Linear Second Order ODE/y'' + y = exp -x cos x
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Theorem
The second order ODE:
- $(1): \quad y'' + y = e^{-x} \cos x$
has the general solution:
- $y = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$
Proof
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
- $y'' + p y' + q y = \map R x$
where:
- $p = 0$
- $q = 1$
- $\map R x = e^{-x} \cos x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $y'' + y = 0$
From Linear Second Order ODE: $y'' + y = 0$, this has the general solution:
- $y_g = C_1 \sin x + C_2 \cos x$
It remains to find a particular solution $y_p$ to $(1)$.
We have that:
- $\map R x = e^{-x} \cos x$
From the Method of Undetermined Coefficients for Exponential of Sine and Cosine:
- $y_p = e^{-x} \paren {A \cos x + B \sin x}$
where $A$ and $B$ are to be determined.
Hence:
\(\ds y_p\) | \(=\) | \(\ds e^{-x} \paren {A \cos x + B \sin x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}'\) | \(=\) | \(\ds -e^{-x} \paren {A \cos x + B \sin x} + e^{-x} \paren {-A \sin x + B \cos x}\) | Product Rule for Derivatives etc. | ||||||||||
\(\ds \) | \(=\) | \(\ds e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}''\) | \(=\) | \(\ds -e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x} + e^{-x} \paren {-\paren {B - A} \sin x - \paren {A + B} \cos x}\) | Product Rule for Derivatives etc. | ||||||||||
\(\ds \) | \(=\) | \(\ds e^{-x} \paren {-2 B \cos x + 2 A \sin x}\) |
Substituting into $(1)$:
\(\ds e^{-x} \paren {-2 B \cos x + 2 A \sin x} + e^{-x} \paren {A \cos x + B \sin x}\) | \(=\) | \(\ds e^{-x} \cos x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -2 B \cos x + A \cos x\) | \(=\) | \(\ds \cos x\) | equating coefficients | ||||||||||
\(\ds 2 A \sin x + B \sin x\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A - 2 B\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds 2 A + B\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 5 A\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds 5 B\) | \(=\) | \(\ds -2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds \dfrac 1 5\) | |||||||||||
\(\ds B\) | \(=\) | \(\ds -\dfrac 2 5\) |
Hence the result:
- $y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x}$
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$
is the general solution to $(1)$.
$\blacksquare$
Sources
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $6$