Solve the ODE $$ y''+ y' = y^2$$ I tried to solve, but I don't know any algorithm except for numerical methods.
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$\begingroup$First, rewrite the ODE as $y''=-y'+y^2=F(y,y')$.
This type of ODE can be reduced to a first order ODE by the following trick. Define $y'(x)=v(y(x)).$ Differentiation of this expression (using the chain rule) with respect to $x$ results in:
$$y''(x)=\dfrac{dv}{dy}\dfrac{dy}{dx}=\dfrac{dv}{dy}y'(x)=\dfrac{dv}{dy}v(y)=\dfrac{dv}{dy}v.$$
Again, using the right-hand side of the ODE we obtain:
$$\dfrac{dv}{dy}v=-(y')+y^2=-(v)+y^2 \implies v+v\dfrac{dv}{dy}=y^2.$$
The resulting ODE is an Abel Differential equation of the second kind. I don't think that there is a closed form solution as Maple couldn't solve it. I also tried to determine the Lie Symmetries of this ODE with the symgen() function that is available in the Maple package DETools which only gave a translational symmetry. You could try other packages for Maple / Mathematica / Mupad / Sympy. There is also a paper by Ali Rostami (2015) that deals with these type of ODEs.
$\endgroup$ $\begingroup$Hint:) Let $y'=u$ and $y''=uu'$ now solve $uu'+u=y^2$.
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