Question

In Problems, determine a region of the xy-plane for which

the given differential equation would have a unique solution whose

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

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\sqrt{\mathbf{x}\mathbf{y}}$

Short answer

The required region where the differential equation have unique solution is $y<0,x<0$ or $y>0,x>0$.

Step-by-step solution

Note the given data

Given differential equation, $\frac{dy}{dx}=\sqrt{xy}$where $f(x,y)=\sqrt{xy}$

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

Finding the continuity of the function

The given function $f(x,y)=\sqrt{xy}$.

The function $f(x,y)=\sqrt{xy}$is continuous on xy-plane where$xy⩾0$.

Finding the partial derivative

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

$\frac{\delta f}{\delta y}=\frac{\delta }{\delta y}\left(\sqrt{xy}\right)$

$$\begin{gathered} =\frac{x}{2\sqrt{xy}} \\ =\frac{1}{2}\sqrt{\frac{x^2}{xy}} \\ =\frac{1}{2}\sqrt{\frac{x}{y}} \end{gathered}$$

This derivative is continuous on $y<0,x<0$or $y>0,x>0$.

Finding the required region

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

So, we get $y_0<0,x_0<0$ or $y_0>0,x_0>0$.

From the theorem of Existence of a unique solution, we get

The region of the unique solution exists of the given differential equation is both the variable are negative or both are positive.

Hence, the region is $y<0,x<0$or $y>0,x>0$