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Problem 23 Easy Difficulty
(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.
Answer
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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Video Transcript
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Topics
Differential Equations
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Anna Marie V.
Campbell University
Caleb E.
Baylor University
Michael J.
Idaho State University