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Quotient of First Order Ordinary Derivatives

  1. Mar 15, 2012 #1
    How do you solve (analytically or numerically) a differential equation of this form,

    $$\frac{\mathrm{d}y(x)/\mathrm{d}x}{\mathrm{d}z(x)/\mathrm{d}x} = a[1-y(x)-z(x)] + b$$

    where a, b are constants. Also,

    $$y(0) = z(0) = 0$$
  2. jcsd
  3. Mar 16, 2012 #2
    Since you have only one equation, but two unknown functions y(x) and z(x) the problem remains undefined. You cannot obtain one function y(x) and one function z(x), but an infinity of couples.
    Nevertheless, tne differential equation allows to express a relationship beteween each couple y(x) and z(x). The relationship is on the form y=function of z.
    Now, how to obtain y = y(z) ?

    dy/dz = a(-y-z)+b
    (dy/dz)+a*y = b -a*z
    (dY/dz)+a*Y = 0 --> Y = c*exp(-a*z)
    c --> f(z)
    y = f(z)*exp(-a*z)
    dy/dz = (df/dz)exp(-a*z)-a*f*exp(-a*z)
    (df/dz)exp(-a*z)-a*f*exp(-a*z) +a*f*exp(-a*z) = b -a*z
    (df/dz)exp(-a*z) = b -a*z
    df = (b-a*z)exp(a*z)*dz
    f = (1/a)*(-a*z+b+1)*exp(a*z)+C
    y = (1/a)*(-a*z+b+1) +C*exp(-a*z)
    y(0)=z(0)=0 --> 0 = (1/a)*(b+1) +C --> C = -(b+1)/a
    y = (1/a)*(-a*z+b+1) -((b+1)/a)*exp(-a*z)

    if you want, you can chose any derivable function z(x). Then, the related function y(x) is :
    y(x) = (1/a)*(-a*z(x)+b+1) -((b+1)/a)*exp(-a*z(x))
    This proves that, due to the incomplete wording, there are an infinity of solutions.
    Last edited: Mar 16, 2012
  4. Mar 16, 2012 #3
    Or did you originally have two equations, one for dy/dx and one for dz/dx and you performed the 'trick' where you divide one by the other to end up with one differential equation?

    You would then probably get an ode like dy/dz = a(1-y-z)+b which is linear.
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