# The differential equation $y'\sin(y')=y\cos(y)$

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?

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.