Question

The solution of the differential equation ${\cos }^{2}x{y}_1-\left(\tan 2x\right)y={\cos }^{4}x,\forall \left|x\right|<{45}^0$ satisfies $y\left(\pi /6\right)=\frac{3\sqrt{3}}{8}$ is given by:

• A. $y=\frac{1}{2}\left(\frac{\sin 2x}{1-{\tan }^{2}x}\right)$
• B. $y=\left(\frac{1-{\tan }^{2}x}{4\sin x\cos x}\right)$
• C. $\sin x\cos x(1-{\tan }^{2}x)$
• D. None of these

Answer 1

The correct option is A $y=\frac{1}{2}\left(\frac{\sin 2x}{1-{\tan }^{2}x}\right)$

Given, ${\cos }^{2}x{y}^{\prime }-\left(\tan 2x\right)y={\cos }^{4}x{y}^{\prime }-\left(\frac{2\tan x}{1-{\tan }^{2}x}{\sec }^{2}x\right)y={\cos }^{2}xI.F.=e^{\int \frac{-2\tan x}{1-{\tan }^{2}x}{\sec }^{2}xdx}=e^{\ln ({\tan }^{2}x-1)}={\tan }^{2}x-1y^{\prime }({\tan }^{2}x-1)+\left(2\tan x{\sec }^{2}x\right)y={\cos }^{2}x({\tan }^{2}x-1)d\left[y\right({\tan }^{2}x-1\left)\right]={\cos }^{2}x({\tan }^{2}x-1)dxy({\tan }^{2}x-1)=\int \frac{{\tan }^{2}x-1}{1+{\tan }^{2}x}dxy({\tan }^{2}x-1)=-\int \cos 2xdxy({\tan }^{2}x-1)=-\frac{\sin 2x}{2}+Cy\left(\frac{\pi }{6}\right)=\frac{3\sqrt{3}}{8}\frac{3\sqrt{3}}{8}(\frac{1}{3}-1)=-\frac{\sqrt{3}}{4}+CC=0y=\frac{\sin 2x}{2(1-{\tan }^{2}x)}$