# from solution to mother equation

#### naviakam

##### New member
Hi
I have got two solutions to unknown equations:
1. power equation in the form y=k x(-n) where k is coefficient and n is the power
1. exponential equation of the form: y=y0+k exp(-(x-x0)/t) Is it possible to find the original equation/s leading to these solutions (kind a reverse engineering!)?

Best

#### Subhotosh Khan

##### Super Moderator
Staff member
Hi
I have got two solutions to unknown equations:
1. power equation in the form y=k x(-n) where k is coefficient and n is the power
1. exponential equation of the form: y=y0+k exp(-(x-x0)/t) Is it possible to find the original equation/s leading to these solutions (kind a reverse engineering!)?

Best
Have you studied Differential Equations?

#### Dr.Peterson

##### Elite Member
Hi
I have got two solutions to unknown equations:
1. power equation in the form y=k x(-n) where k is coefficient and n is the power
1. exponential equation of the form: y=y0+k exp(-(x-x0)/t) Is it possible to find the original equation/s leading to these solutions (kind a reverse engineering!)?

Best
In general, there is no way to determine "THE" original equation; if I gave you the solution x = 6, you can't possibly know what equation I solved.

That's true also of differential equations. But you can find "A" differential equation that yields each of your functions, by taking the derivative and looking for some connection between y' and y. See what you can come up with, and let us know. (Of course, if you are studying differential equations, there is a good chance that you have seen solutions that look like these, so you can have a good idea what to expect.)

#### Subhotosh Khan

##### Super Moderator
Staff member
Hi
I have got two solutions to unknown equations:
1. power equation in the form y=k x(-n) where k is coefficient and n is the power
1. exponential equation of the form: y=y0+k exp(-(x-x0)/t) Is it possible to find the original equation/s leading to these solutions (kind a reverse engineering!)?

Best
What do you get if you differentiate the given expressions of 'y' interms of x?

y = k*x-n $$\displaystyle \ \ \to \ \ \frac{dy}{dx} = ?$$

Similarly for the second problem $$\displaystyle \ \ \ \ \frac{dy}{dx} = ?$$

#### HallsofIvy

##### Elite Member
qThat depends strongly on which of k, t, x, $$\displaystyle y_0$$, and $$\displaystyle x_0$$ are "constants of integration" or parameters or variables! For $$\displaystyle y= kx^{-n}$$, assuming that y and x are the variable, that n is a parameter, and that k is the constant of integration the, differentiating y with respect to x, $$\displaystyle \frac{dy}{dx}= -knx^{-n-1}$$. Dividing that by the original equation to eliminate k, $$\displaystyle \frac{1}{y}\frac{dy}{dx}= \frac{-knx^{-n-1}}{kx^{-n}}$$$$\displaystyle = \frac{-n}{x}$$, $$\displaystyle \frac{dy}{dx}=-\frac{ny}{x}$$ or $$\displaystyle x\frac{dy}{dx}= ny$$.

#### naviakam

##### New member
Thanks.
Ions are accelerated in a potential, the graph for number of ions versus ion energy plotted in origin and fitted. There are two plots: one is fitted with the power and the other with exponential decay in the form mentioned above (y is the number and x is the energy). We've been asked to guess the original function leading to these spectra. Amazingly, both equations seem originating from a single function!

#### Metronome

##### Junior Member
As an alternative way to eliminate the constant of integration (because dividing the ODE by the solution feels less natural to me), you can try to turn it into an additive constant before implicitly differentiating.

$$\displaystyle y = kx^{-n}$$
$$\displaystyle ln(y) = ln(kx^{-n})$$
$$\displaystyle ln(y) = ln(k) + ln(x^{-n})$$
$$\displaystyle ln(y) = k + ln(x^{-n})$$
$$\displaystyle \frac{1}{y}dy = 0 - \frac{n}{x}dx$$
$$\displaystyle \frac{dy}{dx}= - \frac{ny}{x}$$

#### Subhotosh Khan

##### Super Moderator
Staff member
Thanks.
Ions are accelerated in a potential, the graph for number of ions versus ion energy plotted in origin and fitted. There are two plots: one is fitted with the power and the other with exponential decay in the form mentioned above (y is the number and x is the energy). We've been asked to guess the original function leading to these spectra. Amazingly, both equations seem originating from a single function!
How did you deduce that?

#### naviakam

##### New member
How did you deduce that?

there is a single capacitor discharge, resulted in two spectra one fitted by power equation and other exponential decay curve.

#### naviakam

##### New member
As an alternative way to eliminate the constant of integration (because dividing the ODE by the solution feels less natural to me), you can try to turn it into an additive constant before implicitly differentiating.

$$\displaystyle y = kx^{-n}$$
$$\displaystyle ln(y) = ln(kx^{-n})$$
$$\displaystyle ln(y) = ln(k) + ln(x^{-n})$$
$$\displaystyle ln(y) = k + ln(x^{-n})$$
$$\displaystyle \frac{1}{y}dy = 0 - \frac{n}{x}dx$$
$$\displaystyle \frac{dy}{dx}= - \frac{ny}{x}$$
thanks
how the exponential equation is differentiated?

#### Subhotosh Khan

##### Super Moderator
Staff member
thanks
how the exponential equation is differentiated?
Do a Google search and tell us what you found.

#### Subhotosh Khan

##### Super Moderator
Staff member
there is a single capacitor discharge, resulted in two spectra one fitted by power equation and other exponential decay curve.
What are the R^2 values for those curve fits?

#### naviakam

##### New member
Do a Google search and tell us what you found.
There is a y0 in the exponential equation, how it is differentiated?

Staff member

#### Subhotosh Khan

##### Super Moderator
Staff member
There is a y0 in the exponential equation, how it is differentiated?
yo and xo are "generally" the "initial values" - thus treated as constants.

#### naviakam

##### New member
yo and xo are "generally" the "initial values" - thus treated as constants.
then it will be:
y"+(1/t)Y'=0
correct?

#### Subhotosh Khan

##### Super Moderator
Staff member
then it will be:
y"+(1/t)Y'=0
correct?
What is 'x' in the first equation?

What is 'y' & 't' in the second equation?

#### naviakam

##### New member
What is 'x' in the first equation?

What is 'y' & 't' in the second equation?
in both equations y is the number of ions, x is the energy. but t: I just fitted the curve and exponential decay fit was the best. t was given by the fit in the origin software.