Find X in Y=X+e^X: Log It & Solve

[Gelsamel Epsilon]

How do I find X with respect to Y if;

Y = X + e^X?

If I log it then I get Log Y = Log X + X, if this is the correct step how do I isolise X from here?

 

[z-component]

Taking the log of the right-hand side is not log x + x, but instead log(x + e^x).

 

[Gelsamel Epsilon]

****, stupid mistake of mine :-).

I still can't see that helping me at all though :-(.

 

[quasar987]

It's not always possible to explicitely invert a function.

 

[Gelsamel Epsilon]

So I can implicitly get one?

y' = 1+e^x
y'' = e^x

y - y'' = x?

 

[quasar987]

The equation that implicitly defines X as a function of Y is the original equation. Y = X + e^X. The fact that Y(X) is in bijection with its image assures you that X(Y) exists.

It's not the case with say Y(X)=X² because X(Y) would be 2-valued: ±\sqrt{Y}.

But your idea with differentiating bth sides was good. However, I think it's more interesting to acknowledge X(Y) exists, and then differentiate the equation wrt Y! I get

1=\frac{dX}{dY}+\frac{dX}{dY}e^{X(Y)}

\frac{dX}{dY}=\frac{1}{1+e^{X(Y)}}

This way you have a differential equation for X(Y), or inversely, an integral equation.

X(Y)=\int\frac{1}{1+e^{X(Y)}}dY

I don't know if something can be done with those, but there's probably a way to at least extract some information or approximation. For instance, we know from Y(X) that when Y is around 1, Y is small, so you can keep only the first few terms in the series of exp(X(Y)) and integrate to get an approximation of the form of X(Y) in the nbhd of 1.

 

[matt grime]

How do I find X with respect to Y if

what on Earth does that even mean?

 

[Gelsamel Epsilon]

I hope this doesn't get into semantics.

 

[quasar987]

May we know where this problem of yours comes from?P.S.This is nonsense, don't pay attention:

For instance, we know from Y(X) that when Y is around 1, Y is small, so you can keep only the first few terms in the series of exp(X(Y)) and integrate to get an approximation of the form of X(Y) in the nbhd of 1.

 

[Gelsamel Epsilon]

Got to choose from 5 different questions to do animation type exercises in Maple. Chose number 5 but we had problems restricting the domain in the 2nd function to the time in the first function. The first function was in the form of y = t +e^t. We decided it was too hard and started on a different, easier, question. But it still bugged me that we couldn't get a function for t(y). Obviously we weren't supposed to do it like that, good idea we chose an easier one then eh? :-).

 

[Gib Z]

y=x + e^x
dy/dx=1+e^x
dx/dy = 1/(1+e^x)

 

[uart]

1. Subt z=e^x to get : y = z + ln(z)

2. Exponential each side to get : e^y = z e^z

3. From the definition of the LambertW function : z = LambertW( e^y)

4. So x = ln( LambertW( e^y ) )