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A problem with solution |
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| Mar3-12, 06:05 PM | #1 |
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A problem with solution
Have an equation;
d2y(t)/dt2 + 5d2y(t)/dt2 + 4y(t) = 2e-2t Solved the complementary(homogenous) part and the function and got the roots of -1 and -4 so the yh(t) is A1.e-4t + A2.e-t Forcing function is 2.e-2t so yparticular(t) is A.e-2t Am I right here ? Or am I supposed to use Ate-2t Well, if I use the first one, the resultant function doesn't give me the 2.e-2t when I put it into the differential equation, so there is something wrong obviously. However F(t) or one of its derivatives are not identical to terms in the homogenous solution, so I think I have to use the first option, which is A.e-2t After proceeding I ended up with yp(t) = 1/3.e-2t Initial values are y(0) = 0 and y(1)(0) = 0 so, K1 = -1/9 and K2 = -2/9 Still couldn't find where I am wrong Appreciate if you help me. |
| Mar3-12, 07:39 PM | #2 |
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| Mar4-12, 04:36 AM | #3 |
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| Mar4-12, 04:42 AM | #4 |
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A problem with solution
You should double check your calculation for Ae-2t
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| Mar4-12, 05:31 AM | #5 |
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| Mar4-12, 06:37 PM | #6 |
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I got a particular solution of y[itex]_{p}[/itex] = -e[itex]^{-2t}[/itex]. See if you get the same.
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| Mar4-12, 08:20 PM | #7 |
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I get [itex]A_1=\frac{1}{3}[/itex] and [itex]A_2=\frac{3}{2}[/itex]. The particular solution given above is correct.
Double-check your calculations. |
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