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Second Order Differential Initial Value Problem

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data

    y''+4y'+6y
    y(0) = 2; y'(0) = 4

    2. Relevant equations

    [itex]\alpha ± β = e^{x\alpha}(cosβx + sinβx)[/itex]



    3. The attempt at a solution

    Auxilary equation is [itex]r^2+4r+6[/itex], which solves for [itex]-2 ± i[/itex]

    I get the general solution:

    [itex]e^{-2x}(c[/itex]1[itex]cosx + c[/itex]2[itex]sinx)[/itex]

    [itex]y' = -2e^{-2x}(c[/itex]1[itex]cosx + c[/itex]2[itex]sinx) + e^{2x}(c[/itex]2[itex]cosx - c[/itex]1[itex]sinx)[/itex]

    [itex]= c[/itex]1[itex](cosx-sinx) + c[/itex]2[itex](cosx+sinx) = 4[/itex]

    I also have:

    c1 + c2 = 2 from the initial value.

    I now have a system of equations, but don't really know how to solve it without using a computer.
     
  2. jcsd
  3. Oct 11, 2012 #2

    Zondrina

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    Homework Helper

    No no, the system of equations produces infinitely many solutions, but the initial value given gives a specific solution curve when you solve for the constants.
     
  4. Oct 11, 2012 #3
    That's what I was thinking, but I just can't think of how to solve for the constants without using a system of equations at the moment. I know how to solve it when [itex]\alpha = 0[/itex], but not sure how to do this.
     
  5. Oct 11, 2012 #4

    Zondrina

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    Homework Helper

    You have two equations. You get one from solving the initial value problem for y and one for solving the initial value problem for y'.

    They will give you a system of 2 equations in 2 unknowns c1, c2.

    Solving for both unknowns will give you the particular equation for y which will satisfy your initial values.
     
  6. Oct 11, 2012 #5
    Oh, I completely overlooked inputting the x values into the equations for some reason. Thanks though!
     
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