class 12 maths differential equations

$\cfrac{{dy}}{{dx}} = \sqrt {4 - {y^2}} ( - 2 < y < 2)$

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📘 Differential Equations NCERT Ex.9.4,Q.2,Page 395 SA

$\cfrac{{dy}}{{dx}} = \sqrt {4 - {y^2}} ( - 2 < y < 2)$

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.: We have $\cfrac{{dy}}{{dx}} = \sqrt {4 - {y^2}}$ $( - 2 < y < 2)$

…(1)
$\cfrac{{dy}}{{\sqrt {4 - {y^2}} }} = dx$

Integrating (1) both sides,

we get
$\cfrac{d}{{\sqrt {4 - {y^2}} }} = \int {dx} \Rightarrow {\sin ^{ - 1}}\cfrac{y}{2} = x + C \Rightarrow \cfrac{y}{2} = \sin (x + C)$

$\Rightarrow y = 2\sin (x + C)$

which is required solution.

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