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Given two differential equations

1) $\frac{dy}{dx} = x^{\frac{2}{5}}$

2) $\frac{dy}{dx} = y^{\frac{2}{5}}$

with the initial condition $y(0) = 0$ find which lacks uniqueness of the solution, and explain this via the fundamental theorem on existence and uniqueness of solutions of differential equations.

What i get:

On 1) Continuous around $(0,0)$ so there exist at least one solution, but i get the partial derivative of y to be 0 so not a unique solution, but the solution say it has a unique solution

On 2) Continuous around $(0,0)$ so there exist at least on solution, and partial derivate of y is equal to $\frac{y^{\frac{7}{5}}}{7}$ so a unique solution. The solution say this is not a unique solution.

Someone who can help me with this?

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$$y'=f(y,x)$$

For the first case the derivative is zerro and continuous so you have existence and uniqueness of the solution of the IVP.

For the second case, you have continuity of the function and existence but the derivative is not continuous so you don't have uniqueness. You integrated the RHS instead of differentiating $f(y,x)$. $$\dfrac {\partial y^{2/5} }{\partial y}=\dfrac 25 y^{-3/5}$$

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    $\begingroup$ Thank you, for the help :) $\endgroup$
    – magnus
    Apr 30 '20 at 21:57
  • $\begingroup$ You're welcome @magnus $\endgroup$
    – MtGlasser
    Apr 30 '20 at 21:58

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