JEE Main 2026 Jan 22 Shift 2, Mathematics Q6: Methods of Solving First Order, First Degree Differential Equation

Separate variables:

$\int \frac{\cos y}{1+2\sin y}dy=\int \frac{dx}{16\sqrt{x+9\sqrt{x}}\left(4+\sqrt{9+\sqrt{x}}\right)}$.

Let $t=4+\sqrt{9+\sqrt{x}}$. Then $dt=\frac{1}{2\sqrt{9+\sqrt{x}}}\cdot \frac{dx}{2\sqrt{x}}$, so the RHS becomes $\frac{1}{4}\int \frac{dt}{t}$.

Integrating: $\frac{1}{2}\ln |1+2\sin y|=\frac{1}{4}\ln \left|4+\sqrt{9+\sqrt{x}}\right|+C$.

At $(256,\pi /2)$:$\frac{1}{2}\ln 3=\frac{1}{4}\ln 9+C=\frac{1}{2}\ln 3+C\Rightarrow C=0$.

At $x=49$: $\sqrt{x}=7$, $\sqrt{9+7}=4$, so $4+4=8$.

$\frac{1}{2}\ln (2\sin \alpha +1)=\frac{1}{4}\ln 8=\frac{1}{2}\ln 2\sqrt{2}$.

So $2\sin \alpha +1=2\sqrt{2}\Rightarrow 2\sin \alpha =2\sqrt{2}-1$.