For each of the differential equations in find the general soluti… - Vidyadip

Question

For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\sqrt{4-\text{y}^2}(-2<\text{y}<2)$

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Answer

Given: Differential equation $\frac{\text{dy}}{\text{dx}}=\sqrt{4-\text{y}^2} \ \Rightarrow \ \ \text{dy}=\sqrt{4-\text{y}^2}\ \text{dx}$
$\Rightarrow \ \frac{\text{dy}}{\sqrt{4-\text{y}^2}}=\text{dx}$
Integrating both sides, $\int \frac{\text{dy}}{\sqrt{4-\text{y}^2}}\ \text{dy}=\int1\ \text{dx}$
$\Rightarrow \ \ \text{sin}^{-1}\frac{\text{y}}{2}=\text{x}+\text{c}\ \ \Bigg[\therefore\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\ \text{dx}=\text{sin}^{-1}\frac{\text{x}}{\text{a}}\Bigg]$
$\frac{\text{y}}{2}=\text{sin}(\text{x+c)} \ \Rightarrow \ \text{y}=2\text{sin}(\text{x+c)}$

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