Question

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

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

Answer 1

The correct option is A $\frac{1}{3}$

$(2+sinx)\frac{dy}{dx}+(y+1)cosx=0$

$(2+sinx)\frac{dy}{dx}+ycosx+cosx=0$

$(2+sinx)\frac{dy}{dx}+ycosx=-cosx$

Now, Divide by $(2+sinx)$

$\frac{dy}{dx}+\left(\frac{cosx}{2+sinx}\right)y=\frac{-cosx}{(2+sinx)}$

$P=\frac{cosx}{2+sinx}$ $Q=\frac{-cosx}{2+sinx}$

$I.F=e^{\int p.dx}\Rightarrow e^{\int \frac{cosx}{2+sinx}}\Rightarrow e^{log(2+sinx)}$

$I.F\Rightarrow 2+sinx$

$y(I.F)=\int Q.IFdx+c$

$y(2+sinx)=\int -\frac{cosx}{2+sinx}.2+sinxdx$

$(2+sinx)y=-sinx+c$

if $y\left(0\right)=1$

$\Rightarrow (2+sin0)\left(1\right)=-sin\left(0\right)+c$

$2=c$

$y=\frac{2-sinx}{2+sinx}$

if $y=\frac{\pi }{2}\Rightarrow \frac{2-sin\pi /2}{2+sin\pi /2}$

$\Rightarrow \frac{2-1}{2+1}=\frac{1}{3}$

$y\left(\pi /2\right)=\frac{1}{3}$