Differential Equation - Bernoulli Equation

  • Thread starter pat666
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  • #1
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Homework Statement


solve the differential equation
[tex] y'(t)=-4y+6y^3 [/tex]

Homework Equations





The Attempt at a Solution


I'm pretty sure (not positive) that this is a Bernoulli Equation.
I've been following this wiki in an attempt to solve: http://en.wikipedia.org/wiki/Bernoulli_differential_equation
[tex] (y'(t)=-4y+6y^3)/y^3 [/tex]
[tex] y'(t)/y^3=-4/y^2+6[/tex]
[tex] w=1/y^2 [/tex]
[tex] y'(t)*w/y=-4w+6[/tex]

I'm stuck here, also theres no x's or t's here so im not sure that it is a Bernoulli Equation?

thanks for any help
 

Answers and Replies

  • #2
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dy/dt = 6y^3 - 4y. This is a variable separable differential equation, no Bernoulli pangs required. dy/(y)(6y^2 - 4) = dt. Integrate the LHS preferably using partial fractions and voila.
 
  • #3
709
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[tex] dy/dt=-4y+6y^3 [/tex]
[tex]dy/(y(-4+6y^2))=dt[/tex]
[tex] u=y^2, du =2y dy [/tex]
[tex]=1/2 \int 1/(u(6u-4)) [/tex]
etc etc etc
[tex] t=1/8 (ln(3y^2-2)-2ln(y))+c [/tex]
[tex] 8t-c=ln((3y^2-2)/(y^2)) [/tex]
[tex] ce^t=(3y^2-2)/y^2 [/tex]

I am again stuck, how do I get y=----- out of this??

Thanks
 
  • #4
35
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Mate, you need to use partial fractions to separate the denominator and integrate on each term separately.

gif.latex?\frac{1}{y(6y^{2}-4)}%20=%20\frac{A}{y}%20+%20\frac{B}{6y^{2}-4}.gif


Find A and B using the concept of partial fractions (this page is pretty neat...teaches you everything you need to know about them : http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html), then multiply dy on both sides of the equation and integrate term by term.

gif.latex?\int\frac{dy}{y(6y^{2}-4)}%20=%20A\int\frac{dy}{y}%20+%20B\int\frac{dy}{6y^{2}-4}.gif
 
  • #5
HallsofIvy
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[tex] dy/dt=-4y+6y^3 [/tex]
[tex]dy/(y(-4+6y^2))=dt[/tex]
[tex] u=y^2, du =2y dy [/tex]
[tex]=1/2 \int 1/(u(6u-4)) [/tex]
etc etc etc
[tex] t=1/8 (ln(3y^2-2)-2ln(y))+c [/tex]
[tex] 8t-c=ln((3y^2-2)/(y^2)) [/tex]
[tex] ce^t=(3y^2-2)/y^2 [/tex]

I am again stuck, how do I get y=----- out of this??

Thanks

From
[tex]ce^t= \frac{3y^2- 2}{y^2}[/tex]
multiply both sides by [itex]y^2[/itex]:
[tex]ce^ty^2= 3y^2- 2[/tex]

[tex]3y^2- ce^ty^2= 2[/tex]
[tex](3- ce^t)y^2= 2[/tex]

Can you finish that?

(Does the problem require that you solve for x? Solutions to first order equations are often left as implicit functions.)
 
  • #6
709
0
certainly can:
[tex] y=sqrt(2/((3-ce^t))) [/tex]
I think thats right? and it keeps me away from partial fractions which I forgot about 3yrs ago.

Thanks svxx and Hallsofivy

edit:

oops, I used partial fractions in the integrations, guess I remember procedures and not names of said procedures...
 
Last edited:

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