Linear Second Order ODE/y'' - 3 y' + 2 y = 5 exp 3 x
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Theorem
The second order ODE:
- $(1): \quad y - 3 y' + 2 y = 5 e^{3 x}$
has the general solution:
- $y = C_1 e^x + C_2 e^{2 x} + \dfrac {5 e^{3 x} } 2$
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:
- $p = -3$
- $q = 2$
- $\map R x = 5 e^{3 x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $y - 3 y' + 2 y = 0$
From Linear Second Order ODE: $y - 3 y' + 2 y = 0$, this has the general solution:
- $y_g = C_1 e^x + C_2 e^{2 x}$
We have that:
- $\map R x = 5 e^{3 x}$
and so from the Method of Undetermined Coefficients for the Exponential function:
- $y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$
where:
- $K = 5$
- $a = 3$
- $p = -3$
- $q = 2$
Hence:
\(\ds y_p\) | \(=\) | \(\ds \dfrac {5 e^{3 x} } {3^2 - 3 \times 3 + 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {5 e^{3 x} } {9 - 9 + 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {5 e^{3 x} } 2\) |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + \dfrac {5 e^{3 x} } 2$
$\blacksquare$
Sources
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.4$ Particular solution: exponential $\map f x$: Example $3$