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 February 7th, 2008, 09:54 AM #1 Newbie   Joined: Feb 2008 Posts: 2 Thanks: 0 an ordinary differential equation I cannot solve this for the life of me. 3(1+(x^2))(dy/dx) = 2xy((y^3)-1) AT LEAST give me a way to rearrange it to be able to solve it with integrals. thanksss
 February 7th, 2008, 09:59 AM #2 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 2x/(1+x^2)dx=3*dy/(y(y^3)-1)
February 7th, 2008, 11:22 AM   #3
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Quote:
 Originally Posted by milin 2x/(1+x^2)dx=3*dy/(y(y^3)-1)
thank you! I solved it. I actually did do this, but integrated horribly wrong.

 February 8th, 2008, 08:19 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2272 3(1+(x²))(dy/dx) = 2xy((y^3)-1) For y ≠ 0, dividing by (1 + x²)²y^4 and rearranging gives ((1 + x²)(3y^-4)dy/dx - (1 - y^-3)(2x))/(1 + x²)² = 0. Integrating that gives (1 - y^-3)/(1 + x²) = C, where C is a constant. Hence y = 1/³√(1 - C(1 + x²)) (where C is negative or greater than 1, so that y is defined for all values of x) or y ≡ 0.

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### 3(1 x^2)dy/dx=2xy(y^3-1)

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