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anonymous

  • 5 years ago

y''+5y'+6y=2e^t

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  1. anonymous
    • 5 years ago
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    \[yh(t) = \alpha e ^{2t}+\beta e ^{3t}\] \[yp(t) = 1/20e ^{2t}\] Therefore \[Y(t) = \alpha e ^{2t} + \beta e ^{3t} + 1/20e ^{2t}\] Look right?

  2. amistre64
    • 5 years ago
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    this one?

  3. amistre64
    • 5 years ago
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    does it look righht? yes.... can i prove its right...fraid not :)

  4. anonymous
    • 5 years ago
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    That's cool. Thanks for taking a look.

  5. anonymous
    • 5 years ago
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    It does not look right to me.

  6. anonymous
    • 5 years ago
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    The solution of the associated homogeneous equation: \[y''+5y'+6y=0\] can be found using the auxiliary equation: \[m^2+5m+6=0 \implies (m+3)(m+2)=0 \implies m=-3, m=-2\] Hence the complementary solution is: \[y_c=c_1e^{-2t}+c_2e^{-3t}\]

  7. anonymous
    • 5 years ago
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    Let's assume a solution of the form Ae^x for the particular solution, then substituting in the original function gives: \[Ae^t+5Ae^t+6Ae^t=2e^t \implies 12A=2 \implies A={1 \over 6}\] Therefore, the particular solution is y_p=e^t/6 That's the general solution of the differential equation is: \[y=y_c+y_p=c_1e^{-2t}+c_2e^{-3t}+{e^t \over 6}\]

  8. anonymous
    • 5 years ago
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    Ok. Thank you. I think I combined two questions into one in my work, and I always misplace those pesky +/- signs. But, my logic seems correct...just my arithmetic & algebra I get lazy and sloppy with. Thank you very much.

  9. anonymous
    • 5 years ago
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    I can tell that by looking at your answer. You're welcome.

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