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elica85

  • 4 years ago

diff eq..determine the form of a particular solution for the given nonhomogeneous equation y''+2y'=3-4t

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  1. amistre64
    • 4 years ago
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    yp = some educated guess yp=At + B ; if these are not already a linear solution to the homogenous part ...

  2. elica85
    • 4 years ago
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    for fundamental solution set, i have y1=1, y2=e^(-2t) f(t)=3-4t polynomial method so yp=-At+B...check to see if any matches the fund.set. it does so... yp=-At^2+Bt find y'p and y''p and plug in. now solving for A and B is where i'm stuck

  3. amistre64
    • 4 years ago
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    then lets work thru it so I can catch up :)

  4. elica85
    • 4 years ago
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    ok thx

  5. amistre64
    • 4 years ago
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    r^2+2r=0 r=0,-2 right?

  6. elica85
    • 4 years ago
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    at this point, it's just algebra which apparently, i'm very weak at

  7. elica85
    • 4 years ago
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    right

  8. amistre64
    • 4 years ago
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    \[y=c_1e^{0x}+c_2e^{-2x}+y_p\] \[y=c_1+c_2e^{-2x}+y_p\]

  9. amistre64
    • 4 years ago
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    yp = At + B ; but we already have a constant so we need to up the ante yp = At^2 + Bt ; and derive

  10. amistre64
    • 4 years ago
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    yp = At^2 + Bt yp'= 2At + B yp''= 2A agreed?

  11. amistre64
    • 4 years ago
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    yp''+2yp'=3-4t (2A) + 2(2At + B)=3-4t im going to do this in lowercase, just easier to type for me 2a +4at +2b = 3 -4t (4a)t = -4t ; a=-1 (2a+2b) = 3 -2+2b = 3 ; b=5/2

  12. elica85
    • 4 years ago
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    ah ok, i almost did that to solve for a but wasn't sure if it was correct, thank you

  13. amistre64
    • 4 years ago
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    at^2 + bt = -t^2+ 5/2 t\[\] \[y=c_1+c_2e^{-2x}-t^2+\frac 52 t \]

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