Question

If y(x) is a solution of ${\displaystyle \left(\frac{2+\sin x}{1+y}\right)\frac{\text{dy}}{\text{dx}}}$ = – cosx and y (0) = 1, then find the value of ${\displaystyle y\left(\frac{\pi }{2}\right)}$.

Answer 1

Given equation is ${\displaystyle \left(\frac{2+\sin x}{1+y}\right)\frac{\text{dy}}{\text{dx}}}$ = – cosx

⇒ ${\displaystyle \left(\frac{2+\sin y}{\cos x}\right)\frac{\text{dy}}{\text{dx}}}$ = –(1 + y)

⇒ ${\displaystyle \frac{\text{dy}}{(1+y)}=-\left(\frac{\cos x}{2+\sin x}\right)\text{d}x}$

Integrating both sides, we get

${\displaystyle \int \frac{\text{dy}}{1+y}=-\int \frac{\cos x}{2+\sin x}\text{d}x}$

⇒ ${\displaystyle \log |1+y|=-\log |2+\sin x|+\log c}$

⇒ ${\displaystyle \log |1+y|+\log |2+\sin x|}$ = log c

⇒ ${\displaystyle \log (1+y)(2+\sin x)}$ = log c

⇒ ${\displaystyle (1+y)(2+\sin x)}$ = c

Put x = 0 and y = 1, we get

(1 + 1)(2 + sin 0) = c

⇒ 4 = c

∴ Equation is (1 + y)(2 + sinx) = 4

Now put x = ${\displaystyle \frac{\pi }{2}}$

∴ ${\displaystyle (1+y)(2+\sin \frac{\pi }{2})}$ = 4

⇒ (1 + y)(2 + 1) = 4

⇒ 1 + y = ${\displaystyle \frac{4}{3}}$

⇒ y = ${\displaystyle \frac{4}{3}-1}$

⇒ ${\displaystyle \frac{1}{3}}$

So, ${\displaystyle y\left(\frac{\pi }{2}\right)=\frac{1}{3}}$

Hence, the required solution is ${\displaystyle y\left(\frac{\pi }{2}\right)=\frac{1}{3}}$.