Question

Solution of the equation $co{s}^{2}x\frac{dy}{dx}-\left(tan2x\right)y=co{s}^{4}x,|x|<\frac{\pi }{4},$ where $\left(\frac{\pi }{6}\right)=\frac{3\sqrt{3}}{8}$ is

• A. $y=tan2xco{s}^{2}x$
• B. $y=cot2xco{s}^{2}x$
• C. $y=\frac{1}{2}tan2xco{s}^{2}x$
• D. $y=\frac{1}{2}cot2xco{s}^{2}x$

Answer 1

The correct option is C $y=\frac{1}{2}tan2xco{s}^{2}x$

The given differential equation can be written as $\frac{dy}{dx}-\frac{tan2x}{co{s}^{2}x}y=co{s}^{2}x$ which is linear differential equation of first order.
$\int Pdx=\int \frac{-sin2x}{cos2xco{s}^{2}x}dx=-\int \frac{2sin2xdx}{cos2x(1+cos2x)}=\int \frac{dt}{t(1+t)}=\int (\frac{1}{t}-\frac{1}{1+t})dt=log\frac{t}{1+t}wheret=cos2x=log\frac{cos2x}{1+cos2x}[\therefore -\frac{\pi }{2}<2x<\frac{\pi }{2}]e^{\int Pdx}=e^{log\frac{cos2x}{1+cos2x}}=\frac{cos2x}{1+cos2x}=\frac{cos2x}{2co{s}^{2}x}\therefore thesolutionisy\frac{cos2x}{2co{s}^{2}x}=\int \frac{co{s}^{2}xcos2x}{2co{s}^{2}x}dx+c=\frac{1}{4}sin2x+CWhenx=\frac{\pi }{6},y=\frac{3\sqrt{3}}{8}\therefore \frac{3\sqrt{3}}{8}\frac{4}{2\times 2\times 3}=\frac{1}{4}\frac{\sqrt{3}}{2}+C\Rightarrow C=0\therefore y=\frac{1}{2}tan2xco{s}^{2}x$