ODE problem $(1+y)y'=y$ without using Lambert W function Stuck on solving this ODE problem..
I can get answer by using Lambert W function
$$(1+y)y'=y$$
$$\left(\frac{1}{y}+1\right)dy=1dt$$
$$ln(y)+y=t+c$$
$$e^{ln(y)}=e^{t+c-y}$$
$$y=e^{t+c-y}$$
$$ye^y=e^{t+c}$$
$$y=W\left(\frac{1}{e^{-t-c}}\right)$$
But here is the question..
"Is it possible to solve without using Lambert W function?"
If so, how can I solve it and what I`m missing to do??
 A: The line
$$
\ln()+=+
$$
is your answer.  DE textbooks call this an "implicit solution".
As you note, the "explicit solution" involves the Lambert W function.
This can easily happen when solving a DE:  An implicit solution that cannot be solved in closed form to get an explicit solution.
A: Making $y = \sum_{k=0}^n a_k x^k$ and substituting into the DE we have
$$
\left(1+\sum_{k=0}^n a_k x^k\right)\left(\sum_{k=1}^n k a_k x^{k-1}\right)-\sum_{k=0}^n a_k x^k=0
$$
and after equating to zero the coefficients for the powers of $x$ we have a relationship between the $a_k$'s which solved, furnishes an approximation for the series solution. So for $n=4$ we have
$$
\left\{
\begin{array}{rcl}
 a_1&=&\frac{a_0}{a_0+1} \\
 a_2&=&\frac{a_0}{2 \left(a_0+1\right){}^3} \\
 a_3&=&\frac{a_0-2 a_0^2}{6 \left(a_0+1\right){}^5} \\
 a_4&=&\frac{a_0 \left(6 a_0^2-8 a_0+1\right)}{24 \left(a_0+1\right){}^7} \\
\end{array}
\right.
$$
The solution for the DE can be obtained in terms of the Lambert function as
$$
y = W(e^{x+C})
$$
now choosing $y(0) = 1$ we have $C = 1$ which in the series representation is equivalent to $a_0 = 1$ and the series reads
$$
y_4 = 1+\frac x2+\frac{x^2}{16}-\frac{x^3}{192}-\frac{x^4}{3072}+O(x^5)
$$
Note that this is a kind of alternating series which has convergence problems for big $|x|$. Follows a comparison between the closed solution in blue and in red the approximation.

