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Differential equation problem

  1. Jul 28, 2005 #1
    there's an example in my textbook,
    solve the differential equation
    dy/dx + 1/2y =1/2e^(t/3)

    this problem is weird
    because i can't separate the variables on both sides and integrate them...
  2. jcsd
  3. Jul 28, 2005 #2
    Did you have any careful check again your book or possibly think of something that might be incorrect on the right hand side of your equation anyway ?
    If :x: is in place of :t:, it is easily sovable and otherwise, it is going to be different since you have to explain how things are to logically work out in the presence of t.
  4. Jul 28, 2005 #3


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    There are other ways to get the solution. The nice thing about exponentials is that its derivative is also an exponential. Try a function of the form: y(t)=Aexp(t/3) to find a particular solution.
  5. Jul 28, 2005 #4
    Galileo, are you really well today ????????
  6. Jul 28, 2005 #5


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    dy/dx + 1/2y =1/2e^(t/3)

    Are you sure this is the problem??

    Usually for ODE the equation takes this form:

    So you right hand side should be a function of x, not t. Make sure you re-read your problem because otherwise this is not an ODE
  7. Jul 28, 2005 #6


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    I'm fine thanks.

    I assumed the dy/dx is a typo. The equation might have been y'+1/2y=exp(t/3) and by force of habit asdf1 typed dy/dx instead of dy/dt.

    Anyway, assuming a solution of the form Aexp(t/3) will get you a particular solution. Add to that a solution of y'+1/2y=0 for the general one.
  8. Jul 28, 2005 #7
    sorry my mistake~
    dy/dt +1/2y=1/2e^(t/3)
  9. Jul 29, 2005 #8


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    Okay, assuming you mean dy/dt+ (1/2)y= (1/2)e(t/3), that's a linear equation. Not every equation is "separable". If you are doing first order linear equations you should already have learned about exact equations and "integrating factors".

    In this case, the equation is not exact- that is, there is no function f(y) so that d(f(y))/dt= dy/dt+ (1/2)y, but we can find a function u(t) so that, after we multiply the entire equation by u, it is exact. We want to find u(t) so that
    d(u(t)y)/dt= u(t)(dy/dt)+ (1/2)u(t)y. Applying the product rule to the left hand side of that, d(u(t)y)/dt= u(t)(dy/dt)+ (du/dt)y= u(t)(dy/dt)+ (1/2)u(t)y.
    That means we must have du/dt= (1/2)u which is separable: In fact it is easy to see that u(t)= e(1/2)t.

    That tells us that if we multiply the whole equation by e(1/2)t, we get
    e(1/2)t(dy/dt)+ (1/2)e(1/2)ty= e(1/2+ (1/3))t
    The left hand side is just d(e(1/2)ty)/dt and the right hand side is e(5/6)t. In other words, d(e(1/2)ty)/dt= e(5/6)t.

    Integrating both sides, e(1/2)ty= (6/5)e(5/6)t+ C so

    y(t)= (6/5)e(1/3)t+ Ce(-1/2)t.

    Notice the e(1/3)t in there, exactly as asdf1 suggested.
  10. Jul 29, 2005 #9
    you forgot [tex]\frac{1}{2}[/tex] in front of [tex]e^\frac{t}{3}[/tex]. The answer should be [tex]y=\frac{3}{5}e^\frac{t}{3}+Ce^-\frac{t}{2}[/tex]
  11. Jul 29, 2005 #10


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    Your're right, I did. Thanks.
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