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RolyPoly

  • 3 years ago

Solve by Laplace transform: \[\frac{dy}{dx}+3y=10sint\] y(0)=0

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  1. hartnn
    • 3 years ago
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    :D

  2. RolyPoly
    • 3 years ago
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    \[sY - 0+3Y = \frac{10}{s^2+1}\]\[Y = \frac{10}{(s^2+1)(s+3)}\]\[\frac{As+B}{s^2+1}+\frac{C}{s+3}=\frac{10}{(s^2+1)(s+3)}\] A +C = 0 -(1) 3A +B =0 -(2) 3B+C =10 -(3) (3) - (1) = 3B-A = 10 -(4) Solve (2) and (4): A = -1 B =3 C = -A = 1 \[Y = \frac{-s+3}{s^2+1}+\frac{1}{s+3}=\frac{-s}{s^2+1}+\frac{3}{s^2+1}+\frac{1}{s+3}\]\[y=-cost + 3sint + e^{-3t}\] ......

  3. hartnn
    • 3 years ago
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    yes, thats correct.

  4. RolyPoly
    • 3 years ago
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    Ha! That smiley face helps :D

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