Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x/Proof 1

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

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 $(1)$.

$\blacksquare$