Differential Equation and Applications (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question
Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differential Equation and Applications (p-1)1 Mark
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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