# Exact Differential Equation

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• Jul 15th 2009, 03:43 PM
redfox2600
Exact Differential Equation
need your help on this...

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

Answer: x^2 + y^2 = cy
• Jul 15th 2009, 03:51 PM
Krizalid
It's also homogeneous, or you do need to solve it as a exact one?
• Jul 15th 2009, 04:12 PM
Random Variable
That's not an exact equation unless it's $\displaystyle -2xydx +(y^{2}-x^{2}) dy =0$ or $\displaystyle 2xydx +(y^{2}+x^{2}) dy =0$
• Jul 15th 2009, 05:15 PM
redfox2600
thanks.
you're right, it's homogeneous.

a little need help please.

i'm stuck in the ending part.

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

thanks.
• Jul 15th 2009, 05:30 PM
Krizalid
Quote:

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.
• Jul 16th 2009, 03:29 PM
redfox2600
thanks for big help Krizalid.
• Jul 22nd 2009, 01:20 AM
defog171
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?
• Jul 24th 2009, 03:06 PM
redfox2600
yes. thanks for the great help.