Question

In problems 15 and 16 determine by inspection at least two solutions of the given first-order IVP.

$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathit{y}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The two solutions of the initial value problem are $y=0$ and $y=x^3$.

Step-by-step solution

Note the given data 

Given data

$y\left(0\right)=0$ and $y\text{'}=3y^{\frac{2}{3}}$

Divide both sides of the above equation by $y^{\frac{2}{3}}$, we get

$$\begin{gathered} \frac{y\text{'}}{y^{\frac{2}{3}}}=\frac{3y^{\frac{2}{3}}}{y^{\frac{2}{3}}} \\ \Rightarrow \frac{y\text{'}}{y^{\frac{2}{3}}}=3 \end{gathered}$$

Finding solutions by variable separated method

Now we obtain the solution by using variable separated method

$$\begin{gathered} \int \frac{dy}{y^{\frac{2}{3}}}=\int 3dx \\ \Rightarrow \int y^{-\frac{2}{3}}dy=\int 3dx \end{gathered}$$

$\Rightarrow \frac{y^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}=3x+C$, where C is the constant of integration

$\Rightarrow 3y^{\frac{1}{3}}=3x+C$

Determine the solutions by using given initial value

Now, we putting the initial value $y\left(0\right)=0$, we get

$$\begin{gathered} \Rightarrow 0=0+C \\ \Rightarrow C=0 \end{gathered}$$

Then we get one solution is $y=x^3+0$

$y=x^3$

The other solution is $y=0$as the differential equation has two solutions of initial value problem.

Hence, the two solutions of the initial value problem are $\mathit{y}\mathbf{=}\mathbf{0}$and $\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{3}}$.