Finding all functions f satisfying f'(t)=f(t)+\int_{a}^{b}f(t)dt

compagnia04 2022-01-21 Answered
Finding all functions f satisfying
f(t)=f(t)+abf(t)dt
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maul124uk
Answered 2022-01-21 Author has 35 answers
1. How do we know that f(x)=f(x)? Differentiate both sides of
f(x)=f(x)+abf(t)dt.
Remember: abf(t)dt is a number (the net signed area between the graph of y=f(x), the x-axis, and the lines x=a and x=b). So what is its derivative?
Why would you do this? Because that integral is somewhat annoying: if you just had f(x)=f(x), then you would be able to solve the differential equation simply enough (e.g., with the theorem you have).
But since all that is standing in our way is a constant that is adding, differentiating should spring to mind: that will get rid of the constant, and just ''shift'' the problem ''one derivative down'' (to a relation between f′′(x) and f′(x)).
2. Once you know that f(x)=f(x),let g(x)=f(x).
Then we have g(x)=g(x), so the theorem applies to g(x) (exactly what we were hoping for). And you go from there.
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Anzante2m
Answered 2022-01-22 Author has 34 answers
Since abf(t)dt is a constant, we denote it as C, so we get f(t)=f(t)+C. It is df(t)dt=f(t)+C,i.e.df(t)f(t)+C=dt. Integrate it, we get ln(f(t)+C)=t+D, where D is a constant. Thus, f(t)=et+DC. Let eD=K, we get f(t)=KetC.
Now plug it into the original eqution, we get Ket=KetC+ab(KetC)dt. After some manipulation, we have K=ba+1ebeaC. Note here we assume that ba
If b=a, C is obviously 0, so the equations can be simplified to f(t)=f(t). In this case, f(t)=ket, where k is an arbitrary constant and k0.
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RizerMix
Answered 2022-01-27 Author has 438 answers
You know that f(t)=f(t) because you can differentiate the equation you are trying to solve.Since the second term in the right hand side of the equation you are trying to solve is annoying, it is a good idea to try to get rid of it. Since it is constant, then that can be done by differentiating the equality with respect to t.
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