Second Order ODE/x y'' - (2 x + 1) y' + (x + 1) y = 0

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

has the general solution:
 * $y = C_1 e^x + C_2 x^2 e^x$

Proof
Note that:
 * $x - \paren {2 x + 1} + \paren {x + 1} = 0$

so if $y'' = y' = y$ we find that $(1)$ is satisfied.

So:

and so:
 * $y_1 = e^x$

is a particular solution of $(1)$.

$(1)$ can be expressed as:
 * $(2): \quad y'' - \dfrac {2 x + 1} x y' + \dfrac {x + 1} x y = 0$

which is in the form:
 * $y'' + \map P x y' + \map Q x y = 0$

where:
 * $\map P x = -\dfrac {2 x + 1} x$

From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
 * $\map {y_2} x = \map v x \, \map {y_1} x$

where:
 * $\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.

We have that:

Hence:

and so:

From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
 * $y = C_1 e^x + k \dfrac {x^2} 2 e^x$

and so setting $C_2 = \dfrac k 2$:
 * $y = C_1 e^x + C_2 x^2 e^x$