Answer to Question #140581 in Differential Equations for Nikhil Singh

Question

The initial value problem
dy/dx = 3y^2/3, y(0)=0, may not have unique solution in R(|x| < 1,|y| < 1 ).
True or false with correct explanation

Answer 1

The function $3y^{2/3}$ is continuous everywhere in R(|x| < 1,|y| < 1 ), so, according to the Existence and Uniqueness Theorem a unique solution will exist for any initial condition . Let us solve the equation

$\dfrac{dy}{3y^{2/3}} = dx, \quad \int \dfrac{dy}{3y^{2/3}} = x + C, \quad y^{1/3} = x + C.$

According

$y^{1/3} = 0 + C = 0 \Rightarrow C = 0.$

Therefore, the solution will be $y^{1/3}=x$ and it is unique.