# Find the solution to this Differential Equation

1. Aug 23, 2014

### rodolfomz

1. The problem statement, all variables and given/known data

Find the solution to $\frac{dy}{dx} = yLn(y + 1)$

2. Relevant equations

$\frac{dy}{dx} = yLn(y + 1)$

3. The attempt at a solution

$\frac{dy}{dx} = yLn(y + 1)$

$\frac{dy}{yLn(y + 1)} = dx$

but then i cant integrate, any help?

2. Aug 23, 2014

### Staff: Mentor

Using a substitution (u = ln(y + 1)) I was able to get to this integral:
$$\int \frac{du}{u} + \int \frac{du}{u(e^u + 1)}$$
The first integral is very simple. The second might be amenable to another substitution, but I didn't go any further.

3. Aug 23, 2014

### rodolfomz

thats what I have

$∫\frac {dy}{y*ln(y+1)}=∫dx$
changing variables
$u=ln(y+1) ; du=\frac{dy}{y+1}$

$\int \frac{e^u*du}{(u)*(e^u -1)} = x$
which is the same as

$\int \frac{du}{u}*\frac{e^u}{e^u-1} = x$

and then resolving by parts dv= du/u and u=e^u/e^u-1

$\frac{e^u}{e^u-1}+\int \frac{ln(u)*e^udu)}{(e^u-1)^2}$

im stuck here

4. Aug 23, 2014

### LCKurtz

I doubt there is an elementary antiderivative. Where did this problem come from?