Veryfing ODE for complicated y(t)


by rdioface
Tags: complicated, veryfing
rdioface
rdioface is offline
#1
Jan28-12, 01:48 PM
P: 11
1. The problem statement, all variables and given/known data
For the differential equation, verify (by differentiation and substitution) that the given function y(t) is a solution.


2. Relevant equations
[itex] y' - 4ty = 1 [/itex]

[itex] y(t) = \int_{0}^{t} e^{-2(s^{2}-t^{2})} ds[/itex]

3. The attempt at a solution
I attempted to take [itex]\frac{d}{dt}[/itex] of y(t) as usual but
1. if I do not try bringing the [itex]\frac{d}{dt}[/itex] inside the integral I can do nothing because there is no elementary antiderivative of y(t).
2. if I do bring the [itex]\frac{d}{dt}[/itex] inside the integral, I can use the chain rule to get
[itex] y(t) = (-2) \int_{0}^{t} (-2t) e^{-2(s^{2}-t^{2})} ds[/itex]
but since my variable of integration is ds not dt, this doesn't allow me to use a u-substitution as I had hoped nor can I think of a way to relate ds and dt.

More or less I do not know how to take [itex]\frac{d}{dt}[/itex] of y(t) and I do not know any other ways to solve the problem.
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τheory
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#2
Jan28-12, 02:23 PM
P: 43
Do you have to do it that way? Because it looks like you can solve it linearly.
rdioface
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#3
Jan28-12, 02:56 PM
P: 11
Yes, the problem asks to verify that y(t) is a solution to the differential equation y' + 4ty = 1.

Dick
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#4
Jan28-12, 04:34 PM
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P: 25,178

Veryfing ODE for complicated y(t)


Quote Quote by rdioface View Post

[itex] (-2) \int_{0}^{t} (-2t) e^{-2(s^{2}-t^{2})} ds[/itex]
but since my variable of integration is ds not dt, this doesn't allow me to use a u-substitution as I had hoped nor can I think of a way to relate ds and dt.
I don't think you have to be able to do the integral to identify that expression as [itex]4 t y(t)[/itex]. Since you are integrating ds you can factor out the t.
rdioface
rdioface is offline
#5
Jan28-12, 09:30 PM
P: 11
Quote Quote by Dick View Post
I don't think you have to be able to do the integral to identify that expression as [itex]4 t y(t)[/itex]. Since you are integrating ds you can factor out the t.
facepalm.jpg

Thanks!


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