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REPEATER 2024 — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20243 Marks
Question
Solve the differential equation, $d r+(2 r \cot \theta+\sin 2 \theta) d \theta=0$.
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Answer

$d r+(2 r \cot \theta+\sin 2 \theta) d \theta=0$
Diving by $d \theta$, we get,
$\therefore \quad \frac{d r}{d \theta}+(2 r \cot \theta+\sin 2 \theta)=0$
$\therefore \quad \frac{d r}{d \theta}+(2 \cot \theta) r=-\sin 2 \theta\quad\quad\ldots\ldots (1)$
This is the linear differntial equation of the form,
$\frac{d r}{d \theta}+P . r=Q$, where $P=2 \cot \theta$ and $Q=-\sin 2 \theta$
$\therefore \quad$ Integrating factor (I.F.) $=e^{\int P . d x}=e^{\int 2 \cot \theta \cdot d x}=e^{2 \log \sin \theta}=e^{\log \sin ^2 \theta}=\sin ^2 \theta$
$\therefore \quad$ The solution of equation (1) is given by,
$r .(I . F)=.\int Q \cdot(I . F) \cdot d \theta+c$
$\therefore \quad r \cdot \sin ^2 \theta=\int-\sin 2 \theta \cdot \sin ^2 \theta \cdot d \theta+c$
$\therefore \quad r \cdot \sin ^2 \theta=\int-2 \sin \theta \cdot \cos \theta \cdot \sin ^2 \theta \cdot d \theta+c \quad \ldots [\because \sin 2 \theta=2 \sin \theta \cdot \cos \theta]$
$\therefore \quad r \cdot \sin ^2 \theta=-2 \int \sin ^3 \theta \cdot \cos \theta \cdot d \theta+c$
Put $\sin \theta=t$
$\therefore \quad \cos \theta \cdot d \theta=d t$
$\therefore \quad r \cdot t ^2=-2 \int t ^3 \cdot d t+c$
$\therefore \quad r \cdot t ^2=-2 \cdot \frac{ t ^4}{4}+c$
$\therefore \quad r . t ^2=-\frac{ t ^4}{2}+c$
$\therefore \quad r \cdot \sin ^2 \theta=-\frac{\sin ^4 \theta}{2}+c$
$\therefore \quad r \cdot \sin ^2 \theta+\frac{\sin ^4 \theta}{2}=c$
This is the required general solution.

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