Solution of differential equation $(x-3)^2 y'' + (x-1)y' + y = 0$ How do you solve the following differential equation?
\begin{equation}
(x-3)^2 y'' + (x - 1)y' + y = 0, x \ne 3
\end{equation}
 A: I am not proeficient in math but here is my try:
$$
\begin{bmatrix}
y''\\ y'
\end{bmatrix}=
\begin{bmatrix}
-\frac{x-1}{(x-3)^2} & \frac{1}{(x-3)^2}\\ 
1 & 0
\end{bmatrix}*
\begin{bmatrix}
y'\\ y
\end{bmatrix}
$$
Find eigenvalues and eigenvectors if can't compute $e^A$, solution:
$\begin{bmatrix}
y'\\ y
\end{bmatrix} = e^{Ax}*\begin{bmatrix}
y'(0)\\ y(0)
\end{bmatrix}$
$det(A-I \lambda) = (x-3)^2\lambda^2-(x-1)\lambda-1=0$,
$\lambda_{1,2}=\frac{(1-x)\pm\sqrt{5x^2-26x+37}}{2(x-3)^2}$
(MATLAB help)
$V=\begin{bmatrix}
-(x + (5*x^2 - 26*x + 37)^(1/2) - 1)/(2*(x^2 - 6*x + 9)) & ((5*x^2 - 26*x + 37)^(1/2) - x + 1)/(2*(x^2 - 6*x + 9))\\ 
1 & 1
\end{bmatrix}$
then  (use $A^{-1}=\frac{adj(A^T)}{det(A)}$):
$\begin{bmatrix}
y'\\ y
\end{bmatrix} = V\begin{bmatrix}
e^{\lambda_1x} & 0\\ 
0 & e^{\lambda_2x}
\end{bmatrix}V^-1*\begin{bmatrix}
y'(0)\\ y(0)
\end{bmatrix}$
... however i procced with program help, i got (MATLAB help):
$$
e^{Ax}=
\begin{bmatrix}
e^{-(x^2 - x)/(x - 3)^2} & e^{x/(x - 3)^2}\\ 
e^{x} & 1
\end{bmatrix}
$$
I don't know if this is correct, nor how to write the solution family form (I think it should be any linear combination of eigenvectors, but i'm not sure).
