Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations2 Marks
Question
Find the general solution of the differential equation $\frac{d y}{d x}=\sqrt{4-y^{2}}(-2<y<2)$
✓
Answer
Here, we have $ \frac{d y}{d x}=\sqrt{4-y^{2}}$ $\Rightarrow \frac{d y}{\sqrt{4-y^{2}}}$ = dx Integrating throughout, we get $ \int \frac{\mathrm{dy}}{\sqrt{4-\mathrm{y}^{2}}}=\int \mathrm{d} \mathrm{x}$ $\Rightarrow \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}$ $\Rightarrow \sin ^{-1} \frac{y}{2}$ = x + c $\Rightarrow \frac{y}{2}=\sin (x+c)$ $\Rightarrow y=2 \sin (x+c)$ $\therefore ~y=2 \sin (x+c) $is the general solution of the given differential equation.
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