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Undetermined coefficients + initial value problem

  1. Feb 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Find solution for y'' + y' -2y = 2x with initial values of y(0) = 0, y'(0) = 1

    2. Relevant equations
    I have found yc = c1*exp(-2x) + c2*exp(x), but finding yp is what I'm having trouble with... AND THEN I'm not so sure how to go about the initial value.


    3. The attempt at a solution
    Ok, so my "reasonable assumption" :
    yp = Ax
    y'p = A
    y''p= 0

    When I plug into equation, I get:
    0 + A -2(Ax) = 2x... and that's where I'm stuck.

    Any insights?
     
  2. jcsd
  3. Feb 7, 2008 #2

    EnumaElish

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    Is yc the general solution and yp the particular?

    If so, shouldn't you be using the initial conditions to solve for c1 and c2?
     
    Last edited: Feb 7, 2008
  4. Feb 7, 2008 #3
    Edit: general solution y = yc + yp.

    Does that help?

    My strategy is to find the yc part of the general y solution, pretending like the left side of the original eqn is equal to zero. Then to find yp I'm supposed to make a reasonable assumption to find it. All parts of yc and yp must be linearly independent, so yp must be in a similar form of the right side of the original equation. So, I take yp, take 1st and 2nd derivatives and see if whatever I get agrees with 2x. Then when i find what yp is, I add it to yc.

    ...and that's all I know about this equation.
     
  5. Feb 7, 2008 #4

    Dick

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    Your "reasonable assumption" is reasonable, but not general enough. Try yp=Ax+B.
     
  6. Feb 7, 2008 #5

    HallsofIvy

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    In general, for a RHS of xn, you will need yp= Axn+ Bxn-1+ Cxn-2 . . .
     
  7. Feb 7, 2008 #6
    Ahhh... Clever.
     
  8. Feb 7, 2008 #7
    It worked! Thanks much!!

    Now with the initial values:
    I have my general solution y = c1*exp(-2x) + c2*exp(x) -x -1/2

    All I'll do to find c1 and c2 is use y(0)=0 in that equation, then take its derivative and use the y'(0)=1, that's the way to go, correct?
     
    Last edited: Feb 7, 2008
  9. Feb 7, 2008 #8
    Gotcha.
     
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