When y= a constant, how do you find the interval of definition?

Jeff12341234
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I used the linear equation method to solve a D.E. and got y=3/4 at the end. I'm asked to find the interval of definition but I don't know how to do that when Y is just a constant :/
 
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If your solution is y=c then dy/dt must have been 0.
Unless it was specified that dy/dt=0 on some particular interval, then your solution should be valid for all t.
 
So the interval of definition would be (-∞,∞)?

I just don't get how a function can have a domain when it's just a constant...
 
Jeff12341234 said:
So the interval of definition would be (-∞,∞)?

I just don't get how a function can have a domain when it's just a constant...

If your function is y(t) = 3/4, it means that for any t you give the function as an input, the function returns the value 3/4. So, the domain is whatever range of values of t you are allowed to put into your function. It doesn't matter that your function happens to return a constant in this case.
 
thanks
 
Jeff12341234 said:
So the interval of definition would be (-∞,∞)?

I just don't get how a function can have a domain when it's just a constant...
Every function has a domain.
Mute said:
If your function is y(t) = 3/4, it means that for any t you give the function as an input, the function returns the value 3/4. So, the domain is whatever range[/color] of values of t you are allowed to put into your function. It doesn't matter that your function happens to return a constant in this case.
I wouldn't use the word "range" when you're talking about the domain, because of confusing the issue with the function's range.
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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