Question

If $\left(2+\sin x\right)\frac{dy}{dx}+\left(y+1\right)\cos x=0$ and $y\left(0\right)=1,$ then $y\left(\frac{\pi }{2}\right)$ is equal to :-

• A. $\frac{4}{3}$
• B. $\frac{1}{3}$
• C. $-\frac{2}{3}$
• D. $-\frac{1}{3}$

Answer 1

Answer: (B)

$\frac{1}{3}$

$(2+sinx)\frac{\partial y}{\partial x}+(y+1)cosx=0$

$(2+sinx)dy+(y+1)cosxdx=0$

$\downarrow$ $\downarrow$

N M

$\frac{\partial M}{\partial y}=cosx\frac{\partial N}{\partial x}=cosx,$ so $\frac{\partial N}{\partial x}$ ie,

The equation is axact

${\int }_{x}Mdx={\int }_x(y+1)cosxdx$ ( y is treated as constant)

= (y+1) sin x

${\int }_{y}Ndy=\int (2+sinx)dy$

$=(2+sinx)\int dy$ ( x is treated as constant)

$=2y+ysinx$

ignore the term y sin x as it already accurs in $\int Mdx$

General solution is (y+1)sin x+2y=c

$ycos=1\Rightarrow (1+1)0+2=c\Rightarrow c=2$

$\therefore (y+1)sinx+2y=2$

$x=\frac{\pi }{2}\Rightarrow (y+1)1+2y=2$

$\Rightarrow y=\frac{1}{3}$