Question

In Problems 25–28 determine whether Theorem 1.2.1 guarantees that the differential equation $\mathit{y}\mathbf{\text{'}}\mathbf{=}\sqrt{{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{9}}$ possesses a unique solution through the given point $\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}$ .

Short answer

No

Step-by-step solution

Theorem 1.2.1 Existence of a Unique solution

Let R be a rectangular region in the xy-plane defined by $a⩽x⩽b,c⩽y⩽d$ that contains the point $(x_0,y_0)$in its interior. If $f(x,y)$ and $\frac{df}{dy}$are continuous on R, then there exists some interval $I_0:(x_0-h,x_0+h),h>0$ contained in $[a,b]$ and a unique function $y\left(x\right)$, defined on $I_0$, that is a solution of the initial-value problem (2).

Find the partial derivative

Given that, $f(x,y)=\sqrt{y^2-9}$

The function is not continuous for $-3<y<3$because the value of function $f$ would be imaginary, as the square root of a negative number. But the function is continuous for $y⩽-3$and $y⩾3$. Because at the point $(2,-3),y=-3$, this is in the continuous region for the function $f(x,y)$.

Find the partial derivative$\frac{df}{dy}$

$$\begin{gathered} \frac{\partial f}{\partial y}=\frac{1}{2}(y^2-9{)}^{-1/2}\left(2y\right) \\ \frac{\partial f}{\partial y}=\frac{y}{\sqrt{y^2-9}} \end{gathered}$$

Simplify, and observe that unlike the function $f(x,y)$, $\frac{df}{dy}$is not continuous for $-3⩽y⩽3$. In this case, the value of $y=-3$at the point $(2,-3)$, within the region

where the y value for $\frac{df}{dy}$does not exist, so the partial derivative is not continuous at the point.

Therefore, Theorem 1.2.1 does not guarantee that the differential equation localid="1668178167671" $\mathit{y}\mathbf{\text{'}}\mathbf{=}\sqrt{{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{9}}$possesses a unique solution at the point localid="1668178288081" $\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}$ .