MHB Integration - Find all functions f(x)

[Dethrone]

Find all functions $f(x)$ so that $\left(\int \frac{dx}{f(x)}\right)\left(\int f(x) \,dx\right)=c$, constant.

The question says "no guessing". I looked at families of functions, starting with $f(x)=a$, $f(x)=x^n$, and $f(x)=\sin\left({x}\right)$, but they all fail. Any hints? (Wondering)

 

[I like Serena]

Find all functions $f(x)$ so that $\left(\int \frac{dx}{f(x)}\right)\left(\int f(x) \,dx\right)=c$, constant.

The question says "no guessing". I looked at families of functions, starting with $f(x)=a$, $f(x)=x^n$, and $f(x)=\sin\left({x}\right)$, but they all fail. Any hints? (Wondering)

Hmm... how about $f(x)=e^{ax}$? (Wondering)

 

[Dethrone]

That works! :D I mainly posted my question to see if there was a better method of solving the problem instead of going through all elementary functions, but yeah. The question says: "Note: simply guessing an $f(x)$ that satisfies the equation is worth only 1 mark.", so I wasn't sure whether guessing families of $f(x)$ and trying them is considered "guessing". What do you think? (Wondering)

 

[I like Serena]

That works! :D I mainly posted my question to see if there was a better method of solving the problem instead of going through all elementary functions, but yeah. The question says: "Note: simply guessing an $f(x)$ that satisfies the equation is worth only 1 mark.", so I wasn't sure whether guessing families of $f(x)$ and trying them is considered "guessing". What do you think? (Wondering)

I'm not sure. It seems as if we should do something more. (Thinking)

We can at least generalize to $f(x)=ae^{bx}$, which also satisfies the equation with a condition on $b$ that is related to $c$. So at least we're not guessing a single $f(x)$.

The suggestion is there though, that we're supposed to come up with a smarter method to find all f(x), or otherwise that we should prove that we've found all of them.

 

[juantheron]

Given $\displaystyle \int\frac{1}{f(x)}dx\cdot \int f(x)dx =c.........(1)$

Now Differentiate both side w. r to $x$, We Get

$\displaystyle \int \frac{1}{f(x)}dx\cdot f(x)+\frac{1}{f(x)}\cdot \int f(x)dx = 0.............(2)$

Now from equation $(1)$, We get $\displaystyle \int\frac{1}{f(x)}dx = \frac{c}{\int f(x)dx}$

and put into equation $(2)$, We get $\displaystyle \frac{c\cdot f(x)}{\int f(x)dx}+\frac{\int f(x)dx}{f(x)} = 0$

So $\displaystyle \left(\int f(x)dx\right)^2 = -c\cdot \left(f(x)\right)^2$

Now Let $-c=k^2\;,$ Then $\displaystyle \left(\int f(x)dx\right)^2 = k^2\cdot \left(f(x)\right)^2$

So $\displaystyle \int f(x)dx = \pm k\cdot f(x)$

 

[I like Serena]

And there it is! ;)

 

[Dethrone]

Very neat! (Wondering) The answer does not give us a specific type of function, but rather, the relationship that said function must satisfy with its antiderivative. :D

Spoiler

I can only imagine how evil my professor will be...(Dull) for this question is a typical exam question...

 

[I like Serena]

Very neat! (Wondering) The answer does not give us a specific type of function, but rather, the relationship that said function must satisfy with its antiderivative. :D

Can you solve it?

Hint: take the derivative.

 

[Dethrone]

Can we not simply say...

$$f(x)=\frac{\int f(x) \,dx}{\pm k}$$

but:

$$f(x)=\pm kf'(x)$$

I hope this is it, because it seems like it could involve differential equations, which I haven't really studied yet. :(

 

[I like Serena]

Can we not simply say...

$$f(x)=\frac{\int f(x) \,dx}{\pm k}$$

but:

$$f(x)=\pm kf'(x)$$

I hope this is it, because it seems like it could involve differential equations, which I haven't really studied yet. :(

Ah.
Yes, sorry, it does involve a differential equation. (Doh)If you're interested, I'll show you.

If we write $y=f(x)$ and $\d y x = f'(x)$, we get:
\begin{array}{lcl}
y&=&\pm k\d y x \\
\frac {dy}{y} &=& \pm \frac{dx}{k} \\
\int \frac {dy}{y} &=& \int \pm \frac{dx}{k} \\
\ln |y| &= & \pm\frac x k + C_1 \\
|y| &=& e^{\pm x/ k + C_1} \\
y &=& \pm e^{C_1} e^{\pm x/ k} \\
y &=& C_2 e^{\pm x/ k} \\
y &=& C_2 e^{\pm x/ \sqrt{-c}} \\
\end{array}

And this is what we already found by guessing.
Substituing it in the original equation, shows that this entire family of solutions is indeed the solution. (Mmm)

 

[I like Serena]

Alternatively, without resorting to differential equations (well... sort of), we can do:

\begin{array}{lcll}
f(x) &=& \pm k f′(x) \\
\frac{f'(x)}{f(x)} &=& \pm \frac 1 k & \text{This is the derivative of the next line.}\\
\ln(f(x)) &=& \pm \frac x k + C_1 & \text{Actually, I'm leaving out a $\pm$ here, that should be included.}\\
f(x) &=& e^{\pm x/k + C_1}
\end{array}

 

[Dethrone]

Yep, but as you have said, I do smell the differential equation in that one, too. (Wasntme)

This course was meant for students to have already learned differential equations, so I'm sure this is the right way! :D (Yes)