JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 15)
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:
$$2\{\sin (2)+1\}$$
2
1
$$2\{1-\sin (2)\}$$
Explanation
$$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$$
Now both side integrate
$$\begin{aligned} & y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\ & y(x)=\left(x^2-2 x\right) \sin x+\lambda \\ & y(0)=0+\lambda \Rightarrow 2=\lambda \\ & y(x)=\left(x^2-2 x\right) \sin x+2 \\ & y(2)=2 \end{aligned}$$
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