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}<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> now suppose we cannot intergrate p(t), then my professor mentioned that<br /> <br /> we can introduce a dummy variable so that<br /> <br /> the variable becomes somelike s and the<br /> <br /> the upperbound becomes t and the lowerbound becomes $$t_{0}$$.<br /> <br /> <br /> <br /> <br /> <br /> My question is,<br /> <br /> what theorem says that the initial condition value of $$t_{0}$$must be the lowerbound, once we introduce a dummy variable?<br /> <br /> Can we have the upperbound as $$t_{0}$$ and the lowerbound as t?<br /> <br /> if not, why can $$t_{0}$$ be only the lowerbound?<br /> <br /> <br /> <br /> 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.<br /> <br /> I don't want to use dummy variables just because my professor tells me it is okay to do so.<br /> <br /> <br /> Please help me.<br /> <br /> Thank you.[/tex]

 

[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.