1. ## Exact Differential Equation

2xy dx + (y^2 - x^2) dy = 0

Answer: x^2 + y^2 = cy

2. It's also homogeneous, or you do need to solve it as a exact one?

3. That's not an exact equation unless it's $\displaystyle -2xydx +(y^{2}-x^{2}) dy =0$ or $\displaystyle 2xydx +(y^{2}+x^{2}) dy =0$

4. thanks.
you're right, it's homogeneous.

i'm stuck in the ending part.

ln |y| + ln |v^2 + 1| = c
how come that c has a y ( cy )

thanks.

5. Originally Posted by redfox2600

i'm stuck in the ending part.

ln |y| + ln |v^2 + 1| = c
how come that c has a y ( cy )
Suppose that you actually got $\displaystyle \ln y+\ln(v^2+1)=k,$ then put $\displaystyle k=\ln c$ ('cause both of them are constants), thus your equation becomes $\displaystyle y\big(v^2+1\big)=c,$ because of injectivity.

6. thanks for big help Krizalid.

7. the ODE is not exact, so you should find an "intergrating factor" to multiply with the original ODE so that the NEW ODE is solvable. R u interested on how to solve the problem?

8. yes. thanks for the great help.