Solve by the method of variation of parameters y'' + 4y =…

Solve by the method of variation of parameters $y'' + 4y = 2\tan x$.

Step 1: Solve the homogeneous equation

\[ r^2+4=0 \implies r=\pm2i, \qquad y_1=\cos 2x,\ \ y_2=\sin 2x. \]

Step 2: Wronskian

\[ W(y_1,y_2) = \begin{vmatrix}\cos2x & \sin2x \\ -2\sin2x & 2\cos2x\end{vmatrix} = 2\cos^22x+2\sin^22x = 2. \]

Step 3: Set up the variation-of-parameters system

Write $y=u_1\cos2x+u_2\sin2x$, with $f(x)=2\tan x$ and $W=2$. By Cramer's rule: \[u_1' = -\frac{y_2f}{W} = -\sin2x\tan x, \qquad u_2' = \frac{y_1f}{W} = \cos2x\tan x.\]

Step 4: Integrate

Using $\sin2x=2\sin x\cos x$: \[u_1 = -\int\sin2x\tan x\,dx = -2\int\sin^2x\,dx = -\int(1-\cos2x)\,dx = -x+\frac{\sin2x}{2}.\]

Using $\cos2x=1-2\sin^2x$ and $2\sin^3x=2\sin x-2\sin x\cos^2x$: \[u_2 = \int\cos2x\tan x\,dx = \int\!\left(-\tan x+2\sin x\cos x\right)dx = \int(-\tan x+\sin2x)\,dx = \ln|\cos x| - \frac{\cos2x}{2}.\]

Step 5: Form and simplify the particular solution

\[ y^* = u_1\cos2x + u_2\sin2x = \left(-x+\frac{\sin2x}{2}\right)\cos2x + \left(\ln|\cos x|-\frac{\cos2x}{2}\right)\sin2x. \]

The two $\tfrac12\sin2x\cos2x$ terms cancel: $y^* = -x\cos2x + \sin2x\ln|\cos x|.$

Step 6: Verify $y^*$

\[ y^{*\prime} = -\cos2x + 2x\sin2x + 2\cos2x\ln|\cos x| - \sin2x\tan x. \]

\[ y^{*\prime\prime} = 4\sin2x + 4x\cos2x - 4\sin2x\ln|\cos x| - 4\cos2x\tan x - \frac{\sin2x}{\cos^2x}. \]

Adding $4y^*$ cancels the $x\cos2x$ and $\ln|\cos x|$ terms: $y^{*\prime\prime}+4y^* = 4\sin2x - 4\cos2x\tan x - \frac{\sin2x}{\cos^2x}.$

Since $\sin2x/\cos^2x = 2\sin x\cos x/\cos^2x = 2\tan x$, and since $\sin2x\cot x = 2\cos^2x = 1+\cos2x$, so that

\[ 4\sin2x-4\cos2x\tan x = 4\tan x(\sin2x\cot x-\cos2x) = 4\tan x\bigl[(1+\cos2x)-\cos2x\bigr] = 4\tan x, \]

the total is

\[ y^{*\prime\prime}+4y^* = 4\tan x - 2\tan x = 2\tan x.\ \checkmark \]

Step 7: General solution

The solution is recorded in the result below.