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    I must find the general solution to the second order differential equation:
    y"-y'-2y=3x+4

    I know the solution must be:
    y(x) = C1*e^(2x)+C2*e^(-x)-3x/2-5/4

    I'm having trouble finding the particular solution.
    I tried this problem as follows:

    For the complimentary solution:
    r^2-r-2=0
    (r-2)(r+1)=0
    r=2,-1
    yc(x)=C1*e^(2x)+C2*e^(-x)

    For the particular test solution:
    yt(x) = Ax^2+Bx+C
    yt'(x) = 2Ax+B
    yt"(x) = 2A

    yt(x) = (Ax^2+Bx+C)-(2Ax+B)-2(2A)
    yt(x) = Ax^2+Bx+C-2Ax-B-4A
    ^^^ Would I use parenthesis for each term and therefore distribute the negative signs, or should I just plug them in?
    Equations:
    x^2: A = 0
    x: B-2A = 1
    constants: C-B-4A = 1
    ^^^ Equals number of like terms in original DE on RHS?
    So, A=0, B=1, C=2

    yp(x) = x+2

    y(x) = yc(x) + yp(x)
    y(x) = C1*e^(2x)+C2*e^(-x)+x+2

    My complimentary solution is correct, but where am I going wrong with my particular solution?
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    2A - (2Ax+B) - 2(Ax^2+Bx+C) = 3x+4 ...you mean
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    Well, yes, but don't you set up the system of equations based on the number of like terms on the RHS?
    • Thread Starter
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    Oh, I see what happened in the class example, I missed the simplifications. I'll see if I can try this again.
    • Thread Starter
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    Setting up my equations as follows:
    x^2: -A = 0
    x: -2A-B = 3
    const: 2A-B-C = 4
    A=0, B=-3, C=-1

    Which still isn't right...
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    -2a = 0
    -2a-2b = 3
    2a-b-2c = 4
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    Just so you know in future, because the RHS is linear, there is no need to try a quadratic particular integral, you just need to try a linear one.
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    Oh, duh, that time I forgot to distribute the 2 >.<
 
 
 
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