For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to this equation t

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.

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Vivian 3 weeks 2022-01-07T08:36:41+00:00 1 Answer 0 views 0

Answers ( )

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    2022-01-07T08:38:07+00:00

    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<-6 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

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45:7+7-4:2-5:5*4+35:2 =? ( )