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 $\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$ .

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

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

$\frac{df}{dy}=\frac{2y}{2\sqrt{y^2-9}}$

If $y>3$ or $y<-3$, we see that $f$ is not defined at $-3<y<3$, and $\frac{df}{dy}$ is not defined at $-3⩽y⩽3$.

Therefore, the region of the xy-plane where the differential equation would have a solution is as follows.

Hence, Theorem 1.2.1 does not guarantee a unique solution. Since the point $(-1,1)$is not in the region of $y>3$ or $y<-3$,

Then theorem 1.2.1 does not guarantee a unique solution to the differential equation.