Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0/Proof 2

Proof
Note that:


 * $\paren {x^2 + x} + \paren {2 - x^2} - \paren {2 + x} = 0$

so $\map {y_1} x$ such that $y_1 = {y_1}' = {y_1}''$ satisfies $(1)$.

Hence:
 * $y_1 = e^x$

is a particular solution of $(1)$.

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

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

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

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

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:


 * $y = C_1 e^x + \dfrac {C_2} x$