Linear Second Order ODE/y'' - 2 y' = 12 x - 10
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Theorem
The second order ODE:
- $(1): \quad y - 2 y' = 12 x - 10$
has the general solution:
- $y = C_1 + C_2 e^{2 x} + 2 x - 3 x^2$
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:
- $p = 2$
- $q = 0$
- $\map R x = 12 x - 10$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $y - 2 y' = 0$
From Linear Second Order ODE: $y - 2 y' = 0$, this has the general solution:
- $y_g = C_1 + C_2 e^{2 x}$
We have that:
- $\map R x = 12 x - 10$
and it is noted that $12 x - 10$ is not itself a particular solution of $(2)$.
So 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 $(1)$:
\(\ds 2 A_2 - 2 \paren {A_1 + 2 A_2 x}\) | \(=\) | \(\ds 12 x - 10\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -4 A_2 x\) | \(=\) | \(\ds 12 x\) | equating coefficients | ||||||||||
\(\ds 2 A_2 - 2 A_1\) | \(=\) | \(\ds -10\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A_2\) | \(=\) | \(\ds -3\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -6 - 2 A_1\) | \(=\) | \(\ds -10\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -2 A_1\) | \(=\) | \(\ds -4\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A_1\) | \(=\) | \(\ds 2\) |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 + C_2 e^{2 x} + 2 x - 3 x^2$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.18$: Problem $1 \ \text{(h)}$