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

## Theorem

Consider the nonhomogeneous linear second order ODE:

$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = \map R x$

Let $\map {y_g} x$ be the general solution of the homogeneous linear second order ODE:

$(2): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$

Let $\map {y_p} x$ be a particular solution of $(1)$.

Then $\map {y_g} x + \map {y_p} x$ is the general solution of $(1)$.

## Proof

Let $\map {y_g} {x, C_1, C_2}$ be a general solution of $(2)$.

Note that $C_1$ and $C_2$ are the two arbitrary constants that are to be expected of a second order ODE.

Let $\map {y_p} x$ be a certain fixed particular solution of $(1)$.

Let $\map y x$ be an arbitrary particular solution of $(1)$.

Then:

 $\ds$  $\ds \paren {y - y_p} + \map P x \paren {y - y_p}' + \map Q x \paren {y - y_p}$ $\ds$ $=$ $\ds \paren {y + \map P x y' + \map Q x y} - \paren {y_p + \map P x y_p' + \map Q x y_p}$ $\ds$ $=$ $\ds \map R x - \map R x$ $\ds$ $=$ $\ds 0$

We have that $\map {y_g} {x, C_1, C_2}$ is a general solution of $(2)$.

Thus:

$\map y x - \map {y_p} x = \map {y_g} {x, C_1, C_2}$

or:

$\map y x = \map {y_g} {x, C_1, C_2} + \map {y_p} x$

$\blacksquare$