How to solve differential equation dy/dx = k(c - y)?

Just wondering how you would solve this kind of a differential equation.

thanks in advanced <3

3 Answers

  • ?
    Lv 7
    8 years ago
    Favorite Answer

    Assuming that k and c are just constants, the equation is separable, divide both sides by (c - y) and multiply the differential across:

    1/(c - y) * dy = k * dx

    Integrate both sides:

    -ln(c - y) = kx + C

    Make y the subject, divide both sides by -1 and raise both sides to the base of e:

    c - y = e^(-kx) * e^-C

    As e^-C represents a constant, we can just replace it with another constant placeholder:

    c - y = C₁e^(-kx)

    y = c - C₁e^(-kx)

  • 8 years ago

    dy/dx = k(c - y)

    dy/(c-y) = k dx for y≠c (because this would cause division by zero)

    ∫ dy/(c-y) = ∫k dx

    -ln|c-y| = kx + c_1 where c_1 is a real constant

    ln|c-y| = -kx + c_2 where c_2 = -c_1

    |c-y| = e^(-kx + c_2) = c_3*e^(-kx) where c_3 = e^c_2 (c_3 is a positive number)

    c-y = ±c_3*e^(-kx) = c_4 e^-(kx) where c_4 = ± c_3 (c_4 can't be zero)

    y = -c_4 e^(-kx)+c

    If we let c_4 = 0 we also get the valid solution y = c (you can check this by plugging it into the differential equation.)

    final solution:

    y = m e^(-kx) + c where m is a real constant

  • 8 years ago

    dy/dx = k(c-y)

    dy/(c-y) = kdx

    Integrate both sides.

    -ln(c-y) = kx + C

    ln(1/(c-y)) = kx + C

    You can then re-arrange as you wish.


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