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Variation of parameters (1st order)

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of the following equation:

    u(t): u' = u/t + 2t

    2. Relevant equations
    y' + p(x)y = Q(x).............(1)

    yeI = ∫ dx eIQ(x) + constant..............(2)

    3. The attempt at a solution
    I rearranged the equation to give:

    u' - u/t = 2t

    Then I considered the following (homogeneous):

    u' - u/t = 0

    1/u u' = 1/t

    ∫ 1/u du = ∫ 1/t dt

    ln(u) = ln(t) + c

    u(t) = eln(t) + c = tec = At

    Let: I = ln(t)

    eI = eln(t) = t

    u = At

    So: A(t)=u/t

    dA/dt = u//t - u/t2 = A'

    Using equation (2) given above, my solution to this equation was:

    ut = ∫ 2t . t dt = ∫ 2t2 dt = 2t3/3 + constant

    My questions are:
    1. Is this correct?
    2. In equation (2) there is no need for A', so why was this needed?

    Thank you for taking the time to look at these, any help would be much appreciated!
     
  2. jcsd
  3. Oct 15, 2015 #2
    You can always check to see whether your final solution is a solution of the original differential equation yourself by replacing u with the function of t. In this case, we can see that it does not solve the original equation, assuming I read it properly.
    I'm not sure as to what the symbol "l" in your method entails, as I may have learned a different method of variation of parameters. In the method that I learned, the homogeneous solution [itex]u_h(t) = At[/itex] is used to find a particular solution of the form [itex]u_p(t) = p(t)t[/itex], where p(t) is an unknown function of t that we must solve for (the constant "parameter" in the homogeneous solution is replaced by the "variable parameter" p(t)). To find it, we substitute this particular solution into the original differential equation and equate coefficients of like terms. In this case, we get
    [tex]p'(t)t + p(t) = p(t) + 2t[/tex]
    which simplifies to [itex]p'(t) = 2[/itex]. We may then integrate p'(t) to get p(t), and write the general solution as the sum of the homogeneous and particular solutions. You can then verify that the general solution does solve the original equation.

    As an aside, we may also solve this differential equation using a standard integrating factor for first order linear differential equations. If we multiply both sides of the equation
    [tex]u' - \frac{1}{t}u = 2t[/tex]
    by the integrating factor derived from the coefficient of u,
    [tex]e^{\int -\frac{1}{t}\, dt} = \frac{1}{t}[/tex]
    we get
    [tex]\frac{u'}{t} - \frac{u}{t^2} = 2[/tex]
    The left side is the derivative with respect to t of [itex]\frac{u}{t}[/itex], as you should verify. Integrating both sides, we get the same family of solutions.
     
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