Linear Second Order ODE/y'' + 2 y' + 5 y = x sin x

Theorem

The second order ODE:

$(1): \quad y + 2 y' + 5 y = x \sin x$

has the general solution:

$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \ldots$

Proof

It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:

$y + p y' + q y = \map R x$

where:

$p = 2$

$q = 5$

$\map R x = x \sin x$

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

$y + 2 y' + 5 y = 0$

From Linear Second Order ODE: $y + 2 y' + 5 y = 0$, this has the general solution:

$y_g = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$

It remains to find a particular solution $y_p$ to $(1)$.

We have that:

$\map R x = x e^{-x}$

and so from the Method of Undetermined Coefficients for Product of Polynomial and Function of Sine and Cosine, we assume a solution:

\(\ds y_p\)

\(=\)

\(\ds \paren {A_1 \sin x + A_2 \cos x} \paren {B_1 x + B_2}\)

\(\ds \leadsto \ \ \)

\(\ds {y_p}'\)

\(=\)

\(\ds \paren {A_1 \cos x - A_2 \sin x} \paren {B_1 x + B_2} + B_1 \paren {A_1 \sin x + A_2 \cos x}\)

\(\ds \leadsto \ \ \)

\(\ds {y_p}\)

\(=\)

\(\ds \paren {-A_1 \sin x - A_2 \cos x} \paren {B_1 x + B_2} + B_1 \paren {A_1 \cos x - A_2 \sin x} + B_1 \paren {A_1 \cos x - A_2 \sin x}\)

\(\ds \)

\(=\)

\(\ds \paren {-A_1 \sin x - A_2 \cos x} \paren {B_1 x + B_2} + 2 B_1 \paren {A_1 \cos x - A_2 \sin x}\)

Substituting in $(1)$:

\(\ds \)

\(\)

\(\ds \paren {-A_1 \sin x - A_2 \cos x} \paren {B_1 x + B_2} + 2 B_1 \paren {A_1 \cos x - A_2 \sin x}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds 2 \paren {\paren {A_1 \cos x - A_2 \sin x} \paren {B_1 x + B_2} + B_1 \paren {A_1 \sin x + A_2 \cos x} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds 5 \paren {\paren {A_1 \sin x + A_2 \cos x} \paren {B_1 x + B_2} }\)

\(\ds \)

\(=\)

\(\ds x \sin x\)

\(\ds \leadsto \ \ \)

\(\ds \)

\(\)

\(\ds x \sin x \paren {-A_1 B_1 - 2 A_2 B_1 + 5 A_1 B_1}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds x \cos x \paren {-A_2 B_1 + 2 A_1 B_1 + 5 A_2 B_1}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \sin x \paren {-A_1 B_2 - 2 A_2 B_1 - 2 A_2 B_2 + 2 A_1 B_1 + 5 A_1 B_2}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \cos x \paren {-A_2 B_2 + 2 A_1 B_1 + 2 A_1 B_2 + 2 A_2 B_1 + 5 A_2 B_2}\)

\(\ds \)

\(=\)

\(\ds x \sin x\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \)

\(\)

\(\ds x \sin x \paren {4 A_1 B_1 - 2 A_2 B_1}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds x \cos x \paren {4 A_2 B_1 + 2 A_1 B_1}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \sin x \paren {4 A_1 B_2 - 2 A_2 B_1 - 2 A_2 B_2 + 2 A_1 B_1}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \cos x \paren {4 A_2 B_2 + 2 A_1 B_1 + 2 A_1 B_2 + 2 A_2 B_1}\)

\(\ds \)

\(=\)

\(\ds x \sin x\)

Hence by equating coefficients:

\(\ds 4 A_1 B_1 - 2 A_2 B_1\)

\(=\)

\(\ds 1\)

\(\ds 4 A_2 B_1 + 2 A_1 B_1\)

\(=\)

\(\ds 0\)

\(\ds 4 A_1 B_2 - 2 A_2 B_1 - 2 A_2 B_2 + 2 A_1 B_1\)

\(=\)

\(\ds 0\)

\(\ds 4 A_2 B_2 + 2 A_1 B_1 + 2 A_1 B_2 + 2 A_2 B_1\)

\(=\)

\(\ds 0\)

This needs considerable tedious hard slog to complete it. In particular: possibly the least fun you can have while doing mathematics To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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

$y = y_g + y_p = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \ldots$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $10$