# Help with this DE problem

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• February 3rd 2010, 07:20 AM
snaes
ok:
Quote:

Originally Posted by whitepenguin
$y' = \frac{y}{3x - y^2}$
Thank you

$\dfrac{dy}{dx}=\dfrac{y}{3x-y^2}$
$\dfrac{dy}{dx}(3x-y^2) = y$
$(3x-y^2)dy = ydx$
$(3x-y^2)dy-ydx=0$

now this is in the form of what my professor calls "exact" except we still need an integrating factor. They are quite difficult to find, for me, and honestly i cannot get this one to work out so im sorry but, i dont think i can help.
• February 3rd 2010, 01:28 PM
Calculus26
there is no one method of finding int factor but there are special cases

Two things

1)If 1/M (dN/dx - dM/dy) = f(y) then int factor is e^(integralf(y))

2) If 1/N (dM/dy- dN/dx) = f(x) tehn int factor is e^(integralf(x))

In your case 1) applies and 1/y^4 is the int factor
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12