Q250 / 270 · NDA 2022 (Phase 2) · Mathematics

What is the solution of the differential equation $(d y-d x)$ $+\cos x(d y+d x)=0 ?$

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

Solution

$\begin{aligned} & (d y-d x)+\cos x(d y+d x)=0 \\ & \Rightarrow(1+\cos x) d y=(1-\cos x) d x \\ & \Rightarrow \frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}=\tan ^2 \frac{x}{2}=\sec ^2 \frac{x}{2}-1 \\ & \int d y=\int\left(\sec ^2 \frac{x}{2}-1\right) d x \\ & \Rightarrow y=2 \tan \frac{x}{2}-x+C\end{aligned}$

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

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