Solving a differential Eq. by separation of variables

[Duderonimous]

Homework Statement

Find all solutions. Solve explicitly for y.

y$^{'}$=y$^{2}$-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'$\neq$0

y'=y$^{2}$-y

$\frac{1}{y^{2}-y}$y'=1

$\int\frac{1}{y^{2}-y}$y'dt=∫1dt

$\int\frac{1}{y^{2}-y}$dy=t+c

Cant figure our where to go from here.

 

[Karnage1993]

$\displaystyle\frac{1}{y^2 - y} = \frac{1}{y(y-1)}$

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.

 

[Duderonimous]

\displaystyle\frac{A}{y} + \frac{B}{y - 1}

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

$\int\frac{1}{y(y-1)}$=$\int\frac{-1}{y}+\frac{1}{y-1}$

$\int\frac{-1}{y}+\frac{1}{y-1}$=t+c

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

ln($\frac{y-1}{y}$)=t+c

ln(1-$\frac{1}{y}$)=t+c

1-$\frac{1}{y}$=e$^{t+c}$

y=$\frac{1}{1-e^{t+c}}$

I don't think this is right, wolfram alpha says it is y=$\frac{1}{1+e^{t+c}}$

How does the plus sign pop up into the denominator?

 

[VantagePoint72]

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.

 

[Karnage1993]

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.

 

[SammyS]

If 0 < y < 1 , then $\displaystyle 1-\frac{1}{y}<0\,,\$ so that $\displaystyle \left|1-\frac{1}{y}\right|=-\left(1-\frac{1}{y}\right)\ .$

That will give $\displaystyle 1-\frac{1}{y}=-e^{t+C}\ .$

...

 

[HallsofIvy]

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?

 

[Duderonimous]

It won't give the y solution where y(t)=0 so all solutions for y are

y=$\frac{1}{1 - ce^t}$,0

where c is any real number

Right?

 

[HallsofIvy]

Yes, that is correct.