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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution $\large\frac{dy}{dx} $$=\sqrt {4-y^2} \; (-2\lt y \lt 2) $

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  • $\int\large\frac{1}{\sqrt{ a^2 - x^2}}$$ =\sin^{-1} (\large\frac{x}{a})$
Step 1:
Given: $\large\frac{dy}{dx }$$= \sqrt{ 4 - y^2}$
seperating the variables we get,
$\large\frac{dy}{\sqrt{ 4-y^2}} $$= dx$
Step 2:
Integrating on both sides we get,
using the hint from the tool box
$\sin^{-1}(\large\frac{y}{2 }$$)= x + C$
$\large\frac{y}{2 }$$= \sin(x+C)$
$y = 2\sin(x + C)$
This is the required general solution.
answered Aug 15, 2013 by sreemathi.v
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