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Verify that this is the solution to an ordinary differential equation

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NCyellow
#1
Jan13-10, 01:45 PM
P: 22
1. The problem statement, all variables and given/known data
I am given dy/dt -2yt = 1
and y(t) = (e^(t^2))[e^(-s^2)ds] + e^(t^2)
integrate from t to 0 within the brackets.

2. Relevant equations



3. The attempt at a solution
I know that the derivative of y(t) would equal e^(t^2)
However I do not know how I am supposed to solve after I plug all the numbers in. It becomes a huge mess with 2t * the whole mess that is y(t)
Please advise
Thank you!
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Mark44
#2
Jan13-10, 02:10 PM
Mentor
P: 21,215
Is this your y(t)?
[tex]y(t)~=~\int_0^t e^{t^2}(e^{-s^2})ds~+ e^{t^2}[/tex]
NCyellow
#3
Jan13-10, 02:21 PM
P: 22
Quote Quote by Mark44 View Post
Is this your y(t)?
[tex]y(t)~=~\int_0^t e^{t^2}(e^{-s^2})ds~+ e^{t^2}[/tex]
Yes, but the first e^(t^2) is outside and multiplying the integral.

HallsofIvy
#4
Jan13-10, 02:22 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348
Verify that this is the solution to an ordinary differential equation

Quote Quote by NCyellow View Post
1. The problem statement, all variables and given/known data
I am given dy/dt -2yt = 1
and y(t) = (e^(t^2))[e^(-s^2)ds] + e^(t^2)
integrate from t to 0 within the brackets

2. Relevant equations



3. The attempt at a solution
I know that the derivative of y(t) would equal e^(t^2).
So
[tex]y(t)= e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2}[/tex]?
No, the derivative is NOT [itex]e^{t^2}[/itex].
Differentiating,
[tex]y'= 2te^{t^2}\int_t^0 e^{-s^2}ds- e^{t^2}e^{-t^2}+ 2te^{t^2}[/tex]
[tex]= 2t(e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2})- 1= 2ty- 1[/tex].
That's exactly why it satisfies dy/dt- 2ty= 1.
But that is NOT the "general solution". The way you have set it up, when t= 0, the integral is from 0 to 0 and so is 0. [itex]y(0)= 0+ e^{0^2}= 1[/itex]. This is the solution of dy/dx- 2ty= 1 with intial condition y(0)= 1. To get the general solution, add "+ C". Then choose C to give whatever specific value it is supposed to have.

However I do not know how I am supposed to solve after I plug all the numbers in. It becomes a huge mess with 2t * the whole mess that is y(t)
Please advise
Thank you!
Plug what numbers in? I have no idea what you are talking about! Do you mean differentiate it? (As I just did? I used the product rule on [itex]e^{2x}\int e^{-2s}ds[/itex] and the Fundamental theorem of Calculus to differentiate [itex]\int e^{-2s}ds[/itex] itself.)
NCyellow
#5
Jan13-10, 02:31 PM
P: 22
Quote Quote by HallsofIvy View Post
So
[tex]y(t)= e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2}[/tex]?
No, the derivative is NOT [itex]e^{t^2}[/itex].
Differentiating,
[tex]y'= 2te^{t^2}\int_t^0 e^{-s^2}ds- e^{t^2}e^{-t^2}+ 2te^{t^2}[/tex]
[tex]= 2t(e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2})- 1= 2ty- 1[/tex].
That's exactly why it satisfies dy/dt- 2ty= 1.
But that is NOT the "general solution". The way you have set it up, when t= 0, the integral is from 0 to 0 and so is 0. [itex]y(0)= 0+ e^{0^2}= 1[/itex]. This is the solution of dy/dx- 2ty= 1 with intial condition y(0)= 1. To get the general solution, add "+ C". Then choose C to give whatever specific value it is supposed to have.


Plug what numbers in? I have no idea what you are talking about! Do you mean differentiate it? (As I just did? I used the product rule on [itex]e^{2x}\int e^{-2s}ds[/itex] and the Fundamental theorem of Calculus to differentiate [itex]\int e^{-2s}ds[/itex] itself.)
Hi, thank you very much. I guess i misread a rule from the book when I thought that differentiating it would be simple. When i said plugging numbers in I meant plugging the y and y' back into the original equation to make sure it works.
Mark44
#6
Jan13-10, 02:51 PM
Mentor
P: 21,215
y and y' aren't "numbers" to plug in - they are functions. It's very important to communicate clearly, even moreso in the context of a forum where you have to use text only, and there is a lag time between a question and an answer.
dacruick
#7
Jan13-10, 03:03 PM
P: 1,084
when i do this i have the integral of e^(-t^2). which from my understanding is not a very fun integral. I dont understand the method you guys took to doing this. can someone explain??
NCyellow
#8
Jan13-10, 03:12 PM
P: 22
Quote Quote by Mark44 View Post
y and y' aren't "numbers" to plug in - they are functions. It's very important to communicate clearly, even moreso in the context of a forum where you have to use text only, and there is a lag time between a question and an answer.
Hi, Thank you so much for helping me. I will try to be more clear in the future.
HallsofIvy
#9
Jan14-10, 06:30 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348
The derivative of
[tex]\int_0^x e^{-t^2}dt[/tex]
is just [itex]e^{-x^2}[/itex]- that's the "Fundamental Theorem of Calculus". But [itex]\int_0^x e^{x^2}e^{-t^2}dt= e^{x^2}\int_0^t e^{-t^2}dt[/itex] is a product of functions of x and you have to use the chain rule.


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