Question

In Problems, $\mathbf{17}\mathbf{-}\mathbf{24}$determine a region of the xy-plane for which

the given differential equation would have a unique solution whose

graph passes through a point $\mathbf{(}{\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{)}$in the region.

$\left(1+y^3\right)\mathit{y}\mathbf{\text{'}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$

Short answer

The required region where the differential equation have unique solution is everywhere in xy-plane except $y=-1$

Step-by-step solution

Note the given data

Given differential equation, $\left(1+y^3\right)y\text{'}=x^2$

Dividing $1+y^3$both the sides and we get the function $f(x,y)=\frac{x^2}{1+y^3}$

It contains an interior point $(x_0,y_0)$ in the graph

Finding the continuity of the function

The given function $f(x,y)=\frac{x^2}{1+y^3}$

The function $f(x,y)=\frac{x^2}{1+y^3}$is continuous everywhere in xy plane except $1+y^3=0$i.e., the function is continuous everywhere in xy plane except $y=-1$.

Finding the partial derivative

Finding the partial derivative of the given function with respect to y

$\frac{\delta f}{\delta y}=\frac{\delta }{\delta y}\left(\frac{x^2}{1+y^3}\right)$

$$\begin{gathered} =\frac{x^2\times 3y^2}{{(1+y^3)}^2} \\ =\frac{3x^2{y}^2}{{(1+y^3)}^2} \end{gathered}$$

This partial derivative is continuous everywhere in xy plane except $(1+y^3{)}^2=0$

i.e., continuous everywhere in xy plane except$y=-1$

Finding the required region

From given data $(x_0,y_0)$is an interior point.

So, we get $y_0\ne -1$

From thetheorem of Existence of a unique solution, we get

The region of the unique solution exists of the given differential equation is everywhere in xy-plane except $y=-1$