Find the equilibrium solutions of the ordinary differential equationy' = x^2 cos^3yWhich of these solutions are stable?Note: The cosy is raised to the power of 3 (cubed), like the following: (cosy)^3

Answer:[tex]\dfrac{1}{2}\dfrac{sin y}{cos^2y}+\dfrac{1}{4}ln[\dfrac{siny +1}{siny-1}]=\dfrac{x^3}{3}+c[/tex]  Step-by-step explanation:given,y' =  x² (cos y)³solve the equation using variable separable method[tex]\frac{\mathrm{d} y}{\mathrm{d} x} = x^2 cos^3y\\\dfrac{dy}{(cosy)^3}= x^2 dx\\\dfrac{cos\ y}{cos^4 y}\ dy = x^2 dx\\\int \dfrac{cos\ y}{(1-sin^2 y)^2}\ dy = \int x^2dx\\\int \dfrac{1}{(t^2-1)^2}\ dt = \dfrac{x^3}{3}+c[/tex]here sin y  = t     :  cos y = dt[tex]\int(\dfrac{1}{2}{[\dfrac{1}{t-1}-\dfrac{1}{t+1}]}^2 = \dfrac{x^3}{3}+c\\\dfrac{1}{4}\int [\dfrac{1}{(t-1)^2}-\dfrac{1}{t-1}+\dfrac{1}{t-1}+\dfrac{1}{(t+1)^2}]= \dfrac{x^3}{3}+c[/tex]  [tex]\dfrac{1}{2}\dfrac{sin y}{cos^2y}+\dfrac{1}{4}ln[\dfrac{siny +1}{siny-1}]=\dfrac{x^3}{3}+c[/tex]