What is a particular solution to the differential equation dydx=2x1−y2 and y(0)=0?

dydx=2x1−y211−y2dydx=2x∫11−y2dydxdx=∫2xdx∫11−y2dy=∫2xdx⇒sin−1y=x2+Cy=sin⁡(x2+C)y(0)=00=sin⁡(0+C)⇒C=0meaningy=sin⁡x2

Expert answers to "What is a particular solution to the differential..."

Yaritza Estrada

2022-04-19

What is a particular solution to the differential equation

\frac{d y}{d x} = 2 x \sqrt{1 - y^{2}}

and y(0)=0?

Emily Green

Beginner2022-04-20Added 14 answers

\frac{d y}{d x} = 2 x \sqrt{1 - y^{2}}

\frac{1}{\sqrt{1 - y^{2}}} \frac{d y}{d x} = 2 x

\int \frac{1}{\sqrt{1 - y^{2}}} \frac{d y}{d x} d x = \int 2 x d x

\int \frac{1}{\sqrt{1 - y^{2}}} d y = \int 2 x d x

\Rightarrow \sin^{- 1} y = x^{2} + C

y = \sin (x^{2} + C)

y(0)=0

0 = \sin (0 + C) \Rightarrow C = 0

meaning

y = {\sin x}^{2}

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