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Consider the differential equation $$y'\sin(y')=y\cos(y) y(a)=b.$$ The function $f(x)=x\sin(x)$ is invertible in the interval $[0, \frac{\pi}{2}].$

  • What does this mean for the existence and uniqueness of the solution of the equation?

  • Does a unique solution exist in the interval $[0,\frac{\pi}{2}]$?

  • For which pairs $(a,b)$ is the equation uniquely solvable in which interval?

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1 Answer 1

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Given $ b \in [0,\pi/2] $, the invertibility gives $ y'=F (y \cos (y))$. The inverse function theorem and taking a derivative tell you that F is Lipschitz on any $[\delta,\pi/2] $ where $\delta > 0$. At $0$, $ F $ is $0$, so initial conditions there can stay zero for a while before starting to increase.

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