Differential equation using partial fractions

[Shelby]

Homework Statement

I need to integrate this differential equation using partial fractions to obtain an equation for P in terms of t; P(t):

1/P dP/dt = b + aP

Homework Equations

The Attempt at a Solution

So far, this is what I have:
ln /P/ = bP + aP^2/2 +c

Where do I go from here, would I take e from each side and then what would I do? and am I doing this right?
Please help me

 

[christianjb]

Multiply both sides by P?

 

[Curious3141]

Multiply both sides by P?

No sense in doing that, since he still needs to bring the P-terms over to the left hand side.

Bring the P-terms from the RHS to the LHS, then separate variables :

\frac{1}{P(aP+b)}dP = dt

Separate the left hand side using partial fractions (write it as \frac{k_1}{P} + \frac{k_2}{aP+b}, then find the values of k1 and k2 quickly with the "Heaviside cover-up" shortcut, for example) and integrate both sides to solve.

 

[HallsofIvy]

Homework Statement

I need to integrate this differential equation using partial fractions to obtain an equation for P in terms of t; P(t):

1/P dP/dt = b + aP

Homework Equations

The Attempt at a Solution

So far, this is what I have:
ln /P/ = bP + aP^2/2 +c

No. You can't just integrate both sides with respect to P: there is not "dP" on the right side. You can rewrite the equation as
\frac{dP}{P(b+aP)}= dt
and integrate- the left side with respect to P, the right side with respect to t.

As Curious3141 said, you "partial fractions" to integrate the right side.

Where do I go from here, would I take e from each side and then what would I do? and am I doing this right?
Please help me

 

[Shelby]

partial fractions

can you point me in the right direction, with all of these variables I keep getting confused?

 

[HallsofIvy]

What more "pointing" do you want! Use partial fractions to integrate
\frac{dP}{P(b+ aP}

Can you do the "partial fractions" decomposition? You want to find A and B so that
\frac{1}{P(b+aP)}= \frac{A}{P}+ \frac{B}{b+ aP}
Multiply both sides of that by P(b+aP) to get
1= A(b+aP)+ BP
and solve for A and B. (Hint: let P= 0 and P= -b/a)