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Bernoulli Equation

  • Thread starter amcavoy
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  • #1
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I cannot get the correct answer to this for some reason:

[tex]t^2y'+2ty-y^3=0[/tex]

I use the substitution [itex]v=y^{1-n}=y^{-2}\implies y=v^{-\frac{1}{2}}[/itex] and come up with [itex]y'=-\frac{1}{2}v^{-\frac{3}{2}}[/itex] and [itex]y^3=v^{-\frac{3}{2}}[/itex].

[tex]-\frac{1}{2}t^2v^{-\frac{3}{2}}+2tv^{-\frac{1}{2}}-v^{-\frac{3}{2}}=0[/tex]

Then multiplying through by [itex]v^{-2}[/itex] gives:

[tex]-\frac{1}{2}t^2v^{3}+2t-v^{3}=0=2t-\left(\frac{1}{2}t^2+1\right)v^3[/tex]

For which I would say:

[tex]v=\left(\frac{2t}{\frac{1}{2}t^2+1}\right)^{\frac{1}{3}}[/tex]

But according to the back of my book, that answer is incorrect. What did I do wrong?

Thanks for your help.
 

Answers and Replies

  • #2
HallsofIvy
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If [itex] y= v^{-\frac{1}{2}}[/itex] then [itex]y'= -\frac{1}{2}v^{-\frac{3}{2}}v'[/itex].

You seem to have dropped v' throughout. Your differential equation became an algebraic equation!
 
  • #3
saltydog
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And what happen to the constant of integration?

So you got:

[tex]t^2y^{'}+2ty-y^3=0[/tex]

Divide by [itex]t^2[/itex] and then divide by [itex]y^3[/itex], leads to:

[tex]y^{-3}dy+\frac{2}{t}y^{-2}t=\frac{1}{t^2}dt[/tex]

Noting that:

[tex]d(y^{-2})=-2y^{-3}dy[/tex]

and that's what you have up there on the left, we make the substitution:

[tex]v=y^{-2}[/tex]

and:

[tex]dv=-2y^{-3}dy[/tex]

Plug that in up there, rearrange to a first-order ODE in terms of v and t, find the integrating factor, integrate and not forgetting about the constant of integration, solve for v(t), then converting back to y, you should get:

[tex]y(t)=\pm \sqrt{\frac{5t}{2+5ct^5}}[/tex]

Now what?
 
  • #4
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Ahhh, I can't believe I forgot the v'! Thanks for your help.
 

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