Answer to Question #213506 in Differential Equations for Amir

Question

Y"+4y=tg(2x)

Answer 1

Let us solve the differential equation $y''+4y=\tg(2x).$ The characteristic equation $k^2+4=0$ of the homogeneous differential equation has the solutions $k_1=2i,k_2=-2i.$ The solution of the homogeneous equation is $y=C_1\cos 2x+C_2\sin 2x.$ Let us use the method of variation of parameters. We will find the solution of the differential equation in the form: $y=C_1(x)\cos 2x+C_2(x)\sin 2x.$ For this we should to solve the following system:

$\begin{cases}C_1'(x)\cos 2x+C_2'(x)\sin 2x=0\\-2C_1'(x)\sin 2x+2C_2'(x)\cos 2x=\tg 2x\end{cases}$

$\begin{cases}C_1'(x)=-C_2'(x)\tg 2x\\2C_2'(x)\tg 2x\sin 2x+2C_2'(x)\cos 2x=\tg 2x\end{cases}$

$\begin{cases}C_1'(x)=-C_2'(x)\tg 2x\\2C_2'(x)(\frac{\sin^2 2x}{\cos 2x}+\cos 2x)=\tg 2x\end{cases}$

$\begin{cases}C_1'(x)=-C_2'(x)\tg 2x\\2C_2'(x)(\frac{\sin^2 2x+\cos^22x}{\cos 2x})=\tg 2x\end{cases}$

$\begin{cases}C_1'(x)=-C_2'(x)\tg 2x\\2C_2'(x)(\frac{1}{\cos 2x})=\tg 2x\end{cases}$

$\begin{cases}C_1'(x)=-C_2'(x)\tg 2x\\C_2'(x)=\frac{1}{2}\sin 2x\end{cases}$

$\begin{cases}C_1'(x)=-\frac{1}{2}\frac{\sin^2 2x}{\cos 2x}\\C_2'(x)=\frac{1}{2}\sin 2x\end{cases}$

It follows that $C_2(x)=-\frac{1}{4}\cos 2x+C_2$ and $C_1(x)=-\frac{1}{2}\int\frac{\sin^2 2x}{\cos 2x}dx=-\frac{1}{4}\int\frac{\sin^2 2x}{\cos^2 2x}d(\sin 2x) =\frac{1}{4}\int\frac{\sin^2 2x-1+1}{\sin^2 2x-1}d(\sin 2x)= \frac{1}{4}\int d(\sin 2x)+\frac{1}{4}\int\frac{1}{\sin^2 2x-1}d(\sin 2x)= \frac{1}{4}\sin 2x+\frac{1}{8}\int(\frac{1}{\sin2x-1}-\frac{1}{\sin2x+1})d(\sin 2x)= \frac{1}{4}\sin 2x+\frac{1}{8}(\ln|\sin2x-1|-\ln|\sin2x+1|)+C_1= \frac{1}{4}\sin 2x+\frac{1}{8}\ln|\frac{\sin2x-1}{\sin2x+1}|+C_1$

We conclude that the general solution is

$y=(\frac{1}{4}\sin 2x+\frac{1}{8}\ln|\frac{\sin2x-1}{\sin2x+1}|+C_1)\cos 2x+(-\frac{1}{4}\cos 2x+C_2)\sin 2x.$