Linear Second Order ODE/y'' + 2 y' + 5 y = exp -x secant 2 x

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Theorem

The second order ODE:

$(1): \quad y + 2 y' + 5 y = e^{-x} \sec 2 x$

has the general solution:

$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \dfrac {x e^{-x} \sin 2 x} 2 + \dfrac {e^{-x} \cos 2 x \ln \cos 2 x} 4$


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:

$p = 2$
$q = 5$
$\map R x = e^{-x} \sec 2 x$


First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$y + 2 y' + 5 y = 0$

From Linear Second Order ODE: $y + 2 y' + 5 y = 0$, this has the general solution:

$y_g = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$


It remains to find a particular solution $y_p$ to $(1)$.


Expressing $y_g$ in the form:

$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$

we have:

\(\ds \map {y_1} x\) \(=\) \(\ds e^{-x} \cos 2 x\)
\(\ds \map {y_2} x\) \(=\) \(\ds e^{-x} \sin 2 x\)
\(\ds \leadsto \ \ \) \(\ds \map { {y_1}'} x\) \(=\) \(\ds -e^{-x} \cos 2 x - 2 e^{-x} \sin 2 x\) Derivative of Exponential Function
\(\ds \) \(=\) \(\ds -e^{-x} \paren {\cos 2 x + 2 \sin 2 x}\)
\(\ds \map { {y_2}'} x\) \(=\) \(\ds -e^{-x} \sin 2 x + 2 e^{-x} \cos 2 x\) Derivative of Exponential Function
\(\ds \) \(=\) \(\ds e^{-x} \paren {2 \cos 2 x - \sin 2 x}\)


By the Method of Variation of Parameters, we have that:

$y_p = v_1 y_1 + v_2 y_2$

where:

\(\ds v_1\) \(=\) \(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds v_2\) \(=\) \(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\)

where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.


We have that:

\(\ds \map W {y_1, y_2}\) \(=\) \(\ds y_1 {y_2}' - y_2 {y_1}'\) Definition of Wronskian
\(\ds \) \(=\) \(\ds e^{-x} \cos 2 x \paren {e^{-x} \paren {2 \cos 2 x - \sin 2 x} } - e^{-x} \sin 2 x \paren {-e^{-x} \paren {\cos 2 x + 2 \sin 2 x} }\)
\(\ds \) \(=\) \(\ds e^{-2 x} \paren {2 \cos^2 2 x - \cos 2 x \sin 2 x + \sin 2 x \cos 2 x + 2 \sin^2 2 x}\)
\(\ds \) \(=\) \(\ds 2 e^{-2 x} \paren {\cos^2 2 x + \sin^2 2 x}\)
\(\ds \) \(=\) \(\ds 2 e^{-2 x}\) Sum of Squares of Sine and Cosine


Hence:

\(\ds v_1\) \(=\) \(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds \int -\frac {e^{-x} \sin 2 x e^{-x} \sec 2 x} {2 e^{-2 x} } \rd x\)
\(\ds \) \(=\) \(\ds -\frac 1 2 \int \sin 2 x \sec 2 x \rd x\)
\(\ds \) \(=\) \(\ds -\frac 1 2 \int \frac {\sin 2 x} {\cos 2 x} \rd x\) Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds -\frac 1 2 \int \tan 2 x \rd x\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac 1 4 \ln \cos 2 x\) Primitive of $\tan a x$


\(\ds v_2\) \(=\) \(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {e^{-x} \cos 2 x e^{-x} \sec 2 x} {2 e^{-2 x} } \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \int \cos 2 x \sec 2 x \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \int \rd x\) Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds \frac x 2\) Primitive of Constant


It follows that:

\(\ds y_p\) \(=\) \(\ds \paren {\frac 1 4 \ln \cos 2 x} e^{-x} \cos 2 x + \paren {\frac x 2} e^{-x} \sin 2 x\)
\(\ds \) \(=\) \(\ds \frac {x e^{-x} \sin 2 x} 2 + \frac {e^{-x} \cos 2 x \ln \cos 2 x} 4\) rearranging


So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

$y = y_g + y_p = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \dfrac {x e^{-x} \sin 2 x} 2 + \dfrac {e^{-x} \cos 2 x \ln \cos 2 x} 4$

is the general solution to $(1)$.

$\blacksquare$


Sources

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