Linear Second Order ODE/y'' + 4 y = 3 sine x

Theorem

The second order ODE:

$(1): \quad y + 4 y = 3 \sin x$

has the general solution:

$y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$

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:

$p = 0$

$q = 4$

$\map R x = 3 \sin x$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$(2): \quad y + 4 y = 0$

From Linear Second Order ODE: $y + 4 y = 0$, this has the general solution:

$y_g = C_1 \cos 2 x + C_2 \sin 2 x$

We have that:

$\map R x = 3 \sin x$

and it is noted that $\sin x$ is not itself a particular solution of $(2)$.

So from the Method of Undetermined Coefficients for the Sine and Cosine functions:

$y_p = A \sin x + B \cos x$

for some $A$ and $B$ to be determined.

Hence:

\(\ds y_p\)

\(=\)

\(\ds A \sin x + B \cos x\)

\(\ds \leadsto \ \ \)

\(\ds {y_p}'\)

\(=\)

\(\ds A \cos x = B \sin x\)

Derivative of Sine Function, Derivative of Cosine Function

\(\ds \leadsto \ \ \)

\(\ds {y_p}\)

\(=\)

\(\ds -A \sin x - B \cos x\)

Derivative of Sine Function, Derivative of Cosine Function

Substituting into $(1)$:

\(\ds -A \sin x - B \cos x + 4 \paren {A \sin x + B \cos x}\)

\(=\)

\(\ds 3 \sin x\)

\(\ds \leadsto \ \ \)

\(\ds \paren {-A + 4 A} \sin x\)

\(=\)

\(\ds 3 \sin x\)

\(\ds -B \cos x + B \cos x\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds 3 A\)

\(=\)

\(\ds 3\)

\(\ds -B \cos x + 4 B \cos x\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds 1\)

\(\ds B\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds y_p\)

\(=\)

\(\ds \sin x\)

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

$y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.18$: Problem $1 \ \text{(b)}$