# Thread: Ordinary Differential Equations - tough IVP

1. ## Ordinary Differential Equations - tough IVP

Hi, I am supposed to solve the following initial value problem but am stuck with an ugly integral:

y' = y(1-y) y(0)=yo (y subscript 0), where is an arbitrary constant.

I have worked it out as follows:

dy/dt = (y-y)

dy/(y-y) = dt

[ dy/(y-y) ] = [ dt ]

[ dy/(y-y) ] = t + c

But how do you integrate the left side? Or is there another way to solve this for y? Thank you.

2. Originally Posted by thecelticswin
Hi, I am supposed to solve the following initial value problem but am stuck with an ugly integral:

y' = y(1-y) y(0)=yo (y subscript 0), where is an arbitrary constant.

I have worked it out as follows:

dy/dt = (y-y)

dy/(y-y) = dt

[ dy/(y-y) ] = [ dt ]

[ dy/(y-y) ] = t + c

But how do you integrate the left side? Or is there another way to solve this for y? Thank you.
$\int\frac{\,dy}{y-y^2}=\int\frac{\,dy}{y(1-y)}$

You can apply partial fraction decomposition

-OR-

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

So, $\int\frac{\,dy}{y(1-y)}=\int\left[\frac{1}{y}-\frac{1}{y-1}\right]\,dy$

This is a bit easier to integrate now.

Does this make sense?

--Chris

3. thank you, this certainly does make it easier.

4. ## another problem...

I seem to have run into another snag. I solved the integral and have: ln|y|-ln|y-1|=t+c. Solving for y, I obtained:

y= -e^(
t+c) / (1-e^(t+c))

however, the answer that I am supposed to find is y=yo/[yo+(1-yoe^-
t)],
where y(0) = yo. How did this result get obtained from the integral??

5. Originally Posted by thecelticswin
I seem to have run into another snag. I solved the integral and have: ln|y|-ln|y-1|=t+c. Solving for y, I obtained:

y= -e^(
t+c) / (1-e^(t+c))

however, the answer that I am supposed to find is y=yo/[yo+(1-yoe^-
t)],
where y(0) = yo. How did this result get obtained from the integral??
Substitute the initial condition. In other words, when $x=0$, $y=y_0$. Solve for the constant.

6. I understand this, but still cannot get my equation to look like the answer. Could you provide any assistance?

7. or is e^c = yo??

8. Originally Posted by thecelticswin
I seem to have run into another snag. I solved the integral and have: ln|y|-ln|y-1|=t+c. Solving for y, I obtained:

y= -e^(
t+c) / (1-e^(t+c))

however, the answer that I am supposed to find is y=yo/[yo+(1-yoe^-
t)],
where y(0) = yo. How did this result get obtained from the integral??
I just worked out the integral again, and received a different answer...

$\ln |y| - \ln |y-1| = \alpha t + c$
$\ln \left|\frac{y}{y-1}\right | = \alpha t + c$
$\left | \frac{y}{y-1} \right | = e^{\alpha t + c}$
$\left | \frac{y}{y-1} \right | = e^c e^{\alpha t}$
$\frac{y}{y-1} = \pm e^c e^{\alpha t}$
$\frac{y}{y-1} = A e^{\alpha t}$
$1 + \frac{1}{y-1} = A e^{\alpha t}$

Can you go from here?

9. Err, for some reason the y and yo are confusing me. From this point, I did:

1 / (y-1) = -1+Ae^t

y-1 = 1 / (-1+Ae^
t)

y = (1 + Ae^
t - 1) / (Ae^t - 1)

y = (
Ae^t ) / (Ae^t - 1)

when t=0, y=yo

yo = A/(A-1)

yoA-yo = A

-yo = A(1-yo)

A = (1-yo)/-yo = (yo-1)/yo

And now I am stuck again. When I plug this in for A, I still cannot get it to look right..

10. Originally Posted by thecelticswin
Err, for some reason the y and yo are confusing me. From this point, I did:

1 / (y-1) = -1+Ae^t

y-1 = 1 / (-1+Ae^
t)

y = (1 + Ae^
t - 1) / (Ae^t - 1)

y = (
Ae^t ) / (Ae^t - 1)

when t=0, y=yo

yo = A/(A-1)

yoA-yo = A

-yo = A(1-yo)

A = (1-yo)/-yo = (yo-1)/yo

And now I am stuck again. When I plug this in for A, I still cannot get it to look right..
That's because $A = \frac{y_0}{y_0 - 1}$, not $\frac{y_0 - 1}{y_0}$

11. Ok, got it. Thank you everyone.