Differential Equations Question 12
Question 12 - 2024 (30 Jan Shift 1)
Let $y=y(x)$ be the solution of the differential equation $\sec x d y+{2(1-x) \tan x+x(2-x)} dx=0$ such that $y(0)=2$. Then $y(2)$ is equal to :
(1) 2
(2) $2{1-\sin (2)}$
(3) $2{\sin (2)+1}$
(4) 1
Show Answer
Answer (1)
Solution
$$ \frac{d y}{d x}=2(x-1) \sin x+\left(x^{2}-2 x\right) \cos x $$
Now both side integrate
$$ \begin{aligned} & y(x)=\int 2(x-1) \sin x d x+\left[\left(x^{2}-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\ & y(x)=\left(x^{2}-2 x\right) \sin x+\lambda \\ & y(0)=0+\lambda \Rightarrow 2=\lambda \\ & y(x)=\left(x^{2}-2 x\right) \sin x+2 \\ & y(2)=2 \end{aligned} $$
