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Bernoulli ODE

  • Thread starter cue928
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  • #1
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(x^2)y' + 2xy = 5y^3
y' + (2/x)y = 5(x^-2)(y^3) v=y^-2, y=v^(-1/2), y' = -(1/2)v^(-3/2)
-(1/2)v^(-3/2)v' + (2/x)v^(-1/2) = 5x^-2(v^-3/2)
v' - (4/x)v = -10x^-2 Integrating factor: x^-4
(skipping a few mechanical steps)
vx^-4 = 2x^-5 + C
v = 2x^-1 + Cx^4
I come up with y = (2x^-1 + Cx^4)^-1/2

Problem is the book shows y^2 = x/(2+Cx^5)? What am I missing?
 

Answers and Replies

  • #2
fzero
Science Advisor
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y = (2x^-1 + Cx^4)^-1/2 and y^2 = x/(2+Cx^5) are equivalent. Multiply the first expression by [tex]\sqrt{x/x}[/tex].
 
  • #3
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OMG, I am an idiot. I should have seen that.
 

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