Linear Second Order ODE/y'' + y = K

Theorem
The second order ODE:
 * $(1): \quad y'' + y = K$

has the general solution:
 * $y = C_1 \sin x + C_2 \cos x + K$

Proof
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
 * $y'' + p y' + q y = \map R x$

where:
 * $p = 0$
 * $q = 1$
 * $\map R x = K$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $(2): \quad y'' + y = 0$

From Second Order ODE: $y'' + y = 0$, this has the general solution:
 * $y_g = C_1 \sin x + C_2 \cos x$

We have that:
 * $\map R x = K$

So from the Method of Undetermined Coefficients for Polynomial:
 * $y_p = A K$

where $A$ is to be determined.

Hence:

Substituting into $(1)$:

Our work is done.

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


 * $y = y_g + y_p = C_1 \sin x + C_2 \cos x + K$