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

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

has the general solution:
 * $y = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$

Proof
$(1)$ can be manipulated into the form:
 * $y'' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = x + 1$

It can be seen that this is a nonhomogeneous linear second order ODE in the form:
 * $y'' + P \left({x}\right) y' + Q \left({x}\right) y = R \left({x}\right)$

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

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $\left({x^2 + x}\right) y'' + \left({2 - x^2}\right) y' - \left({2 + x}\right) y = 0$

From Second Order ODE: $\left({x^2 + x}\right) y'' + \left({2 - x^2}\right) y' - \left({2 + x}\right) y = 0$, this has the general solution:
 * $y_g = C_1 e^x + \dfrac {C_2} x$

It remains to find a particular solution $y_p$ to $(1)$.

Expressing $y_g$ in the form:
 * $y_g = C_1 y_1 \left({x}\right) + C_2 y_2 \left({x}\right)$

we have:

By the Method of Variation of Parameters, we have that:


 * $y_p = v_1 y_1 + v_2 y_2$

where:

where $W \left({y_1, y_2}\right)$ is the Wronskian of $y_1$ and $y_2$.

We have that:

Hence:

It follows that:

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:


 * $y = y_g + y_p = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$

is the general solution to $(1)$.