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

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:

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


 * $y_p = v_1 y_1 + v_2 y_2$

where:

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

We have that:

Hence:

It follows that:

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 $(1)$.