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Second Order ODE - Initial Value Problem

  1. Dec 16, 2007 #1
    Solve the initial value problem y''+3y'+2y = 3e[tex]^{2t}[/tex]+1 with initial values y(0) = 1, y'(0) = 1.

    I am unsure if I am going about the solution correctly.

    1.) Find the characteristic equation.
    r[tex]^{2}[/tex]+3r+2=0 [tex]\Rightarrow[/tex] (r + 1)(r + 2) = 0
    Therefore, y = c1•e[tex]^{-t}[/tex]+c2•e[tex]^{-2t}[/tex]

    2.) Use method of undetermined coefficients for RHS. Attempt Y = Ae[tex]^{2t}[/tex]+B. This means Y' = 2Ae[tex]^{2t}[/tex] and Y'' = 4Ae[tex]^{2t}[/tex]

    Substituting it all in and simplifying, I get...

    12Ae[tex]^{2t}[/tex]+2B = 3e[tex]^{2t}[/tex]+1

    Comparing coefficients, I get the following...

    12A = 3 => A = 1/4
    2B = 1 => B = 1/2

    So this would yield a particular solution of:
    y = c1•e[tex]^{-2t}[/tex]+c2•e[tex]^{-t}[/tex]+(1/4)e[tex]^{2t}[/tex]+(1/2)

    4.) Next, I take the derivative of the above equation, and then substitute the initial conditions. When I do this, I get a system of equations that yields the following constants. C1 = 3/4, C2 = -1/2.

    5.) Then I just replace C1 and C2 in the equation for the particular solution.

    Could somebody please tell me where I am going astray? Thank you VERY much!!!
    Last edited: Dec 16, 2007
  2. jcsd
  3. Dec 16, 2007 #2


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    Homework Helper

    Well that is basically the correct method...so why do you think you are wrong?
  4. Dec 16, 2007 #3


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    Staff Emeritus
    Science Advisor

    Unfortunately, although you are referring to an "initial value problem", you never bothered to tell us what those initial values are!
  5. Dec 16, 2007 #4
    Ooops...added them now. Sorry! But thanks for looking!
  6. Dec 17, 2007 #5


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    Staff Emeritus
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    Recheck you calculations for [itex]C_1[/itex] and [itex]C_2[/itex]. I get [itex]C_1= -\frac{3}{4}[/itex] which also changes [itex]C_2[/itex].
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