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Math Help - Stuck on a diff EQ: Undetermined Coefficient

  1. #1
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    Stuck on a diff EQ: Undetermined Coefficient

    Find this general solution:

    y" + 3y' + 2y = 36xe^x

    Here's my work:

    1. r^2 + 3r + 2 = 0; r=-1, -2
    2. yc = c1e^-x + c2^e^-2x

    3. yp=Axe^x
    y'p=Axe^x + Ae^x
    y"p=Axe^x + 2Ae^x

    4. (Axe^x + 2Ae^x) + 3(Axe^x + Ae^x) +2(Axe^x) = 36xe^x

    5. 6Axe^x + 5Ae^x = 36xe^x

    And here's where I get fuzzy. I need to find this value of A and then put in the form y = yc + yp. The answer in the back of the book is:

    y= c1e^-x + c2^e^-2x + (6x-5)e^x

    Any suggestions? Thanks, JN
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Jim Newt View Post
    Find this general solution:

    y" + 3y' + 2y = 36xe^x

    Here's my work:

    1. r^2 + 3r + 2 = 0; r=-1, -2
    2. yc = c1e^-x + c2^e^-2x

    3. yp=Axe^x
    y'p=Axe^x + Ae^x
    y"p=Axe^x + 2Ae^x

    4. (Axe^x + 2Ae^x) + 3(Axe^x + Ae^x) +2(Axe^x) = 36xe^x

    5. 6Axe^x + 5Ae^x = 36xe^x

    And here's where I get fuzzy. I need to find this value of A and then put in the form y = yc + yp. The answer in the back of the book is:

    y= c1e^-x + c2^e^-2x + (6x-5)e^x

    Any suggestions? Thanks, JN
    Since you have a polynomial multiplied by an exponential function your particular solution should look like

    y_p=(Ax+B)e^x

    Take a few derivatives and plug into the original equation to to solve for A and B.

    Good luck
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    Since you have a polynomial multiplied by an exponential function your particular solution should look like

    y_p=(Ax+B)e^x

    Take a few derivatives and plug into the original equation to to solve for A and B.

    Good luck
    Thanks for the input EmptySet. So from this point:

    1. yp = (Ax + B)e^x
    y'p= Ae^x + (Ax + B )e^x
    y"p=2Ae^x + Axe^x + Be^x

    2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x

    So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x

    how do I make the jump to yp = (6x-5)e^x?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Jim Newt View Post
    Thanks for the input EmptySet. So from this point:

    1. yp = (Ax + B)e^x
    y'p= Ae^x + (Ax + B )e^x
    y"p=2Ae^x + Axe^x + Be^x

    2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x

    So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x

    how do I make the jump to yp = (6x-5)e^x?
    5Ae^x + 6Axe^x + 6Be^x = 36xe^x<br />

    Rewritting as (adding a 0e^{x}) may help you see it

    So then
    (5A + 6B)e^x + 6Axe^x = 36xe^x + 0e^{x}<br />

    We get a "system" of equations for A and B

    5A+6B=0
    6A=36

    So we can see A=6 and B=-5.

    I hope this helps.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jim Newt View Post
    Thanks for the input EmptySet. So from this point:

    1. yp = (Ax + B)e^x
    y'p= Ae^x + (Ax + B )e^x
    y"p=2Ae^x + Axe^x + Be^x

    2. yp= 2Ae^x + Axe^x + Be^x +3Ae^x + 3xAe^x + 3Be^x + 2Axe^x + 2Be^x 2Be^x

    So then 5Ae^x + 6Axe^x + 6Be^x = 36xe^x

    how do I make the jump to yp = (6x-5)e^x?
    Compare the coefficients:

    \begin{aligned}<br />
5A+6B&=0 \\<br />
6A&=36<br />
\end{aligned}<br />

    This implies that A=6 and B=-5

    Therefore, y_p=(6x-5)e^x.

    Hope this clarified things
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