# How do you find the differential equation in terms of P when 1/P*dP/dt = b+aP?

P(0) = 3.9

a = -0.000091695

b = 0.028659

C = 76.8997

Update:

The equation must be in the form P(t)=......

Relevance
• 10 years ago

1/P dP/dt = b+aP

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

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

Integrate both sides

∫ 1/(P(b+aP)) dP = ∫ dt

∫ ((1/b)/P - (a/b)/(b+aP)) dP = ∫ dt

(1/b) ln(P) - (a/b) (1/a) ln(b+aP) = t + C₀

(1/b) ln(P) - (1/b) ln(b+aP) = t + C₀

ln(P) - ln(b+aP) = bt + bC₀

ln(P/(b+aP)) = bt + C₁ . . . . . . . . . . where C₁ = bC₀

P / (b+aP) = e^(bt + C₁)

P / (b+aP) = C₂e^(bt) . . . . . . . . . . . where C₂ = e^C₁

(b+aP)/P = Ce^(-bt) . . . . . . . . . . . . where C = 1/C₂

b/P + a = Ce^(-bt)

b/P = Ce^(-bt) - a

P = b / (Ce^(-bt) - a)

P(0) = 3.9

b / (C - a) = 3.9

C - a = b/3.9

C = b/3.9 + a

C = 0.028659/3.9 - 0.000091695

C = 0.0072568

____________________ ____________________

Check:

P = b / (Ce^(-bt) - a)

dP/dt = Cb²e^(-bt) / (Ce^(-bt) - a)²

Differential equation:

1/P*dP/dt = b+aP

dP/dt = P(b+aP)

Cb²e^(-bt) / (Ce^(-bt) - a)²

= P(b+aP)

= b/(Ce^(-bt)-a) (b + ab/(Ce^(-bt)-a)

= b/(Ce^(-bt)-a) * b(1 + a/(Ce^(-bt)-a)

= b²/(Ce^(-bt)-a) ((Ce^(-bt)-a + a)/(Ce^(-bt)-a))

= b²/(Ce^(-bt)-a) (Ce^(-bt))/(Ce^(-bt)-a)

= Cb²e^(-bt) / (Ce^(-bt)-a)²

ok

• 10 years ago

1/P*dP/dt = b+aP

separating variables

dP/P(b+aP) = dt

split LHS into partial fractions

[(1/b)(1/P) - (a/b)(1/(b + aP) ]dP = dt

integrating

(1/b) ln I P I - (1/b)ln I b + aP I = t + c

(1/b) ln [P / (b + aP ] = t + c

P / (b + aP) = Ce^(bt)

since P(0) = 3.9

3.9 /(b + 3.9a) = C

P /(b + aP) = 3.9 e^(bt) /(b + 3.9a)