91Ó°ÊÓ

Clearly, ${\mathrm{Ï•}}_1\left(\mathrm{x}\right)=0$as y is a solution to the given initial value problem.

Now taking $\mathrm{y}={\mathrm{Ï•}}_2\left(\mathrm{x}\right)={\left(\mathrm{x}-2\right)}^3$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=3{\left(x-2\right)}^2\\ \frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{y}}^{\frac{2}{3}}\end{array}$

(Because, $\mathrm{y}={\left(\mathrm{x}-2\right)}^3$, i.e., ${\mathrm{y}}^{\frac{1}{3}}=\left(\mathrm{x}-2\right)$)

which is identical to the given differential equation. So, ${\mathrm{Ï•}}_2\left(\mathrm{x}\right)={\left(\mathrm{x}-2\right)}^3$is a solution to the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{y}}^{\frac{2}{3}}$.

Hence, this initial value problem has multiple solutions.

Here,$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=3{\mathrm{y}}^{\frac{2}{3}}$

$\begin{array}{l}\frac{∂\mathrm{f}}{∂\mathrm{y}}=2{\mathrm{y}}^{\frac{-1}{3}}\\ \frac{∂\mathrm{f}}{∂\mathrm{y}}=\frac{2}{{\mathrm{y}}^{\frac{1}{3}}}\end{array}$

which is continuous in any rectangle containing the point $\left(0,{10}^{-7}\right)$.

Now from Step 1, we find that both of the functions$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$and$\frac{∂f}{∂y}$are continuous in any rectangle containing the point $\left(0,{10}^{-7}\right)$, so the hypotheses of Theorem 1 are satisfied.

It then follows from the theorem that the given initial value problem has a unique solution in an interval about $x=0$of the form $\left(-\mathrm{δ},\mathrm{δ}\right)$, where$\mathrm{δ}$is some positive number.

Therefore, the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathbf{y}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{,}\mathbf{}\mathbf{y}\left(0\right)\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}$ have a unique solution in a neighbourhood of $\mathbf{x}\mathbf{=}\mathbf{0}$.