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

Let $y=y(x)$ be the solution of the differential equation $\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0$ such that $y(0)=2$. Then $y(2)$ is equal to:

  1. A $2\{\sin (2)+1\}$
  2. B 2 Correct answer
  3. C 1
  4. D $2\{1-\sin (2)\}$

Solution

<p>$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$</p> <p>Now both side integrate</p> <p>$$\begin{aligned} & y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\ & y(x)=\left(x^2-2 x\right) \sin x+\lambda \\ & y(0)=0+\lambda \Rightarrow 2=\lambda \\ & y(x)=\left(x^2-2 x\right) \sin x+2 \\ & y(2)=2 \end{aligned}$$</p>

About this question

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

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