[Big Toosie]

I have work this problem (1/P)dP/dt=b+aP to the point that I am confused on the partial fraction part

(1/P)dP/dt=b+aP

1/dt=bP+aP^2/dp this where Iam stuck to finish the equation I must do partial fraction.:eek:

[asterisk_man]

Can you state the original problem? What are you trying to solve for?

Is your dP/dt implying a derivative?
I assume it must because otherwise your left hand side the d/d and the P/P would cancel to give 1/t
But then I'm still confused because you're taking the derivative of P with respect to t... but you don't have any t anywhere.

More info on the problem will help us help you... unless someone comes by and tells me I'm an idiot... which is possible!

[Big Toosie]

The Original Problem is

1/P dP/dt=b+aP <<Edit by capuchin, you didnt mean to put the ^2 here right?

Now multiplying P to both sides gives

dP/dt=bP+aP^2

Now dividing by dP to get dt on one and dP on one side

1/dt=(bP+aP^2)/dp

This where I am suck I have to use Partial fraction to finish the problem.

[Capuchin]

Why do you move the dp over?

Why don't you integrate wrt t?

[asterisk_man]

but that's where I'm confused. Since there isn't any t in the equation bP+aP^2 is a constant so just replace it with C and you've got dP/dt=C and the integral is P=Ct=(bP+aP^2)t
then I guess you can solve for P if you'd like

P/(P(b+aP))=t
1/(b+aP)=t
b+aP=t
aP=t-b
P=(t-b)/a

but I'm just saying this because I'm not sure where else to go with this

[malijfar]

Gerald and Michelle went on a 24-mile bike ride. By lunchtime, they had ridden 5/8 of the total distance. How many miles did they have left to ride after lunch?

[galactus]

This is a logistic equation.



Separate and take partial fractions:





Integrate:



Multiply both sides by b and use law of logs



Now, can you finish?