Find the general solution of the differential | vedclass

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Question

(A) The given differential equation is: $\frac{dy}{dx} = \sqrt{4 - y^2}$.Separating the variables,we get:$\frac{dy}{\sqrt{4 - y^2}} = dx$.Integrating

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given differential equation is: $\frac{dy}{dx} = \sqrt{4 - y^2}$.Separating the variables,we get:$\frac{dy}{\sqrt{4 - y^2}} = dx$.Integrating both sides with respect to their variables:$\int \frac{dy}{\sqrt{2^2 - y^2}} = \int dx$.Using the standard integral formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,we get:$\sin^{-1}(\frac{y}{2}) = x + C$.Taking the sine of both sides:$\frac{y}{2} = \sin(x + C)$.Thus,the general solution is:$y = 2 \sin(x + C)$.

Find the general solution of the differential equation: $\frac{dy}{dx} = \sqrt{4 - y^2}$ where $-2 < y < 2$.

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