Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
Let $y = y(x)$ be the solution of the differential equation $2\cos x\,\dfrac{dy}{dx} = \sin 2x - 4y\sin x$, $x\in\left(0,\dfrac{\pi}{2}\right)$. If $y\!\left(\dfrac{\pi}{3}\right) = 0$, then $y'\!\left(\dfrac{\pi}{4}\right)+y\!\left(\dfrac{\pi}{4}\right)$ is equal to ____.
Step-by-Step Solution
Key Concept: Rewrite as $\dfrac{dy}{dx}+2y\tan x = \sin x$; the integrating factor is $\sec^2 x$, giving $\dfrac{d}{dx}(y\sec^2 x)=\tan x\sec x$.
Rewrite: $\dfrac{dy}{dx}+2y\tan x = \sin x$. I.F. $= e^{2\int\tan x\,dx} = \sec^2 x$.
$\dfrac{d}{dx}(y\sec^2 x) = \sin x\sec^2 x = \tan x\sec x$.
$y\sec^2 x = \sec x + C$.
At $x=\pi/3$: $y=0 \Rightarrow 0 = 2+C \Rightarrow C=-2$.
$y = \cos x-2\cos^2 x$.
$y(\pi/4) = \tfrac{1}{\sqrt{2}}-1$.
$y'(x) = -\sin x+4\cos x\sin x \Rightarrow y'(\pi/4) = -\tfrac{1}{\sqrt{2}}+2$.
$$y'\!\left(\frac{\pi}{4}\right)+y\!\left(\frac{\pi}{4}\right) = \left(-\frac{1}{\sqrt{2}}+2\right)+\left(\frac{1}{\sqrt{2}}-1\right) = 1.$$
Correct Answer: 1