Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x

Theorem

The second order ODE:

$(1): \quad y + 4 y = 8 x^2 - 4 x$

has the general solution:

$y = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$

Proof 1

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:

$p = 0$

$q = 4$

$\map R x = 8 x^2 - 4 x$

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:

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

From the Method of Undetermined Coefficients for Polynomials:

$y_p = A_0 + A_1 x + A_2 x^2$

where $A$ and $B$ are to be determined.

Hence:

\(\ds y_p\)

\(=\)

\(\ds A_0 + A_1 x + A_2 x^2\)

\(\ds \leadsto \ \ \)

\(\ds {y_p}'\)

\(=\)

\(\ds A_1 + 2 A_2 x\)

Power Rule for Derivatives

\(\ds \leadsto \ \ \)

\(\ds {y_p}\)

\(=\)

\(\ds 2 A_2\)

Power Rule for Derivatives

Substituting into $(1)$:

\(\ds 2 A_2 + 4 \paren {A_0 + A_1 x + A_2 x^2}\)

\(=\)

\(\ds 8 x^2 - 4 x\)

\(\ds \leadsto \ \ \)

\(\ds 2 A_2 + 4 A_0\)

\(=\)

\(\ds 0\)

equating coefficients

\(\ds 4 A_1\)

\(=\)

\(\ds -4\)

\(\ds 4 A_2\)

\(=\)

\(\ds 8\)

\(\ds \leadsto \ \ \)

\(\ds A_1\)

\(=\)

\(\ds -1\)

\(\ds A_2\)

\(=\)

\(\ds 2\)

\(\ds A_0\)

\(=\)

\(\ds -1\)

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

$y = y_g + y_p = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$

is the general solution to $(1)$.

$\blacksquare$

Proof 2

Taking Laplace transforms:

$\laptrans {y + 4 y} = \laptrans {8 x^2 - 4 x}$

We have:

\(\ds \laptrans {y + 4 y}\)

\(=\)

\(\ds \laptrans {y} + 4 \laptrans y\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds s^2 \laptrans y - s \map y 0 - \map {y'} 0\)

Laplace Transform of Second Derivative

We also have:

\(\ds \laptrans {8 x^2 - 4 x}\)

\(=\)

\(\ds 8 \laptrans {x^2} - 4 \laptrans x\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds \frac {8 \cdot 2!} {s^3} - \frac {4} {s^2}\)

Laplace Transform of Power

\(\ds \)

\(=\)

\(\ds \frac {16 - 4 s} {s^3}\)

So:

$\paren {s^2 + 4} \laptrans y = s \map y 0 + \map {y'} 0 + \dfrac {16 - 4 s} {s^3}$

Giving:

\(\ds \laptrans y\)

\(=\)

\(\ds \map y 0 \frac s {s^2 + 4} + \map {y'} 0 \frac 1 {s^2 + 4} + \frac {16 - 4 s} {s^3 \paren {s^2 + 4} }\)

\(\ds \)

\(=\)

\(\ds \map y 0 \frac s {s^2 + 4} + \map {y'} 0 \frac 1 {s^2 + 4} + \frac 4 {s^3} + \frac {s + 1} {s^2 + 4} - \frac 1 {s^2} - \frac 1 s\)

partial fraction expansion

\(\ds \)

\(=\)

\(\ds \paren {\map y 0 + 1} \frac s {s^2 + 4} + \paren {\map {y'} 0 + 1} \frac 1 {s^2 + 4} + \frac 4 {s^3} - \frac 1 {s^2} - \frac 1 s\)

So:

\(\ds y\)

\(=\)

\(\ds \invlaptrans {\paren {\map y 0 + 1} \frac s {s^2 + 4} + \paren {\map {y'} 0 + 1} \frac 1 {s^2 + 4} + \frac 4 {s^3} - \frac 1 {s^2} - \frac 1 s}\)

Definition of Inverse Laplace Transform

\(\ds \)

\(=\)

\(\ds \paren {\map y 0 + 1} \invlaptrans {\frac s {s^2 + 2^2} } + \frac {\paren {\map {y'} 0 + 1} } 2 \invlaptrans {\frac 2 {s^2 + 2^2} } + 2 \invlaptrans {\frac {2!} {s^3} } - \invlaptrans {\frac {1!} {s^2} } - \invlaptrans {\frac {0!} s}\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds \paren {\map y 0 + 1} \invlaptrans {\laptrans {\cos 2 x} } + \frac {\paren {\map {y'} 0 + 1} } 2 \invlaptrans {\laptrans {\sin 2 x} } + 2 \invlaptrans {\laptrans {x^2} } - \invlaptrans {\laptrans x} - \invlaptrans {\laptrans 1}\)

Laplace Transform of Cosine, Laplace Transform of Sine, Laplace Transform of Power

\(\ds \)

\(=\)

\(\ds \paren {\map y 0 + 1} \cos 2 x + \paren {\frac {\paren {\map {y'} 0 + 1} } 2 } \sin 2 x - 1 - x + 2 x^2\)

Definition of Inverse Laplace Transform

Setting $C_1 = \dfrac {\paren {\map {y'} 0 + 1} } 2$ and $C_2 = \map y 0 + 1$ gives the result.

$\blacksquare$