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Homework Help: Second Order Differential Equations

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve the following initial value problem:

    [itex]\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = t[/itex]

    [itex]y(0) = 1, \ y'(0) = 0[/itex]

    The correct answer must be: [itex]y(t)=3e^{-t} + 2 t e^{-t} + t -2[/itex]

    3. The attempt at a solution

    Where did they get the term 2te-t from?

    Here's how I've done it so far:

    We can find the general solution of the corresponding homogeneous system:

    [itex]\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = 0[/itex]

    [itex]\frac{d^2 e^{st}}{dt^2} + 2 \frac{de^{st}}{dt}+e^{st} = (s^2 + 2s +1) e^{st} =0[/itex]

    [itex](s+1)(s+1) \implies s=-1[/itex]

    So we have a repeated eigenvalue. Solution to the homogeneous equation is

    [itex]y_h= C_1 e^{-t}[/itex]

    Now we must find a particular solution, I chose yp=t-2, since we it is a solution of the nonhomogeneous equation. Therefore the solution is

    y = yh + yp = C1 e-t + t - 2

    Now I can use the given initial values to work out C1:

    y' = -C1e-t + 1

    y(0) = C1 - 2 = 1 ⇒ C1 = 3

    y'(0) = -C1+1 = 0

    I get y = 3e-t + t - 2, but there must be another coefficient so we can solve two equations in two unknowns. But how do we get the term 2te-t? Even if I wrote the repeated eigenvalue twice C1e-t + C2e-t, I still don't get the term 2te-t. So what's the problem?

    Any help is greatly appreciated.
  2. jcsd
  3. Jun 2, 2012 #2


    Staff: Mentor

    No, this is only one solution, and you need two. In cases like this where there is a repeated root of the characteristic equation, you get another solution by multiplying by t.

    So two solutions of the homogeneous equation are
    [itex]\{e^{-t}, te^{-t}\}[/itex]

    The solution to the homogeneous equation would be all linear combinations of the above; ##y_h= C_1 e^{-t} + C_2t e^{-t}##
    A better particular solution to the nonhomogeneous problem would be yp = A + Bt
  4. Jun 3, 2012 #3
    Oh I see, that makes perfect sense to me know. Thank you so much for your help, I really appreciate it.
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