Question

Find the general solution of the equation:
(y^2 −2xy)dx+(2xy−x^2)dy=0

Answer 1

Look for a function F such that
\[ \partial F/\partial x = y^2 - 2xy \]
and
\[ \partial F/\partial y = 2xy - x^2 \]
Then your equation is an expression for \( dF \)

Answer 2

I just wanna check my answer.. i put it into wolfram alpha and for some reason it gave me e^(-c) which is just a constant so i wasnt sure why it would put it like that as opposed to just putting an arbitrary constant C

Answer 3

my answer that i got was
y= \[x^2\pm \sqrt{x^4-4xC}\] ------------------
2x

Answer 4

hey

Answer 5

The DE we have is exact because \({\partial \over \partial x}(2xy−x^2)=2y-2x={\partial \over \partial y}(y^2 −2xy)=2y-2x.\) So there exists a solution f(x,y) such that:
1) \(\large f_x=y^2-2xy \text{ and } f_y=2xy−x^2.\)

Answer 6

By integrating both sides of the first equation with respect to \(x\) we get:
\(f(x,y)=y^2x-x^2y+g(y)\) \( (*)\)
Now differentiate both sides of \((*)\) with respect to y:
\(f_y=2xy-x^2+g'(y)=2xy-x^2 \implies g'(y)=0 \implies g(y)=c\). Hence \(f(x,y)=y^2x-x^2y+c\) is a solution of the given differential equation.

Answer 7

shatoor xD

Answer 8

I think it's better to write the solution as \(y^2x-x^2y=c\). And Shokran Lana, menkom net3allam! :D

Answer 9

Oh I might add one thing, you can use the quadratic formula to solve for y. :D

Answer 10

The quadratic equation is \(\large y^2x-x^2y+c=0 \implies y=\frac{x^2\pm \sqrt{x^4-4cx}}{2x}.\)