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:

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!

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.

Why do you move the dp over?

Why don't you integrate wrt t?

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

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?

This is a logistic equation.

\frac{dP}{dt}=P(b+aP)

Separate and take partial fractions:

\frac{dP}{P(a+bP)}=dt

\left(\frac{\frac{1}{b}}{P}-\frac{\frac{a}{b}}{b+aP}\right)dP=dt

Integrate:

\frac{1}{b}ln|P|-\frac{1}{b}ln|b+aP|=t+c

Multiply both sides by b and use law of logs

ln|\frac{P}{b+aP}|=bt+bc

Now, can you finish?