Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/y(1) = 3, y'(1) = 5

Theorem
The second order ODE:
 * $(1): \quad x^2 y'' - 2 x y' + 2 y = 0$

with initial conditions:
 * $\map y 1 = 3$
 * $\map {y'} 1 = 5$

has the particular solution:
 * $y = x + 2 x^2$

Proof
From Second Order ODE: $x^2 y'' - 2 x y' + 2 y = 0$, the general solution of $(1)$ is:
 * $y = C_1 x + C_2 x^2$

Differentiating $x$:
 * $y' = C_1 + 2 C_2 x$

Thus for the initial conditions:

and:

Hence the result.