compagnia04
2022-01-21
Answered

Finding all functions f satisfying

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

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maul124uk

Answered 2022-01-21
Author has **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

Answered 2022-01-22
Author has **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

Answered 2022-01-27
Author has **438** 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.

asked 2022-05-13

In a course on partial differential equations I came through this theorem about the general solution of a first order quasi-linear partial differential equation.

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

asked 2021-11-17

Compute $\Delta y$ and dy for the given values of x and

$dx=\mathrm{\u25b3}x$

$y={x}^{2}-5x,\text{}x=4,\text{}\mathrm{\u25b3}x=0.5$

asked 2022-01-18

Obtaining Differential Equations from Functions

is a first order ODE,

is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for

using the orginal function and (1). Finally,

which is the required differential equation.

Similarly, if the function is

following similar steps as above.

asked 2021-12-31

Find the solution of the following Differential Equations by Exact $(2y+xy)dx+2xdy=0,y\left(3\right)=\sqrt{2}$

asked 2021-02-02

Solve the following differential equations using the Laplace transform and the unit step function

$y"+4y=g(t)$

$y(0)=-1$

${y}^{\prime}(0)=0,\text{where}g(t)=\{\begin{array}{ll}t& ,t\le 2\\ 5& ,t2\end{array}$

$y"-y=g(t)$

$y(0)=1$

${y}^{\prime}(0)=2,\text{where}g(t)=\{\begin{array}{ll}1& ,t\le 3\\ t& ,t3\end{array}$

asked 2021-10-17

Find r(t) if $r}^{\prime}\left(t\right)=ti+{e}^{t}j+t{e}^{k$ and $r\left(0\right)=i+j+k$

asked 2020-10-28

Use the Laplace transform to solve the given initial-value problem.

$y{}^{\u2033}-6{y}^{\prime}+13y=0,\text{}\text{}\text{}y\left(0\right)=0,\text{}\text{}\text{}{y}^{\prime}\left(0\right)=-9$