CalculusHardDifferential Equations ShareDifferential Equations · Hard problemSolve the IVP dydx=y2sinx\dfrac{dy}{dx}=y^2\sin xdxdy=y2sinx, y(0)=1y(0)=1y(0)=1, and state where the solution is valid.A∣x∣<π3|x|<\tfrac{\pi}{3}∣x∣<3πB∣x∣<π4|x|<\tfrac{\pi}{4}∣x∣<4πC∣x∣<2π|x|<\tfrac{2}{\pi}∣x∣<π2D∣x∣<π2|x|<\tfrac{\pi}{2}∣x∣<2πCheck Show answer Walk through it step by step Practice more like this Review this concept