Question

For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to this equation through the point
1. (−4,6)

2. (2,−6)

3. (−5,39)

4. (−1,45)

Please explain the process of obtaining the answer.

Answer 1

Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45)  there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

$F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }$

Now, let's find the domain of $F(x,y)$, due to the square root:

$y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6$

So the domain of the function is:

$y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6$

Now, due to the fraction $\frac{\partial F}{\partial y}$ the denominator must be also different from 0, so:

$y^2-36\neq0\\\\y \neq \pm6$

So, the theorem  tells us that for each $y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0$ there exists a  unique solution defined in an open interval around $x_0$.

1. (-4,6)  there is no a solution to the equation through this point because $y_0=6$

2. (2,−6)  there is no a solution to the equation through this point because

$y_0=-6$

3. (−5,39) there is a solution to the equation through this point because

$y_0>6$

4. (−1,45)  there is a solution to the equation through this point because

$y_0>6$