# Homework Help: How to solve the differential equation e^y-1/y'=1 ?

1. Jun 22, 2010

### NegativeGPA

Does anyone know how to solve the differential equation

e^y-1/y'=1

?

2. Jun 22, 2010

### Staff: Mentor

Re: e^y-1/y'=1

Welcome to the PF. Could you please use parens or Latex to make the equation clearer for everybody? Also, you are required to show a little work on the problem before we can help you (see the PF Rules link at the top of the page). Do you have an idea of what first steps you should take to work toward a solution?

Is it: $$e^y -\frac{1}{\frac{dy}{dt}} = 1$$ ?

3. Jun 22, 2010

### NegativeGPA

Re: e^y-1/y'=1

My original problem was trying to find an inverse of y=e^x+x and so I differentiated it and got that equation. So far the closest I've gotten is y=ln(x-1) but then i realized that gave me 1=0, so that obviously doesn't work.

I started trying plugging different values in to make some directional fields but it's not doing much towards giving me a solid answer, and I haven't learned more than the basic techniques to solve differential equations, and none of them apply to this, so I was seeing if anyone else had any ideas.

4. Jun 22, 2010

### Dickfore

Re: e^y-1/y'=1

Standard separation of variables should do the trick. The problem is, you arrive at a solution which cannot be solved explicitly in terms of y, just as your original problem was unsolvable (in terms of elementary functions) for x.

5. Jun 22, 2010

### HallsofIvy

Re: e^y-1/y'=1

Multiplying through by y', $y'e^y- 1= y'$ so $y'(e^y- 1)= 1$ and that results in $(e^y- 1)dy= dx$.

That's not at all hard to integrate.

6. Jun 23, 2010

### Mute

Re: e^y-1/y'=1

Differentiating your equation and then trying to integrate it will just give you back your original equation:

$$x = e^y + y \Rightarrow 1 = (e^y +1)\frac{dy}{dx}$$

which you rearranged into $1/y'(x) -e^{y(x)} = 1$. (or close to it - it looks like there's a sign mistake). Solving this equation, however, will just give you back your original equation $x = e^y + y$.

To solve this in terms of a known function that you can use in Mathematica, exponentiate both sides:

$$e^x = e^y e^{e^y}$$

This is now in a form which defines the Lambert W function: $z = W(z)e^{W(z)}$. Hence,

$$e^{y(x)} = W(e^x) \Rightarrow y(x) = \ln W(e^x)$$

From the definition of the Lambert W function, we can also write $\ln z = \ln W(z) + W(z)$, so using this we may write the solution as

$$y(x) = \ln e^x - W(e^x) = x - W(e^x)$$

A note on the Lambert W function: this is a multivalued function, so there are really infinitely many branches, denoted by n: $W_n(z)$, where z can be complex. The principal branch, n = 0, and the branch n = -1 are real valued for real valued inputs.

Last edited: Jun 23, 2010