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First order differentials

  1. Nov 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve: (2t+x) dx/dt + t = 0

    2. Relevant equations
    y' +p(X)y = q(x)
    and y(x) = ([tex]\int[/tex]u(x)q(x) + c)/u(x)
    where u(x) = e[tex]\int[/tex]p(x)dx
    Note this u(x) is 2 to the power of the integral of p(x)

    3. The attempt at a solution
    (2t+x) dx/dt + t = 0 becomes:
    dx/dt + t/(2t+x) = 0 by dividing through by (2t+x)

    However, now I don't know how to separate t and x in the second term.
  2. jcsd
  3. Nov 25, 2008 #2
    It looks like it is first-order homogeneous. Make the substitution z=x/t by dividing by t throughout your t/(2t+x), and then x=tz and x'=z+tz' by the product rule. Now it's in a form you want it for the relevant equations you listed. Just make the substitution the other way when you've solved for z.
  4. Nov 25, 2008 #3


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    These apply only to a linear first order d.e. And this d.e. is NOT linear because of the x multiplying dx/dt.

  5. Nov 25, 2008 #4
    Okay so does that mean that what jeffreydk would not work? If so what can i use instead? I have only been taught about linear first order d.e so really don't know what to do!
  6. Nov 25, 2008 #5
    Yea it is non-linear I didn't notice that. What I said wont work.
  7. Nov 25, 2008 #6


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    No, being "homogeneous" has nothing to do with being linear. I would consider it very strange that you have "only been taught about linear first order d.e.". Usually separable d.e. comes before that.

    In any case, as jeffreydk said (and I won't allow him to take it back!) making the substitution z= x/t so that tz= x and z+ t dz/dt= dx/dt. Dividing the equation by t gives (2+ x/t)dx/dt+ 1= 0.

    (2+ z)(z+ t dz/dt)+ 1= 0 or t(2+z) dz/dt= 1-(2+ z2)= z2- 1 which is a separable equation.
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