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{\bf Question}
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Find the solution of the initial value problem

$\displaystyle \frac{dx}{dt} = \frac{2}{3} \frac{1}{x^2} =
\frac{2}{3} t \times \frac{1}{x^2}$; with $x(0) = 0$

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{\bf Answer}

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$\displaystyle \frac{dx}{dt} = \frac{2}{3} \frac{1}{x^2} =
\frac{2}{3} t \times \frac{1}{x^2}$

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Separable: \begin{eqnarray*} \Rightarrow x^2 \, dx & = &
\frac{2}{3} t \, dt \\ \Rightarrow \int x^2 \, dx & = &
\frac{2}{3} \int t \, dt \\ \Rightarrow \frac{1}{3}x^3 & = &
\frac{1}{3}t^2 + \mathrm{constant} \\ \Rightarrow x^3  & = & t^2 +
\mathrm{constant}
\end{eqnarray*}
Now $x(0) = 0$, so constant is zero

Solution is $\displaystyle x^3 = t^2 \Rightarrow \ \ \ x =
t^{\frac{2}{3}}$




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