💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # (a) Solve the differential equation $y' = 2x \sqrt {1 - y^2}.$(b) Solve the initial-value problem $y' = 2x \sqrt {1 - y^2}, y(0) = 0,$ and graph the solution.(c) Does the initial-value problem $y' = 2x \sqrt {1 - y^2}, y(0) = 2,$ have a solution? Explain.

## a) $\sin ^{-1} y=x^{2}+C$b) $$\sin ^{-1} y=x^{2} \text { and } y=\sin \left(x^{2}\right) \text { for }-\sqrt{\pi / 2} \leq x \leq \sqrt{\pi / 2}$$c) No since $C$ is undefined $y(x)$ does not have a particular solution at $y(0)=2$

#### Topics

Differential Equations

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##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University   ##### Michael J.

Idaho State University

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### Video Transcript

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#### Topics

Differential Equations

##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University   ##### Michael J.

Idaho State University

Lectures

Join Bootcamp