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

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

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

Proof
Note that:
 * $x - \left({2 x + 1}\right) + \left({x + 1}\right) = 0$

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:
 * $y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

where:
 * $P \left({x}\right) = - \dfrac {2 x + 1} x$

From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
 * $y_2 \left({x}\right) = v \left({x}\right) y_1 \left({x}\right)$

where:
 * $\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \, \mathrm d x} \, \mathrm d x$

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