JEE Main 2025 Jan 24 Shift 2, Mathematics Q23: Methods of Solving First Order, First Degree Differential Equation

Divide through by $2\cos x$ and use $\sin 2x=2\sin x\cos x$:

$\frac{dy}{dx}+2y\tan x=\sin x$

This is linear. Integrating factor: $e^{\int 2\tan xdx}=e^{-2\ln |\cos x|}={\sec }^{2}x$.

$\frac{d}{dx}(y{\sec }^{2}x)=\sin x{\sec }^{2}x=\tan x\sec x$

Integrate: $y{\sec }^{2}x=\sec x+C\Rightarrow y=\cos x+C{\cos }^{2}x$.

Apply $y(\pi /3)=0$: $\frac{1}{2}+C\cdot \frac{1}{4}=0\Rightarrow C=-2$.

So $y(x)=\cos x-2{\cos }^{2}x$, and

$y^{\prime }(x)=-\sin x+4\cos x\sin x=-\sin x+2\sin 2x$

Evaluate at $x=\pi /4$:

$y(\pi /4)=\frac{1}{\sqrt{2}}-2\cdot \frac{1}{2}=\frac{1}{\sqrt{2}}-1$

$y^{\prime }(\pi /4)=-\frac{1}{\sqrt{2}}+2\cdot 1=-\frac{1}{\sqrt{2}}+2$

Sum: $\left(\frac{1}{\sqrt{2}}-1\right)+\left(-\frac{1}{\sqrt{2}}+2\right)=1$.

Answer: 1