Differential Equation - Bernoulli Equation

[pat666]

Homework Statement

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

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
(y'(t)=-4y+6y^3)/y^3
y'(t)/y^3=-4/y^2+6
w=1/y^2
y'(t)*w/y=-4w+6

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

thanks for any help

 

[SVXX]

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.

 

[pat666]

dy/dt=-4y+6y^3
dy/(y(-4+6y^2))=dt
u=y^2, du =2y dy
=1/2 \int 1/(u(6u-4))
etc etc etc
t=1/8 (ln(3y^2-2)-2ln(y))+c
8t-c=ln((3y^2-2)/(y^2))
ce^t=(3y^2-2)/y^2

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

Thanks

 

[SVXX]

Mate, you need to use partial fractions to separate the denominator and integrate on each term separately.

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.

 

[HallsofIvy]

dy/dt=-4y+6y^3
dy/(y(-4+6y^2))=dt
u=y^2, du =2y dy
=1/2 \int 1/(u(6u-4))
etc etc etc
t=1/8 (ln(3y^2-2)-2ln(y))+c
8t-c=ln((3y^2-2)/(y^2))
ce^t=(3y^2-2)/y^2

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

Thanks

From
ce^t= \frac{3y^2- 2}{y^2}
multiply both sides by y^2:
ce^ty^2= 3y^2- 2

3y^2- ce^ty^2= 2
(3- ce^t)y^2= 2

Can you finish that?

(Does the problem require that you solve for x? Solutions to first order equations are often left as implicit functions.)

 

[pat666]

certainly can:
y=sqrt(2/((3-ce^t)))
I think that's 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...