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:
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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