compagnia04

2022-01-21

Finding all functions f satisfying

${f}^{\prime}\left(t\right)=f\left(t\right)+{\int}_{a}^{b}f\left(t\right)dt$

maul124uk

Beginner2022-01-21Added 35 answers

1. How do we know that $f{}^{\u2033}\left(x\right)={f}^{\prime}\left(x\right)?$ Differentiate both sides of

${f}^{\prime}\left(x\right)=f\left(x\right)+{\int}_{a}^{b}f\left(t\right)dt$ .

Remember:${\int}_{a}^{b}f\left(t\right)dt$ is a number (the net signed area between the graph of $y=f\left(x\right)$ , the x-axis, and the lines $x=a\text{}\text{and}\text{}x=b)$ . So what is its derivative?

Why would you do this? Because that integral is somewhat annoying: if you just had${f}^{\prime}\left(x\right)=f\left(x\right)$ , 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$f{}^{\u2033}\left(x\right)={f}^{\prime}\left(x\right),\text{let}\text{}g\left(x\right)={f}^{\prime}\left(x\right)$ .

Then we have${g}^{\prime}\left(x\right)=g\left(x\right)$ , so the theorem applies to g(x) (exactly what we were hoping for). And you go from there.

Remember:

Why would you do this? Because that integral is somewhat annoying: if you just had

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

Then we have

Anzante2m

Beginner2022-01-22Added 34 answers

Since ${\int}_{a}^{b}f\left(t\right)dt$ is a constant, we denote it as C, so we get ${f}^{\prime}\left(t\right)=f\left(t\right)+C$ . It is $\frac{df\left(t\right)}{dt}=f\left(t\right)+C,i.e.\frac{df\left(t\right)}{f\left(t\right)+C}=dt$ . Integrate it, we get $\mathrm{ln}(f\left(t\right)+C)=t+D$ , where D is a constant. Thus, $f\left(t\right)={e}^{t+D}-C$ . Let ${e}^{D}=K$ , we get $f\left(t\right)=K{e}^{t}-C$ .

Now plug it into the original eqution, we get$K{e}^{t}=K{e}^{t}-C+{\int}_{a}^{b}(K{e}^{t}-C)dt$ . After some manipulation, we have $K=\frac{b-a+1}{{e}^{b}-{e}^{a}}C$ . Note here we assume that $b\ne a$

If$b=a$ , C is obviously 0, so the equations can be simplified to ${f}^{\prime}\left(t\right)=f\left(t\right)$ . In this case, $f\left(t\right)=k{e}^{t}$ , where k is an arbitrary constant and $k\ne 0$ .

Now plug it into the original eqution, we get

If

RizerMix

Expert2022-01-27Added 656 answers

You know that ${f}^{\u2033}(t)={f}^{\prime}(t)$ 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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inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$