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Differential Equation Initial Value Problem

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data

    I managed to work this problem all the way through, but I am in no way certain of my answer. I'd greatly appreciate any insight!

    Find the solution of the initial value problem.

    y'''+4y'=x, y(0)=y'(0)=0, y''(0)=1

    2. Relevant equations

    Just for clarification purposes, my professor uses yc for the general homogeneous solution; I've seen yh used and didn't want to cause confusion.

    fn(x)y(n)+fn-1(x)y(n-1)+...+f1(x)y'+f0(x)y=0

    General solution: yc=c1y1(x)+c2y2(x)+...+cnyn(x)

    3. The attempt at a solution

    Note: m is used as the variable for characteristic equations rather than r, which I've also seen.

    First, to find yc, I set the left side of the equation equal to zero and found the characteristic equation m3+4m=0, from which I found roots 0, 2i, -2i. I converted them into real solutions and ended up with yc=c1+c2cos(2x)+c3sin(2x).

    To find c1 and c2, I took the first and second derivatives of yc and plugged in the initial values as given. I ended up with yc=(1/4)-(1/4)cos(2x).

    Next, to find yp, I guessed yp=Ax3+Bx2+Cx+D and plugged its first and third derivatives into the original differential equation. I ended up finding A=0, B=1/8, C=-1/2. That gave me yp=(1/8)x2-(1/2)x.

    I got the general solution to the heterogeneous differential equation y=yc+yp:
    y=(1/4)-(1/4)cos(2x)+(1/8)x2-(1/2)x.
     
  2. jcsd
  3. Apr 14, 2014 #2

    LCKurtz

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    I didn't check your work, but I can tell you that you must first write the general solution ##y = y_c + y_p## before you evaluate the constants with the initial conditions.
     
  4. Apr 14, 2014 #3
    I thought that might be the case, but I tried it and came up with a bunch of long, crazy rational expressions with several radicals. I'll try again.

    If anyone reading knows how to do it correctly, I'd be eternally grateful for your help!
     
  5. Apr 14, 2014 #4

    LCKurtz

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    Your ##y_p## isn't correct. You should get ##y_p = \frac{x^2} 8##. You can check that it works. And you won't get radicals anywhere.
     
  6. Apr 14, 2014 #5
    I'll start working on getting it right now; thank you so much!
     
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