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General Solution to the Diff. Eq. dy/dt = ty

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of the differential equation specified.

    [tex]\frac{dy}{dt}[/tex] = ty.

    3. The attempt at a solution

    I already know the answer to be ke[tex]^{t^{2}/2}[/tex], but can't figure out how it got here. I'm rusty with my integrals and am just really starting diff eqs, but I think I'm just supposed to take the integral of ty (which is (t^2y)/2 i believe) and add a + c to the end, c being some constant, so I'm confused...
     
  2. jcsd
  3. Jan 11, 2009 #2

    jgens

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    I think this might be a logical first step: dy/y = t dt. Now integrate that.
     
  4. Jan 11, 2009 #3
    dy/y = t dt

    The integral of that would be...

    dy = t^2/2

    So

    (t^2/2)/dt = ty

    t^2/2 = ty dt

    Where do I go from here?
     
  5. Jan 11, 2009 #4

    jgens

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    No. You might want to brush up on your integrals. The integral of dy/y is not dy but rather log(y). Therefore, your equation should be?
     
  6. Jan 11, 2009 #5
    log(y) = t^2/2

    So y = e^(t^2/2)

    Alright. And the k is because it's a general equation and k needs to be there so it can be solved for any initial condition. But why is the k being multiplied and not added?
     
  7. Jan 11, 2009 #6

    jgens

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    Well, the equation should read: log(y) = t^2/2 + C; therefore, y = e^(t^2/2 + C) = e^(t^2/2)e^C. We let e^C = k and the equation reads: y = ke^(t^2/2)
     
  8. Jan 12, 2009 #7
    Just to be sure, is the basic starting step for all problems of this type getting dy/y?

    Like if I have the same question with dy/dt = t/(t^2y + y), I would try to get dy/y on one side?
     
  9. Jan 12, 2009 #8

    Mark44

    Staff: Mentor

    The basic goal is to get y and dy on one side, and t and dt on the other, which is called separation. Your example differential equation is not separable, if I understand what you wrote. Is the first term in the denominator t^(2y) or is it t^2 * y? If it's the first, your example isn't separable.
     
  10. Jan 12, 2009 #9
    Oops, my bad. It's the 2nd one.

    dy/dt = t/((t^2)y + y)
     
  11. Jan 12, 2009 #10
    Lesse

    dy/dt = t/((t^2)y + y)

    dy/y = t/((t^2)y^2 + y^2) dt

    dy/y * y^2 + y^2 = t/(t^2) dt

    dy/y * 2(y^2) = 1/t

    ln(y) * 2(y^3/3) = 1/t

    Ugh...I dunno, something in that ballpark? I'm confused...
     
  12. Jan 12, 2009 #11

    jgens

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    I don't believe that's quite right. Your initial equation is: dy/dt = t/(y(t^2 + 1) correct? Then we claim that y dy = (t dt)/(t^2 + 1) which we then integrate.
     
  13. Jan 12, 2009 #12
    These are called SEPARABLE DEs. Separable because we can separate them into f(y) dy = g(t) dt. And then integrate both sides. This is also why DE requires you to do well in Calculus classes so you don't have to think about how to integrate :) This is also why separable DEs are introduced in Calc 2(I guess), to give you a flavor of what to come.
     
  14. Jan 12, 2009 #13
    y dy = (t dt)/(t^2 + 1)

    y^2/2 = (t dt/t^2) + t dt (If I understand correctly, this is separating?)

    y^2/2 = ((t^2)/2)/t^2 + t^2/2

    y^2/2 = 1/2 + t^2/2

    y^2 = 1/4 + t^2/4

    y = sqrt(k(.25 + t^2/4))

    Eh?
     
  15. Jan 12, 2009 #14
    No the separating was the process of writing dy/dt = g(t)/f(y) as f(y) dy = g(t) dt.

    a/(b+c) is definitely not a/b + a/c, you should remember that from algebra.

    You should know how to integrate t dt/(t^2 + 1) from calculus.
     
  16. Jan 12, 2009 #15

    jgens

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    In the case that you're uncertain how to integrate t dt/(t^2 + 1) use the substitution u = t^2 + 1. However, you really should learn how to do these steps and properly apply elementary algebra - especially is you're working with DEs.
     
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