1. ## 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.

2. $ydy - 4xdx = 0$

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

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

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

3. 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

4. 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

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

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

Sorry, I had mixed up the signs.