Question
Solve the following differential equations:$dr =\operatorname{ar} d \theta-\theta dr$

Answer

$
\begin{aligned}
& dr = ar d \theta-\theta dr \\
& \therefore dr +\theta dr = ar d \theta \\
& \therefore(1+\theta) dr = ar d \theta \\
& \therefore \frac{d r}{r}=\frac{a d \theta}{1+\theta}
\end{aligned}
$
On integrating, we get
$
\begin{aligned}
& \int \frac{d r}{r}=a \int \frac{d \theta}{1+\theta} \\
& \therefore \log |r|=a \log |1+\theta|+C
\end{aligned}
$
This is the general solution.

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