Your differential equation is \(\displaystyle \frac{d^2y}{dx^2}- x- 2x^3= 0\) and you are asked to "sketch the paths in the xy-plane"?
I would just write the equation as \(\displaystyle \frac{d^2y}{dx^2}= 2x^3+ x\) and integrate twice:
\(\displaystyle \frac{dy}{dx}= \frac{1}{2}x^4+ \frac{1}{2}x^2+ C\)
\(\displaystyle y(x)= \frac{1}{10}x^5+\frac{1}{6}x^3+ Cx+ D\)
Each choice of C and D gives a different curve.