Does Theorem 1 Guarantee a Unique Solution for Given Differential Equations?

[cbarker1]

Hello,

In my book on Differential Equations, There is a Theorem that states: "Consider the IVP
$\d{y}{x}=f(x,y), y(x_0)=y_0$

If $f(x,y)$ and $\pd{f}{y}$ are continuous in some $a<x<b$, $c<y<d$ containing the point $(x_0,y_0)$, then the IVP has a unique solution $y=\phi(x)$ in some Interval $x_0-\delta<x<x_0+\delta$.

The question:

Consider the DE: $\d{y}{x}=\sqrt{y^2-9}$, $y(x_0)=y_0$. Determine whether Theorem 1 guarantees that this DE possesses a unique solution through the following points:
a) (1,4)
b) (5,3)
c) (2,-3)
d) (-1,1)

Work

The function $f(x,y)=\sqrt{y^2-9}$ is continuous everwhere except when
$y^2-9\ge0$
$y^2\ge9$
$y\ge\left| 3 \right|$
Therefore, the function does not exist when $-3>y>3$.

$\pd{f}{y}=\frac{1}{2}\frac{2y}{\sqrt{y^2-9}}=\frac{y}{\sqrt{y^2-9}}$
The function is continuous everywhere except when
$y^2-9>0$
$y^2>9$
$y>\left| 3 \right|$
Therefore the function is continuous when y<-3 and y>3.

The conclusion with these points:
a) The Theorem applies to the point.
b) The Theorem applies to the point.
c) the Theorem applies to the point.
d.) Theorem does not apply to the point.Thank you for your help,

Cbarker1

 

[Ackbach]

Hello,

In my book on Differential Equations, There is a Theorem that states: "Consider the IVP
$\d{y}{x}=f(x,y), y(x_0)=y_0$

If $f(x,y)$ and $\pd{f}{y}$ are continuous in some $a<x<b$, $c<y<d$ containing the point $(x_0,y_0)$, then the IVP has a unique solution $y=\phi(x)$ in some Interval $x_0-\delta<x<x_0+\delta$.

The question:

Consider the DE: $\d{y}{x}=\sqrt{y^2-9}$, $y(x_0)=y_0$. Determine whether Theorem 1 guarantees that this DE possesses a unique solution through the following points:
a) (1,4)
b) (5,3)
c) (2,-3)
d) (-1,1)

Work

The function $f(x,y)=\sqrt{y^2-9}$ is continuous everwhere except when
$y^2-9\ge0$

Actually, the function is continuous precisely there! If $y^2-9<0$, then the function does not exist, at least not in the real world.

$y^2\ge9$
$y\ge\left| 3 \right|$
Therefore, the function does not exist when $-3>y>3$.

$\pd{f}{y}=\frac{1}{2}\frac{2y}{\sqrt{y^2-9}}=\frac{y}{\sqrt{y^2-9}}$
The function is continuous everywhere except when
$y^2-9>0$

Again, this is where the partial derivative actually does exist.

$y^2>9$
$y>\left| 3 \right|$
Therefore the function is continuous when y<-3 and y>3.

The conclusion with these points:
a) The Theorem applies to the point.
b) The Theorem applies to the point.
c) the Theorem applies to the point.
d.) Theorem does not apply to the point.Thank you for your help,

Cbarker1

You've got the right idea. Just carry those corrections down.