Q76 / 170 · Manipal (MET) 2020 · Mathematics

Solution of the equation $\cos ^2 x \frac{d y}{d x}-(\tan 2 x) y=\cos ^4 x,|x| \lt \frac{\pi}{4}$, where $y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}$, is given by

  • A
    $y \frac{\tan 2 x}{1-\tan ^2 x}=0$
  • B
    $y\left(1-\tan ^2 x\right)=C$
  • C
    $y=\sin 2 x+C$
  • Correct answer
    D
    $y=\frac{1}{2} \cdot \frac{\sin 2 x}{1-\tan ^2 x}$

Solution

Given differential equation can be written as $\frac{d y}{d x}-\left(\frac{\tan 2 x}{\cos ^2 x}\right) y=\cos ^2 x$ Here, $P=-\frac{\tan 2 x}{\cos ^2 x}=\frac{-\sin 2 x}{\cos 2 x\left(\frac{\cos 2 x+1}{2}\right)}$ and $Q=\cos ^2 x$ $\because \quad \mathrm{IF}=e^{\int P d x}=e^{-\int \frac{2 \sin 2 x}{\cos 2 x(\cos 2 x+1)} d x}$ Put $\cos 2 x=t \Rightarrow-2 \sin 2 x d x=d t$ $\begin{aligned} \therefore \mathrm{If} & =e^{\int \frac{1}{t}\left(\frac{1}{t+1}\right) d t} \\ & =e^{\int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t} \\ & =e^{\{\log t-\log (t+1)]}=e^{\log \frac{t}{t+1}} \\ & =e^{\log \frac{\cos 2 x}{\cos 2 x+1}}=\frac{\cos 2 x}{\cos 2 x+1}\end{aligned}$ Now, solution is $\begin{aligned} y \times \frac{\cos 2 x}{\cos 2 x+1} & =\int \frac{\cos 2 x}{\cos 2 x+1} \times \cos ^2 x d x+C \\ & =\int \frac{\cos 2 x}{2 \cos ^2 x} \times \cos ^2 x d x+C \\ & =\frac{1}{2} \int \cos 2 x d x+C \\ & =\frac{1}{2} \times \frac{\sin 2 x}{2}+C\end{aligned}$ $\Rightarrow y \frac{\cos 2 x}{\cos 2 x+1}=\frac{1}{4} \sin 2 x+c$ ....(i) But $y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}$ $\begin{array}{ll}\therefore \frac{3 \sqrt{3}}{8} \times \frac{\cos \left(2 \times \frac{\pi}{6}\right)}{\cos \left(2 \times \frac{\pi}{6}\right)+1}=\frac{1}{4} \sin 2\left(\frac{\pi}{6}\right)+C \\ \Rightarrow & \frac{\frac{3 \sqrt{3}}{8} \times \frac{1}{2}}{\frac{1}{2}+1}=\frac{1}{4} \times \frac{\sqrt{3}}{2}+C \\ \Rightarrow & \frac{3 \sqrt{3}}{2 \times 8 \times \frac{3}{2}}=\frac{\sqrt{3}}{8}+C \\ \Rightarrow & \frac{\sqrt{3}}{8}=\frac{\sqrt{3}}{8}+C \Rightarrow C=0\end{array}$ From Eq. (i), we get $\begin{aligned} y \frac{\cos 2 x}{\cos 2 x+1} & =\frac{1}{4} \sin 2 x+0 \\ \Rightarrow \quad y=\frac{1}{4} \frac{\sin 2 x}{\frac{\cos 2 x}{\cos 2 x+1}} & =\frac{1}{4} \frac{\sin 2 x}{\frac{\cos ^2 x-\sin ^2 x}{2 \cos ^2 x}} \\ & =\frac{1}{2} \cdot \frac{\sin 2 x}{1-\tan ^2 x}\end{aligned}$

Hence, the correct answer is option: 'D'.

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