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Ode

Hi , I have this equation y Prime=-y+t+1
my question am not sure if we integrate how did it become y=t+e^-t


y prime is dy/dt

this question is from euler's method

thanks
(edited 9 years ago)
Original post by meme12
Hi , I have this equation y Prime=-y+t+1
my question am not sure if we integrate how did it become y=t+e^-t


y prime is dy/dt

this question is from euler's method

thanks
Thread will be moved to the maths study help sub-forum where you will get a much faster answer.

Thanks. :smile:
Original post by meme12
Hi , I have this equation y Prime=-y+t+1
my question am not sure if we integrate how did it become y=t+e^-t


y prime is dy/dt

this question is from euler's method

thanks

Head moved for the above reason :smile:
Reply 3
Removed, been completely dumb here using the wrong method, I'm not too knowledgeable about Euler's Method, but I'm sure someone here can help you, sorry :smile:!

Euler's Method, however is an approximation, you've given the exact solution, which I assumed was found via solving it using an integrating factor.

Here's what I proviously wrote:

Spoiler

(edited 9 years ago)
Reply 4
Original post by meme12
Hi , I have this equation y Prime=-y+t+1
my question am not sure if we integrate how did it become y=t+e^-t


y prime is dy/dt

this question is from euler's method

thanks

If you want to find an exact solution, then use an integrating factor and solve the O.D.E normally, then I assume there are some initial conditions, which enable you to deduce that the constant of integration is 1.

Euler's method involves setting up an iterative formula to give an approximation to the solution, this one's a little more complicated than other examples I've seen, but I'll give it a shot. . .
If you write Δy=yn+1yn, Δt=tn+1tn\Delta y = y_{n+1}-y_n , \ \Delta t= t_{n+1}-t_n then you can show that
yyn+1ynhy' \approx \dfrac{y_{n+1}-y_n}{h}
This assumes that you use a constant change in time, h=tn+1tnh=t_{n+1}-t_n for each step.
You then have an iterative formula in the form:
yn+1=yn+f(yn,tn)y_{n+1}=y_n+f(y_n,t_n)
Using any initial conditions you have, you may then be able to write yny_n in terms of y0y_0, and find any point specified, without further information, there's not a lot I can say :redface:
(edited 9 years ago)
Reply 5
Thanks for the try , I am not sure how did we get that answer but I will try to ask
Reply 6
Original post by meme12
Thanks for the try , I am not sure how did we get that answer but I will try to ask


The answer is exact, as I and joostan mentioned. Euler's method is an approximation, it's not exact. Hence its just solved a first order DE, by the integrating factor. I assumed it gave you some initial conditions, which would need to be y(0)=1y(0) = 1 or such. Can you post the full question?
(edited 9 years ago)
Original post by meme12
Thanks for the try , I am not sure how did we get that answer but I will try to ask


I'm guessing you want to know how to get the exact answer without using Euler's method?

For the general method, you can see the notes here: http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx

First you solve the homogeneous part of the equation y=y y'=-y to get the complimentary function as yC(t)=Aety_C(t)=Ae^{-t}

Next you look for a particular solution of the form yP(t)=Bt+Cy_P(t)=Bt+C. Plug this into your original equation to get B+Bt+C=t+1B+Bt+C=t+1. From inspection of equating coefficients, we see that B=1B=1 and C=0C=0.

The general solution of the ode is given by adding the complimentary and the particular solution to give y(t)=Aet+ty(t)=Ae^{-t} + t. To find the value of the constant AA you need an initial condition which I assume you were given as y(0)=1y(0)=1.
Reply 8
Ok what about solving it using first order did equa that what my tutor said what is the difference?
Reply 9
Original post by meme12
Ok what about solving it using first order did equa that what my tutor said what is the difference?

Thanks
Reply 10
This way seem more easier , I will try this way any idea what is the dif between using this way and solve first order differential equation way under linear equation


thank you
Reply 11
Will you do this same way . I will try tonputcas much as I can from question
y prime =1/t(y^2-y)
y(1)=1/2

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