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)=......

2 Answers

Relevance
  • 10 years ago
    Favorite Answer

    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)

Still have questions? Get your answers by asking now.