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Verify that this is the solution to an ordinary differential equation 
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#1
Jan1310, 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! 


#2
Jan1310, 02:10 PM

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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] 


#3
Jan1310, 02:21 PM

P: 22




#4
Jan1310, 02:22 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348

Verify that this is the solution to an ordinary differential equation
[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. 


#5
Jan1310, 02:31 PM

P: 22




#6
Jan1310, 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.



#7
Jan1310, 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??



#8
Jan1310, 03:12 PM

P: 22




#9
Jan1410, 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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