Q.14 Find the particular solution of : rdr/dθ+cosθ=5 at r=sqrt2 and θ=0

Answer: r²=10θ-2sinθ+2 r dr=(5-cosθ) dθ . Integrating, r²/2=5θ-sinθ+c . At θ=0 , r=sqrt2 : 2/2=0-0+c⇒ c=1 . Particular solution: r²/2=5θ-sinθ+1 , i.e. r²=10θ-2sinθ+2 .

Chapter: Differential Equations· Mathematics & Statistics Part 2

Find the particular solution of : \(r\dfrac{dr}{d\theta}+\cos\theta=5\) at \(r=\sqrt2\) and \(\theta=0\)

Q.14 Find the particular solution of : \(r\dfrac{dr}{d\theta}+\cos\theta=5\) at \(r=\sqrt2\) and \(\theta=0\)

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \(r^2=10\theta-2\sin\theta+2\)

\(r\,dr=(5-\cos\theta)\,d\theta\). Integrating, \(\dfrac{r^2}{2}=5\theta-\sin\theta+c\).

At \(\theta=0\), \(r=\sqrt2\): \(\dfrac{2}{2}=0-0+c\Rightarrow c=1\).

Particular solution: \(\dfrac{r^2}{2}=5\theta-\sin\theta+1\), i.e. \(r^2=10\theta-2\sin\theta+2\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Find the particular solution of : \(r\dfrac{dr}{d\theta}+\cos\theta=5\) at \(r=\sqrt2\) and \(\theta=0\)

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