$$y'+\lambda \ y^4 +y=0$$ Where $\lambda$ is very small and $y(1)=-0.5$.
I've tried to solve it by Substituting: $y=y_0+\lambda y_1 +\lambda^2 y_2+\cdots$
but I had problems with the integration constants that appeared after solving the other partial differential equation when substituting: $y=y_0+\lambda y_1 +\lambda^2 y_2+\cdots$ in the main equation.
Just solve it directly by rearrangement. You need to evaluate the integral $\int_1^y \frac{dz}{\lambda z^4+z}$. Notice that:
$$\frac{1}{\lambda z^4+z}=\frac{1}{z}-\frac{\lambda z^2}{1+\lambda z^3},$$
which you hopefully know how to integrate.