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 Linear 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:

\(\ds y_p\)

\(=\)

\(\ds A K\)

\(\ds \leadsto \ \ \)

\(\ds {y_p}'\)

\(=\)

\(\ds 0\)

Derivative of Constant

\(\ds \leadsto \ \ \)

\(\ds {y_p}\)

\(=\)

\(\ds 0\)

Derivative of Constant

Substituting into $(1)$:

\(\ds A K\)

\(=\)

\(\ds K\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds 1\)

Our work is done.

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

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

$\blacksquare$