Question

How to solve: \(x = p(1-e^{-kx})^2\) where k and p are constant
Please, help

Answer 1

@dan815

Answer 2

What are you solving for?

Answer 3

He said k , p are constants so no other variable than x

Answer 4

for x , I believe

Answer 5

Thank's looks like a lot of algebra Need to get some paper.

Answer 6

please, try to use the Taylor expansion of \(\Large e^{-kx}\):

\[\huge {e^{ - kx}} \approx 1 - kx + \frac{{{k^2}{x^2}}}{2}\]

Answer 7

If so, we use power series for \(e^{-kx}= \sum_{n=0}^\infty \dfrac{(-kx)^n}{n!}\) . Is it a mess?

Answer 8

That is an excellent idea, because simple algebra won't solve this equation.

Answer 9

Ithink it is a way in order to get an algebraic equation, which can be easily solved

Answer 10

oops.. I think...

Answer 11

\(1- e^{-kx} )^2 = 1 - 2e^{-kx}+e^{-2kx}= e^{-kx}(e^{kx}-2+e^{-kx})\)
and we know that \(2cos(kx) = e^{kx}+e^{-kx}\)
hence it becomes \(2e^{-kx}(cos(kx) -1)\)

Answer 12

A quick guess is x=0

Answer 13

That's right!! @Kainui

Answer 14

You can confirm that it's the only solution by taking the derivative and seeing that it's going to be the only solution.

Answer 15

I mean because the slopes of the functions will have only one intersection and all that... I'm a little tired haha hard to explain myself.

Answer 16

furthermore, since we are searching for an intersection between a straight line \(y=x\) and the exponential function, we can consider only the linear term of the Taylor expansion, so we get the subsequent equation:
\[\huge x = p\left( {{k^2}{x^2}} \right)\]

Answer 17

I got what you mean. Thanks a lot.

Answer 18

therefore, we get the subsequent solutions:
\[\huge x = 0,\quad x = \frac{1}{{p{k^2}}}\]

Answer 19

Here's a trick to remove a constant:

\(y=-kx\)

\[y=-kp(1-e^y)^2\]

\[-kp=c\] is the new arbitrary constant:

\[y=c(1-e^y)^2\]

Answer 20

and then?