Questions regarding dummy variables

[Math451]

I am sorry guys I don't know how to edit latex stuff.



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Is there anyone who can tell me why we are justified in using dummy variables when we are dealing with some inital problems?

suppose, some differential equation has a solution and has the intial condition.


$$y(t) = \int (p(t)) dt$$

[tex]y(t_{0}) = y_{0}








now suppose we cannot intergrate p(t), then my professor mentioned that

we can introduce a dummy variable so that

the variable becomes somelike s and the

the upperbound becomes t and the lowerbound becomes $$t_{0}$$.





My question is,

what theorem says that the initial condition value of $$t_{0}$$must be the lowerbound, once we introduce a dummy variable?

Can we have the upperbound as $$t_{0}$$ and the lowerbound as t?

if not, why can $$t_{0}$$ be only the lowerbound?



The pace in my differential equation course is really fast, so we are far passed the point of thinking about the legitimacy of dummy variables in class but I am so frustrated that I don't quite have a firm understanding on this.

I don't want to use dummy variables just because my professor tells me it is okay to do so.


Please help me.

Thank you.

 

[LCKurtz]

Your post is basically unreadable and I can only guess at what is really bothering you. Since I think I know, I will take a shot at it. Apparently you have a differential equation with a solution written in the form:

$$y(t) = \int\ p(t)\, dt$$

and which has initial condition:

$$y(t_0) = y_0$$

and you are wondering why the solution can be written as:

$$y(t) = \int_{t_0}^t p(u)\, du + y_0$$

Remember the fundamental theorem of calculus that says if F is any antiderivative of f, then:

$$F(b) - F(a) = \int_a^b\ f(u)\, du$$

In the first equation above, y(t) is an antiderivative of p(t), so by the fundamental theorem of calculus:

$$y(t) - y(t_0) = \int_{t_0}^t\ p(u)\, du$$

Since y(t₀) = y₀, put that in and solve for y(t) to get your equation.