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April 12th, 2007, 10:20 PM   #1
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Differential equation

somebody that can help me solve this?

find the general solution

dy/dt=-2y

I know that the preposition dy/dt=ay gives y=ce^at and the answer is supposed to be y=ac^-2t, but proving this was worse...
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April 13th, 2007, 05:09 AM   #2
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In other words, you are saying that:

y' + 2y = 0

If we had initial values, we could solve this by Laplace transformations, but since we don't, we'll have to go by a different route. We'll have to say that r^2 + 2 = 0 and then solve for the complex number r, and then use the two real parts of e^r as the terms that will give us the solution to the problem.
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April 14th, 2007, 01:43 AM   #3
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Multiply by e^(2t), then integrate.
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May 6th, 2007, 03:29 PM   #4
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dy/dt = -2y

multiply with dy and divide with y and you get:

dy/y=-2dt

integrate and:

ln(y)-ln(y0)=-2t

ln(y)=-2*t+ln(y0)

y=exp(-2t+ln(y0))

y=y0*exp(-2t)

where y0 is the initial value!
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May 6th, 2007, 11:36 PM   #5
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Quote:
Originally Posted by Infinity
If we had initial values, we could solve this by Laplace transformations, . . .
One can write y0 for y(0) and use the Laplace transform method. Don't expect to get y=ac^-2t, since the correct solution is y = y0e^(-2t).

Alternatively, one can solve the characteristic equation, r + 2 = 0 (not r^2 + 2 = 0), and hence write down the answer.
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