In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem $\frac{dy}{dx}=f(x,y).y(x_0)=y_0.$ If fand $\frac{d}{\text{d}\text{e}\text{l}\text{y}}$ are continuous functions in some rectangle $(x_0,y_0)(x)\frac{dy}{dx}=y^{\frac{1}{3}}y=0,y(0)=0,\frac{dy}{dx}=0y^{\frac{1}{3}}=0y=0\frac{dy}{dx}=y^{\frac{1}{3}}y(0)=0f(x,y)=x_0-,$ where $is$ a positive number. The method for separable equations can give a solution, but $it$ may not give alt the solutions. $To$ ilustrate this, consider the equation $\frac{dy}{dx}=y^{\frac{1}{3}}.$ Answer parts $(a)$ through $(d).$ solution.

For $y=0,y(0)=0,\frac{dy}{dx}=0,$ and $y^{\frac{1}{3}}=0.$ Thus, $it$ holds that the constant function $y=0$ satisties the inital value protiem $\frac{dy}{dx}=y^{\frac{1}{3}}$ with $y(0)=0.$

$(d)$ Finally, show that the conditions for the theorem are not satisfied for the initial value problem $in$ part $(b).$ Begin $by$ identifying $f(xy)in$ the initial value problem $in(b).$

$f(x,y)=$

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