Question

If y(x) is a solution of $\left(\frac{2+sinx}{1+y}\right)\frac{dy}{dx}=-cosxandy\left(0\right)=1,$then find the value of $y\left(\frac{\pi }{2}\right).$

Answer 1

Given that, $\left(\frac{2+sinx}{1+y}\right)\frac{dy}{dx}=-cosx\Rightarrow \frac{dy}{1+y}=-\frac{cosx}{2+sinx}dx$
On integarting both sides, we get
$\int \frac{1}{1+y}dy=-\int \frac{cosx}{2+sinx}dx\Rightarrow log(1+y)=-log(2+sinx)+logC$
$\Rightarrow log(1+y)+log(2+sinx)=logC\Rightarrow log\left[\right(1+y\left)\right(2+sinx\left)\right]=logC\Rightarrow (1+y)(2+sinx)=C\Rightarrow 1+y=\frac{C}{2+sinx}\Rightarrow y=\frac{C}{2+sinx}-1$ ...(i)
When x=0 and y=1, then
$1=\frac{C}{2}-1\Rightarrow C=4$
On putting C=4 in Eq. (i), we get
$y=\frac{4}{2+sinx}-1y\left(\frac{\pi }{2}\right)=\frac{4}{2+sin\frac{\pi }{2}}-1=\frac{4}{2+1}-1=\frac{4}{3}-1=\frac{1}{3}$