A problem with solution


by zoom1
Tags: solution
zoom1
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#1
Mar3-12, 06:05 PM
P: 23
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.
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gulfcoastfella
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#2
Mar3-12, 07:39 PM
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Quote Quote by zoom1 View Post
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
For the differential equation you posted, the above roots are not correct. Did you intend for there to be two second order terms in the equation when you posted it?
zoom1
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#3
Mar4-12, 04:36 AM
P: 23
Quote Quote by gulfcoastfella View Post
For the differential equation you posted, the above roots are not correct. Did you intend for there to be two second order terms in the equation when you posted it?
Ohh, pardon me, the second term is not the second derivative, it had to be first derivative.

Office_Shredder
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#4
Mar4-12, 04:42 AM
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A problem with solution


You should double check your calculation for Ae-2t
zoom1
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#5
Mar4-12, 05:31 AM
P: 23
Quote Quote by Office_Shredder View Post
You should double check your calculation for Ae-2t
Got it! Thank you ;)
gulfcoastfella
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#6
Mar4-12, 06:37 PM
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I got a particular solution of y[itex]_{p}[/itex] = -e[itex]^{-2t}[/itex]. See if you get the same.
dash-dot
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#7
Mar4-12, 08:20 PM
P: 2
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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