=C(2x-x^2)e^x
$$
After division by $e^{2x}$ we get
$$
\frac{y'_2(x)e^x-y_2(x)e^x}{e^{2x}}=C(2x-x^2)e^{-x}
$$
which is equivalent to
$$
\left(\frac{y_2(x)}{e^x}\right)'=C(2x-x^2)e^{-x}
$$
It is remains to integrate
$$
\frac{y_2(x)}{e^x}=Cx^2 e^{-x}+D
$$
and write down the answer
$$
y_2(x)=Cx^2+D e^{x}
$$
In fact this is a general solution of original equation.
A: $$
(x^2 - 2x)y'' - (x^2 - 2)y' + (2x - 2)y = 0
$$
Let's first notice that $c y_1(x)= c e^x$ is also a solution.
To find other solutions let's suppose that $c$ depends of $x$ (this method is named 'variation of constants') :
If $y(x)= c(x) e^x$ then your O.D.E. becomes :
$$
(x^2 - 2x)(c''+c'+c'+c)e^x - (x^2 - 2)(c'+c)e^x + (2x - 2)ce^x = 0
$$
$$
(x^2 - 2x)(c''+2c'+c) - (x^2 - 2)(c'+c) + (2x - 2)c = 0
$$
Of course the $c$ terms disappear and we get :
$$
(x^2 - 2x)(c''+2c') - (x^2 - 2)c' = 0
$$
Let's set $d(x)=c'(x)$ then :
$$
(x^2 - 2x)d' = (x^2 - 2)d-(x^2 - 2x)2d
$$
$$
(x^2 - 2x)d' = (-x^2 +4x- 2)d
$$
$$
\frac{d'}d = \frac{-x^2 +4x- 2}{x^2 - 2x}
$$
I'll let search the integral at the right, the answer should be ($C_0$, $C_1$, $C_2$ are constants) :
$$
\ln(d)=\ln(x^2-2x)-x+C_0
$$ 
$$
d=(x^2-2x)e^{-x}C_1
$$
but $c'=d$ so that
$$
c=C_2+C_1\int (x^2-2x)e^{-x} dx 
$$
$$
c=C_2-C_1x^2e^{-x}
$$
And we got the wished general solution :
$$
y(x)=c(x)e^x=C_2e^x-C_1x^2
$$
