Solution of the equation ² x \;d y/d x - ( 2x) y = ^4 x, \;\; |x|

Correct Answer is : y=sin⁡2x+Cy = 2x + Cy=sin2x+C

Correct Answer is : y=sin⁡2x+Cy = 2x + Cy=sin2x+C

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CGPET 2014 Solved Paper

Section: Mathematics

Question : 138 of 150

Marks: +1, -0

\cos^2 x \;\frac{d y}{d x} - (\tan 2x) y = \cos^4 x, \;\; |x| < \frac{\pi}{4}

y \left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{8}

Solution:

Given differential equation can be written as

\frac{d y}{d x} - \left(\frac{\tan 2x}{\cos^2 x}\right) y = \cos^2 x

Here,

P = -\frac{\tan 2x}{\cos^2 x} = \frac{-\sin 2x}{\cos 2x \left(\frac{\cos 2x+1}{2}\right)}

and

Q = \cos^2 x

\because \;\; \mathbb{F} = e^{\int P dx} = e^{-\int \frac{2\sin 2x}{\cos 2x (\cos 2x+1)} dx}

Put

\cos 2x = t \Rightarrow -2 \sin 2x \, dx = dt

\therefore \;\; IF = e^{\int \frac{1}{t} \left(\frac{1}{t+1}\right) dt}

= e^{\int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt}

= e^{\left[ \log t - \log(t+1) \right]}

= e^{\log \frac{t}{t+1}}

= e^{\log \frac{\cos 2x}{\cos 2x+1}}

= \frac{\cos 2x}{\cos 2x+1}

Now, solution is

y \times \frac{\cos 2x}{\cos 2x+1} = \int \frac{\cos 2x}{\cos 2x+1} \times \cos^2 x \, dx + C

= \int \frac{\cos 2x}{2 \cos^2 x} \times \cos^2 x \, dx + C

= \frac{1}{2} \int \cos 2x \, dx + C

= \frac{1}{2} \times \frac{\sin 2x}{2} + C

\Rightarrow y \frac{\cos 2x}{\cos 2x+1} = \frac{1}{4} \sin 2x + C

\text{ But } y \left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{8}

\therefore \frac{3\sqrt{3}}{8} \times \frac{\cos \left(2 \times \frac{\pi}{6}\right)}{\cos \left(2 \times \frac{\pi}{6}\right) + 1} = \frac{1}{4} \sin \left(2 \cdot \frac{\pi}{6}\right) + C

\Rightarrow \frac{ \frac{3\sqrt{3}}{8} \times \frac{1}{2} }{ \frac{1}{2} + 1 } = \frac{1}{4} \times \frac{\sqrt{3}}{2} + C

\Rightarrow \frac{3\sqrt{3}}{2 \times 8 \times \frac{3}{2}} = \frac{\sqrt{3}}{8} + C

\Rightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8} + C \Rightarrow C = 0

From Eq. (i), we get

\frac{\cos 2x}{\cos 2x+1} = \frac{1}{4} \sin 2x + 0

\Rightarrow y = \frac{1}{4} \frac{\sin 2x}{ \frac{\cos 2x}{\cos 2x+1} }

= \frac{1}{4} \frac{\sin 2x}{ \frac{\cos^2 x - \sin^2 x}{2\cos^2 x} }

= \frac{1}{2} \cdot \frac{\sin 2x}{1 - \tan^2 x}

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