Another one

[kedaman]

y'=2+e^y

and i don't know how to solve it so i really appreciate help!

[HarryW]

Beats me. I get

y = 2y + e^y

and then just get sort of stuck.

ln(-y) = y ?

*Shrug*

[kedaman]

as guv said, were more programmers here than mathematicians :(

[Guv]

Kedaman, integration is one of my weaker areas of knowledge. I did a bit of research in a book with a table of integrals. I used a couple techniques I looked up in an Advanced Calculus text.

I found nothing and got nowhere on my own with integral( e^Y)

Since this is weak area for me, there is still some hope, but I would not bet a lot of money on there being an analytical solution.

I do not know how to begin to find a numerical solution.

Good luck.

BTW: What is the context here?

[parksie]

I have a nasty feeling complex numbers are involved here, as when I rearranged the integral, I got

y - e^y = 0

And when y was 0, the f(y) = -1, at the cusp of the curve.

Plotting a graph of the function gave this:

[parksie]

Well...I put the whole lot into Derive, and it gave me an answer of -0.567143.

[Guv]

HarryW: You should have gotten y = 2 * x + Integral( e ^ y dx )

Parksie: What the hell is Derive? How could it possibly come up with a number when a differential equation defines (or implies) a function? The solution to a differential equation like Kedaman's is supposed to look like y = Function( x )

For example: y' = 3 * x ^ 2 has the solution y = x ^ 3

How did you get: y - e ^ y = 0 ?

BTW: the above function can never equal zero for real y, but has the following complex root (I hope I did not make a typo here)
.31813 15052 04764 + 1.33723 57014 30690iI doubt that it has other complex roots, but perhaps it does. My intuition here is not reliable, and I did not look for other roots.

Kedaman: See above comment for HarryW. I started thinking about this a bit. To solve this differential equation, it, it seems to me that you have to integrate (e^y) with respect to x, which just does not look possible.

I tried integration by parts and got nowhere. There is nothing which seems pertinent in a table of integrals I have.

There seems to be something fundamentally wrong with this differential equation. If not wrong, then just intractable.

There is a lot I do not know about differential equations, so I could be wrong.

Duh, duh, duh!!! The following just occurred to me, and might be tractable.

Y' is notation for dy / dx

dy / dx = 2 + e ^ y

dy = (2 + e ^ y )dx

dx = dy / (2 + e ^ y)

x = Integral( dy / (2 + e ^ y) )

There might be something you can do with the above. I did not try because I am weak on integration. I can often manage numerical integration of definite integrals, but indefinite integrals usually give me a headache if they are not in a table of known integrals.

If this one is in a standard table, you can find it as easily as I can. If not, good luck!!

[parksie]

Derive is an algebraic manipulation package.....I think I was getting confused by then (look at the time of the post).

Oh well....shows how little I know :D

[kedaman]

Hey don't look at me all ppl, why otherways would i post this qwestion, i dont' know the answer!! I've dealed with implicit derivation but i dunno about this case, it's just too confusing not having x around.

[HarryW]

Oh, well looking back I think I must have been integrating with respect to y, which is a pretty stupid thing to do :rolleyes: Thank you for the correction.

I would have thought integrating e^y with respect to x would simply treat e^y as a constant, although that wouldn't do much for you in terms of getting a reasonable solution :)

What is the benefit of solving for x? Would you try to rearrange it, once solved, to y=f(x) ?

Ah well, as I said, I'm stuck :)

[HarryW]

Oh hang on, I was thinking of differentiation... oops. If you integrate e^y with respect to x you should get (xe^y + c) shouldn't you? Or perhaps I'm having another fit of mathematical imcompetence ;)

[kedaman]

yeah thats right but what do you do about the 2 and what happens with the y?

[HarryW]

Well if you integrate (2 + e^y) with respect to x then you get (2x + xe^y) or (x(2+e^y)). So..

y = x(2+e^y)

x = y/(2+e^y) <-- this is pretty much what Guv said

I don't know where to go from there really. The exponential makes me want to use logs but I don't really know how to rearrange things....

y = 2x + xe^y

xe^y = y - 2x

ln(x) + y = ln(y - 2x)

y = ln(y - 2x) - ln(x)


Which all looks okay but then how do you get that y out of the right hand side? Beats me.

[Guv]

Kedaman: It was in my standard table of integrals!! I did not expect it to be so I did not look prior to committing my previous post.

x = Integral( dy / (2 + e ^ y) ) Id est: x = Integral[ 1 / (2 + e^y) ] with respect to y. From previous post.

x = y / 2 - ( log(2 + e^y) ) / 2 From a handbook of mathematical formulae.

I like it better when you can get it in the form y = f( x ), but the above is not terrible to work with. The variables are not in some germixen form, and you can plot x versus y.

There seems to be some erroneous integration in previous posts by others. For example the following is incorrect.Well if you integrate (2 + e^y) with respect to x then you get (2x + xe^y) or (x(2+e^y))In general: You cannot integrate f(y) with respect to x.

[HarryW]

Why is that? I thought that the y terms would simply be treated as constant. I am not saying you're wrong, I would just appreciate if you could explain it to me :)

[Guv]

HarryW: It is easy to confuse the concepts involving single, double, & triple integrals.

Double/triple integrals involve 2/3 integrations using 2/3 variables of integration. The notation is hard to show here. The idea is that you have a function of 2/3 variables which is integrated 2/3 times. When doing the first integration of a multiple integral, only the variable of integration is treated as a variable; The other variables are treated as constants. Similarly for subsequent integrations. This concept cannot be applied to single integrals, but can lead to misunderstandings.

A single integral must only involve one variable. Sometimes confusing notation is used. You often see Integral[ ydx ] instead of Integral[ F(x)dx ], where y = F(x). Deferential equations problems often involve notation like the following. dy/dy = Function(x, y) dy = Function(x, y)dx Integral[ dy ] = Integral[ Function(x,y)dx ]However, to actually do the integration involved in the last equation above, you must somehow eliminate variable y from the integrand on the right of the equals sign. Sometimes this is done by substitution for y in Function(x, y) Algebraic manipulation of dy = Function(x, y)dx, might result in Integral[ F(y)dy ] = Integral[ G(x)dx ]

[noble]

kedamn,
are you trying to solve this equation for y?

[kedaman]

I was trying to solve the differential, thanks for all your support guys, especially guv :) but it was a bit too late, maybe next time i have something to bring up that could be solved in two days..

[noble]

I believe Guv to be right but i think what needs to be
mentioned is that if the other variable in the problem is
not a function of x then it can be treated as a constant.

For example:
y' = 2 + e^z
where z is not a function of x

Also, it needs to be noted that this is a non-seperable
differential equation. There is no integrating factor
which can be used to get this in the proper form of a first
order differential equation. When i get some free time
at work, I will pursue this problem to it's extent.

[noble]

also, you can not use double (multiple) integration here
as the problem does not call for it. It is a single
integration.

dy/dx = 2 + e^y

dy = (2 + e^y) dx

that's one integral, therefore this is not a multiple
integration

[noble]

ok, after spending a few minutes on this problem, let
me tell u what i think

I don't think there is a closed form solution to this
problem. What you'll find is that numerical methods
or a series solution must be used to solve it.

Consider treating the problem as a riccati equation:
if you tried to find the particular solution you will
find that y = ln(-2) which can not be solved.

This is the first approach I used against this problem
and i will have to guess that it can not be solved with any closed form approach.

[Guv]

Noble: The differential equation which was the original subject of this thread has been solved. See a post of mine several posts back.

You said the following.I believe Guv to be right but i think what needs to be
mentioned is that if the other variable in the problem is
not a function of x then it can be treated as a constant.

For example:
y' = 2 + e^z
where z is not a function of xIn the above, z must either be a constant or have a functional relationship with x or y or both. If the above equation is to have a meaning, one of the following must be defined as valid. z is a constant, leading to y = (2 + e^z) * x y = F[ x ] and z = G [ x ] y = F[ x ] and z = F[ y ] y = F[ x ] and y = G[ z ] F[ x, y, z ] = 0I may have missed a possibility in the above. Except for z being a constant, all of the above imply a functional relationship between x and z. I think that all of the above are variations on F[ x, y, z ] = 0, with some of the forms indicating that one variable can be eliminated by substitution.

Given any of the above, I do not think there is enough data to solve the equation, except when z is a constant.