Differential equation $y''+y=\tan(x)$

Question

Given the following differential equation: \begin{align} y''+y=\tan(x) \end{align} I tried this way: \begin{align} y_H(x)=A\cos(x)+ B\sin(x)\\ A,B\in \Re \\ y_p(x)=A(x)\cos(x)+B(x)\sin(x) \end{align} I continue with the Wronski method \begin{align} W(F)= \begin{bmatrix} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{bmatrix}\\\\ det \ W(F) = 1 \neq 0 \end{align} So is invertible \begin{align} W(F)^{-1}= \begin{bmatrix} \cos(x) & -\sin(x)\\ \sin(x) & \cos(x) \end{bmatrix}\\\\ \end{align} Now since: \begin{align} W(F) \cdot\begin{bmatrix} A'(x) \\ B'(x) \end{bmatrix}= \begin{bmatrix} 0 \\ b(x) \end{bmatrix}\\\\ \begin{bmatrix} A'(x) \\ B'(x) \end{bmatrix}= \begin{bmatrix} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix} \cdot \begin{bmatrix} 0 \\ \tan(x) \end{bmatrix}\\\\ B'(x)= \cos(x) \tan(x)=\sin(x) \\ B(x) = -\cos(x)\\\\ A'(x) = -\sin(x)\tan(x)\\ A(x) = -\int \sin(x) \tan(x) dx = \cos(x)\tan(x) - \int \frac{1}{\cos(x)} dx= \\ = \sin(x) - \frac{1}{2} \log{\frac{1+sin(x)}{1-\sin(x)}}\\\\ y_p(x)=A(x)\cos(x) + B(x)\sin(x) = \\\\ = \sin(x)\cos(x) - \frac{1}{2}\cos(x) \log{\frac{1+sin(x)}{1-\sin(x)}} + -\cos(x)\sin(x)= \\\\ - \frac{1}{2}\cos(x) \log{\frac{1+sin(x)}{1-\sin(x)}} \end{align} So now the linear combination of the two solution should be \begin{align} y(x)= A\cos(x) + B\sin(x) - \frac{1}{2}\cos(x) \log{\frac{1+sin(x)}{1-\sin(x)}} \end{align} but the correct solution is \begin{align} y(x)= A\cos(x) +B\sin(x) + \cos(x) \log\left(\frac{\cos(\frac{x}{2})-\sin(\frac{x}{2})}{cos(\frac{x}{2})+sin(\frac{x}{2})}\right) \end{align} Is my solution wrong? If yes where I did the mistake? Is my solution right? How can I get the other form? What steps I need to do?

Answer 1

Your solution is correct just note that $(\sin(x/2)\pm \cos+x/2))^2=1\pm \sin(x)$. Now use the log property that $\log(a^b)=b\log(a)$ and you are done. And if you are confused with the discrepancy of plus and minus sign before the log expression we have $\log(\frac{1}{x})=-\log(x)$