Question

The differential equation $\frac{d^{2}y}{d{x}^2}+y+{\cot }^{2}x=0$ must be satisfied by

• A. $y=2+c_1\cos x+\sqrt{c_2}\sin x$
• B. $y=\cos x\cdot \ln \left(\tan \frac{x}{2}\right)+2$
• C. $y=\sin x+\cos x$
• D. all the above

Answer 1

Answer: (B)

(1) $\frac{dy}{dx}=-c_1\sin x+\sqrt{c_2}\cos x$
$\frac{d^{2}y}{d{x}^2}=-c_1\cos x-\sqrt{c_2}\sin x=2-y$
$\frac{d^{2}y}{d{x}^2}+y-2=0$
(2) $\frac{dy}{dx}=\cos x\frac{{\sec }^{2}x/2}{2\tan \frac{x}{2}}-\sin x\ln \left(\tan \frac{x}{2}\right)$
$\frac{dy}{dx}=\cot x-\sin x\ln \left(\tan \frac{x}{2}\right)$
$\frac{d^{2}y}{d{x}^2}=-{\mathrm{cosec}}^{2}x-\left(\sin x\cdot \frac{1}{\sin x}+\cos x\ln \left(\tan \frac{x}{2}\right)\right)$
$\frac{d^{2}y}{d{x}^2}=-{\cot }^{2}x-2-\cos x\ln \left(\tan \frac{x}{2}\right)$
$\frac{d^{2}y}{d{x}^2}+y+{\cot }^{2}x=0$
(3) $\frac{dy}{dx}=\cos x-\sin x$
$\frac{d^{2}y}{d{x}^2}=-\cos x-\sin x$
$\frac{d^{2}y}{d{x}^2}+y+{\cot }^{2}x={\cot }^{2}x$