Let y=y(x) be the solution of the differential equation [ x ~d y+2(1-x) x+x(2-x) d x=0] su…

Question

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:

Accepted Answer

$$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$$

Now both side integrate

$$\begin{aligned}
& y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \
& y(x)=\left(x^2-2 x\right) \sin x+\lambda \
& y(0)=0+\lambda \Rightarrow 2=\lambda \
& y(x)=\left(x^2-2 x\right) \sin x+2 \
& y(2)=2
\end{aligned}$$

Solutions · 1