Solving an ordinary differential equation.

[theBEAST]

Homework Statement

Here is an example from the book:

https://dl.dropbox.com/u/64325990/MATH%20255/EX%204.PNG


Does anyone know why they chose t to be equal to 0?

Thanks!

 

[Bohrok]

I think it's just convention. Once my DE book used the general x₀ as the lower limit (where the independent variable was x).

 

[Ray Vickson]

Homework Statement

Here is an example from the book:

https://dl.dropbox.com/u/64325990/MATH%20255/EX%204.PNG


Does anyone know why they chose t to be equal to 0?

Thanks!

The point t=0 is special, because the initial condition was given there. However, in the integral you could choose any specific lower limit you want; that would just change the needed value of c. For example, we could choose the lower limit = 2. Then we would have
e^{t^2/4}y= \int_2^t e^{s^2/4}\, ds + c,
so we would determine c by equating e^0 y(0) = y(0) = 1 to
\int_2^0 e^{s^2/4}\, ds + c, to get
c = 1 - \int_2^0 e^{s^2/4}\, ds = 1 + \int_0^2 e^{s^2/4}\, ds, and this gives
y = e^{-t^2/4} \left[1 + \int_0^2 e^{s^2/4}\, ds + \int_2^t e^{s^2/4}\, ds \right]\\<br /> \;\;\; = e^{-t^2/4}\left[1 + \int_0^t e^{s^2/4}\, ds \right]. You would have gotten this right away if you had used 0 in the lower limit right from the start.

RGV

 

[LCKurtz]

To add to what others have said, you can write the step before you integrate as$$
(e^{\frac {t^2} 4}y)' = e^{\frac {t^2} 4}$$
Now by the fundamental theorem of calculus if you integrate from 0 to t you get $$
e^{\frac {t^2} 4}y(t)-e^0y(0)=\int_0^te^{\frac {s^2} 4}ds$$The point of using 0 for the lower limit is you know $y(0)=1$, and it is a simple step to solve for $y(t)$.

 

[Curious3141]

I find the neatest way to impose the initial conditions is by using definite integration of both LHS and RHS. The lower bounds correspond to the initial conditions, the upper are left variable. This way you never have to bother to calculate the constant of integration explicitly.