ODE problem $(1+y)y'=y$ without using Lambert W function

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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??

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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.

enter image description here

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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.