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February 7th, 2008, 09:54 AM   #1
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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
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February 7th, 2008, 09:59 AM   #2
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2x/(1+x^2)dx=3*dy/(y(y^3)-1)
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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.
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February 8th, 2008, 08:19 AM   #4
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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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