IIT-JEE 2004 ScreeningQuestion 1
If $y=y(x)$ and $\dfrac{2+\sin x}{y+1}\left(\dfrac{dy}{dx}\right)=-\cos x$, $y(0)=1$, then $y\left(\dfrac{\pi}{2}\right)$ equals
Differential Equation
Solution
Concept. Separate the variables; both sides integrate to logarithms. Formulas used. (1) $\int\dfrac{\cos x\,dx}{2+\sin x}=\ln(2+\sin x)$Step 1 - Integrate. $\dfrac{dy}{y+1}=-\dfrac{\cos x}{2+\sin x}\,dx$, so by (1), $(y+1)(2+\sin x)=C$. Step 2 - Use $y(0)=1$. $C=4$. Step 3 - At $x=\frac\pi2$. $3(y+1)=4$, so $y=\frac13$. Answer: (A)