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Math Help - How do I "guess" the solution to y to this D.E.

  1. #1
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    How do I "guess" the solution to y to this D.E.

    y"-4y'+3y = 8xex

    I guessed that the solution to y was Axex but when I plugged it into y it did not get me the correct answer. Please don't work out the problem for me, right now I just need help understanding how I would guess y in this situation. Much appreciated.
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  2. #2
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    Re: How do I "guess" the solution to y to this D.E.

    Quote Originally Posted by fb280 View Post
    y"-4y'+3y = 8xex

    I guessed that the solution to y was Axex but when I plugged it into y it did not get me the correct answer. Please don't work out the problem for me, right now I just need help understanding how I would guess y in this situation. Much appreciated.
    Since the complimentary solution is y_c=c_1e^{x}+c_2e^{3x}

    You would normally gues the solution is

    y=Ae^{x}+Bxe^{x}

    But since the first term is part of the complimentary solution we need to multiply the particular soluiton by x to get the guess

    y_p=Axe^{x}+Bx^2e^{x}
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  3. #3
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    Re: How do I "guess" the solution to y to this D.E.

    why would the guess normally be Aex for one part instead of Axex? And why is there a B? If there was a B I would have thought that it would be Bex and not Bxex
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  4. #4
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    Re: How do I "guess" the solution to y to this D.E.

    On the right side of the ODE you have the product of a linear function and an exponential function, so you would expect the particular solution would take the form:

    y_p=(Ax+B)e^x

    Note: while I have switched A and B from the form The EmptySet used, they are equivalent.

    As stated, you don't want any of the terms of your particular solution to be part of the complementary solution, so you multiply the particular solution by x^s where s is the smallest integer so that no terms of the particular solution are part of the complementary solution. In this case, s=1. Hence:

    y_p=x(Ax+B)e^x=Ax^2e^x+Bxe^x
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    Re: How do I "guess" the solution to y to this D.E.

    Okay I think I understand what I was confused on before. But when I plug in Ax2ex+Bxex I got stuck at the part where it turns out to be

    -2A2xex+A2ex+Bx4ex-4Bx2ex+3Bxex=8xex

    Ax2ex-4Ax2ex+3Ax2ex cancels out so I did not include this part above
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: How do I "guess" the solution to y to this D.E.

    Using:

    y_p=Ax^2e^x+Bxe^x

    we find:

    y_p'=e^x(Ax^2+2Ax+Bx+B)

    y_p''=e^x(Ax^2+4Ax+Bx+2A+2B)

    and substituting into the ODE, we get:

    e^x(Ax^2+4Ax+Bx+2A+2B)-4e^x(Ax^2+2Ax+Bx+B)+3e^x(Ax^2+Bx)=8xe^x

    Divide though by e^x, distribute constants and collect like terms:

    Ax^2+4Ax+Bx+2A+2B-4(Ax^2+2Ax+Bx+B)+3(Ax^2+Bx)=8x

    Ax^2+4Ax+Bx+2A+2B-4Ax^2-8Ax-4Bx-4B+3Ax^2+3Bx=8x

    (-4A)x+(2A-2B)=8x+0

    Equating coefficients yields the system:

    -4A=8

    2A-2B=0

    Now, solve this system to determine A and B, and then you have your particular solution.
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  7. #7
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    Re: How do I "guess" the solution to y to this D.E.

    okay I understand where I went wrong now. thanks for replying, this really helped me a lot.
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