Linear Second Order ODE/y'' + 4 y = tan 2 x

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Theorem

The second order ODE:

$(1): \quad y + 4 y = \tan 2 x$

has the general solution:

$y = C_1 \cos 2 x + C_2 \sin 2 x - \dfrac 1 4 \cos 2 x \map \ln {\sec 2 x + \tan 2 x}$


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:

$p = 0$
$q = 4$
$\map R x = \tan 2 x$


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

$y + 4 y = 0$

From Linear Second Order ODE: $y + 4 y = 0$, this has the general solution:

$y_g = C_1 \cos 2 x + C_2 \sin 2 x$


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:

\(\ds \map {y_1} x\) \(=\) \(\ds \cos 2 x\)
\(\ds \map {y_2} x\) \(=\) \(\ds \sin 2 x\)
\(\ds \leadsto \ \ \) \(\ds \map { {y_1}'} x\) \(=\) \(\ds -2 \sin 2 x\) Derivative of Sine Function
\(\ds \map { {y_2}'} x\) \(=\) \(\ds 2 \cos 2 x\) Derivative of Cosine Function


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
\(\ds \) \(=\) \(\ds \cos 2 x \paren {2 \cos 2 x} - \sin 2 x \paren {-2 \sin 2 x}\)
\(\ds \) \(=\) \(\ds 2 \paren {\cos^2 2 x + \sin^2 2 x}\)
\(\ds \) \(=\) \(\ds 2\)


Hence:

\(\ds v_1\) \(=\) \(\ds \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds \int -\frac {\sin 2 x \tan 2 x} 2 \rd x\)
\(\ds \) \(=\) \(\ds -\frac 1 2 \int \frac {\sin^2 2 x} {\cos 2 x} \rd x\)
\(\ds \) \(=\) \(\ds -\frac 1 2 \paren {-\frac {\sin 2 x} 2 + \frac 1 2 \ln \map \tan {x + \frac \pi 4} }\) Primitive of $\dfrac {\sin^2 a x} {\cos a x}$
\(\ds \) \(=\) \(\ds \frac 1 4 \paren {\sin 2 x - \map \ln {\sec 2 x + \tan 2 x} }\) Tangent of Half Angle plus $\dfrac \pi 4$


\(\ds v_2\) \(=\) \(\ds \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {\cos 2 x \tan 2 x} 2 \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \int \sin 2 x \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {-\frac {\cos 2 x} 2}\) Primitive of $\sin a x$
\(\ds \) \(=\) \(\ds -\frac 1 4 \cos 2 x\)


It follows that:

\(\ds y_p\) \(=\) \(\ds \frac 1 4 \paren {\sin 2 x - \map \ln {\sec 2 x + \tan 2 x} } \cos 2 x - \frac 1 4 \cos 2 x \sin 2 x\)
\(\ds \) \(=\) \(\ds -\frac 1 4 \cos 2 x \map \ln {\sec 2 x + \tan 2 x}\) simplifying


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

$y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x - \dfrac 1 4 \cos 2 x \map \ln {\sec 2 x + \tan 2 x}$

is the general solution to $(1)$.

$\blacksquare$


Sources

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