differential equations

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August 2nd 2007, 06:42 PM
davecs77

differential equations

How would you solve this:
ydy - 4xdx = 0
The answer in the book is:
y^2 = 4x^2 + C
I looked through the chapter and my notes from class, but do not see any examples that are similar to this. What is the problem asking and how do you know what form to put it in? I am confused on how to do these...thanks for the help and tutorial.
August 2nd 2007, 07:33 PM
tukeywilliams

$ydy - 4xdx = 0$

$\int y \ dy - \int 4x \ dx = c_1$ .

$\frac{1}{2}y^2 - \frac{4}{2}x^2 = c_1$

$\frac{1}{2}y^2 - 2x^2 = c_1$ where $c_1 \geq 0$ which is a family of hyperbolas.

Multiplying by $2$ and rearranging we get $y^2 = 4x^2 + c$ . Its basically separation of variables.
August 2nd 2007, 07:42 PM
davecs77

Quote:

Originally Posted by tukeywilliams

$ydy + 4xdx = 0$

$\int y \ dy + \int 4x \ dx = c_1$ .

$\frac{1}{2}y^2 + \frac{4}{2}x^2 = c_1$

$\frac{1}{2}y^2 + 2x^2 = c_1$ where $c_1 \geq 0$ which is a family of ellipses.

Multiplying by $2$ and rearranging we get $y^2 = 4x^2 + c$ .

How would you get y^2 = 4x^2 + C
I can get it in the form for y^2 + 4x^2 = C by multiplying every term by 1/2
subtracting a 4x^2 in that would make it: y^2 = -4x^2 + c
August 2nd 2007, 07:45 PM
CaptainBlack

Quote:

Originally Posted by davecs77

How would you get y^2 = 4x^2 + C
I can get it in the form for y^2 + 4x^2 = C by multiplying every term by 1/2
subtracting a 4x^2 in that would make it: y^2 = -4x^2 + c

Because there is a mistake in the sign of the 4x dx term it was -ve not positive. Make that correction and you will get the given answer.

RonL
August 2nd 2007, 07:47 PM
tukeywilliams

$y^2 - 4x^2 = c_1$

So $y^2 = 4x^2 + c_1$ .

Sorry, I had mixed up the signs.

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