Medium MCQ +4 / -1 PYQ · JEE Mains 2024

A function $y=f(x)$ satisfies $f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0$ with condition $f(0)=0$. Then, $f\left(\frac{\pi}{2}\right)$ is equal to

  1. A 2
  2. B 1 Correct answer
  3. C $-$1
  4. D 0

Solution

<p>$$\begin{aligned} & \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\ & \text { I.F. }=1+\cos ^2 x \\ & y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\ & =-\cos x+C \\ & x=0, C=1 \\ & y\left(\frac{\pi}{2}\right)=1 \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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