Thread: 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.

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.

You can apply partial fraction decomposition

-OR-



So,

This is a bit easier to integrate now.

Does this make sense?

--Chris

thank you, this certainly does make it easier.

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??

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 , . Solve for the constant.

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

or is e^c = yo??

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...









Can you go from here?

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..

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 , not

Ok, got it. Thank you everyone.