Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 3
Jump to navigation
Jump to search
Theorem
The second order ODE:
- $(1): \quad y - 2 y' + y = 2 x$
has the general solution:
- $y = C_1 e^x + C_2 x e^x + 2 x + 4$
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:
- $p = -2$
- $q = 1$
- $\map R x = 2 x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $(2): \quad y - 2 y' + y = 0$
From Linear Second Order ODE: $y - 2 y' + y = 0$, this has the general solution:
- $y_g = C_1 e^x + C_2 x e^x$
We have that:
- $R \left({x}\right) = 2 x$
So from the Method of Undetermined Coefficients for Polynomial:
- $y_p = A_0 + A_1 x$
Hence:
\(\ds y_p\) | \(=\) | \(\ds A_0 + A_1 x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}'\) | \(=\) | \(\ds A_1\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}\) | \(=\) | \(\ds 0\) | Derivative of Constant |
Substituting into $(1)$:
\(\ds 0 - 2 A_1 + \paren {A_0 + A_1 x}\) | \(=\) | \(\ds 2 x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -2 A_1 + A_0\) | \(=\) | \(\ds 0\) | equating coefficients | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds A_1\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A_0\) | \(=\) | \(\ds 4\) |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$
is the general solution to $(1)$.
$\blacksquare$