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Differential Equations
NCERT Class 12
CBSE
Grade 12
Question:
Find the particular solution of the linear differential equation $\dfrac{dy}{dx} + y \cot x = 4x \csc x$ ($x eq 0$), given that $y = 0$ when $x = \pi/2$.
Step-by-Step Solution
$\text{I.F.} = e^{\log\sin x} = \sin x$. [1.5 Marks] $y \sin x = \int 4x dx = 2x^2 + C$. [1.5 Marks] At $x = \pi/2, y = 0 \Rightarrow 0 = 2(\pi^2/4) + C \Rightarrow C = -\dfrac{\pi^2}{2}$. [1.0 Mark] $y \sin x = 2x^2 - \dfrac{\pi^2}{2} \Rightarrow y = \left(2x^2 - \dfrac{\pi^2}{2}\right) \csc x$. [1.0 Mark]
--- 🎯 Official CBSE Marking Scheme: Evaluating I.F. $= \sin x$: 1.5 Marks Evaluating general solution $y\sin x = 2x^2 + C$: 1.5 Marks Evaluating constant $C = -\pi^2/2$: 1.0 Mark Evaluating particular solution $y = (2x^2 - \pi^2/2)\csc x$: 1.0 Mark
Correct Answer:
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