Solving a differential Eq. by separation of variables

  • #1

Homework Statement


Find all solutions. Solve explicitly for y.

y[itex]^{'}[/itex]=y[itex]^{2}[/itex]-y

Homework Equations





The Attempt at a Solution


Case where y'=0

0=y(y-1) y=0,1 when y(t)=0

Case where y'[itex]\neq[/itex]0

y'=y[itex]^{2}[/itex]-y

[itex]\frac{1}{y^{2}-y}[/itex]y'=1

[itex]\int\frac{1}{y^{2}-y}[/itex]y'dt=∫1dt

[itex]\int\frac{1}{y^{2}-y}[/itex]dy=t+c

Cant figure our where to go from here.
 
Last edited:

Answers and Replies

  • #2
133
1
##\displaystyle\frac{1}{y^2 - y} = \frac{1}{y(y-1)}##

Use partial fraction decomposition to split the above into the form ##\displaystyle\frac{A}{y} + \frac{B}{y - 1}##. Then, integrating will be a breeze.
 
  • #3
\displaystyle\frac{A}{y} + \frac{B}{y - 1}

1=A(y-1)+By
1=Ay-A+By
A=-1
B=1

[itex]\int\frac{1}{y(y-1)}[/itex]=[itex]\int\frac{-1}{y}+\frac{1}{y-1}[/itex]

[itex]\int\frac{-1}{y}+\frac{1}{y-1}[/itex]=t+c

-ln|y|+ln|y-1|=t+c

ln([itex]\frac{y-1}{y}[/itex])=t+c

ln(1-[itex]\frac{1}{y}[/itex])=t+c

1-[itex]\frac{1}{y}[/itex]=e[itex]^{t+c}[/itex]

y=[itex]\frac{1}{1-e^{t+c}}[/itex]

I dont think this is right, wolfram alpha says it is y=[itex]\frac{1}{1+e^{t+c}}[/itex]

How does the plus sign pop up into the denominator?
 
  • #4
You've assumed |y-1| = y-1. Is this consistent with the range of y you get when you push on to the solution? The absolute value matters, you can't just drop it and move on.
 
  • #5
133
1
I believe it's due to the fact that the ##|y - 1|## can also be written as ##|1 - y|##, perhaps that is what Wolfram used. In any case, note that ##e^{t+c} = e^ce^t = ce^t## because ##e^c## is still a constant. So ##\displaystyle y = \frac{1}{1 - ce^t}## would be the best way to write the answer, since if ##c## is indeed negative, it would look like what Wolfram has.
 
  • #6
SammyS
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If 0 < y < 1 , then [itex]\displaystyle 1-\frac{1}{y}<0\,,\ [/itex] so that [itex]\displaystyle \left|1-\frac{1}{y}\right|=-\left(1-\frac{1}{y}\right)\ .[/itex]

That will give [itex]\displaystyle 1-\frac{1}{y}=-e^{t+C}\ . [/itex]

...
 
  • #7
HallsofIvy
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You were asked to find all solutions. Your formula, once you have solved for y, will give all except one solution. What is the solution you are missing?
 
  • #8
It wont give the y solution where y(t)=0 so all solutions for y are

y=[itex]\frac{1}{1 - ce^t}[/itex],0

where c is any real number

Right?
 
  • #9
HallsofIvy
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Yes, that is correct.
 

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