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As per my knowledge and current understanding of differential equations, we may solve these kind of equations by solving the homogeneous part guessing the non-homogeneous part of the solution. I get the general solution for the homogeneous part, but when I try to make an ansatz for the non-homogeneous part to calculate the particular solution, I fail to eliminate the constant. So, how do I eliminate the constant, and get to the solution of the equation?

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Write $y''+y=Cy^{-3}$ and let $y'=u$ then $y''=uu'$ with independent variable $y$, then $2uu'=2Cy^{-3}-2y$ concludes $$u^2=-\dfrac{C}{y^2}-y^2+C_0$$ thus $$y'=\dfrac{\sqrt{-C-y^4+C_0y^2}}{y}$$ or $$\dfrac{y}{\sqrt{-C-y^4+C_0y^2}}dy=dx$$ gives us the general solution $\color{blue}{\dfrac12\arcsin\dfrac{2y^2-C_0}{\sqrt{C_0^2-4C}}=x+C_1}$.

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