Question

Find the equilibrium solutions of the ordinary differential equation

Answer 1

Answer:

$\dfrac{1}{2}\dfrac{sin y}{cos^2y}+\dfrac{1}{4}ln[\dfrac{siny +1}{siny-1}]=\dfrac{x^3}{3}+c$  

Step-by-step explanation:

given,

y' =  x² (cos y)³

solve the equation using variable separable method

$\frac{\mathrm{d} y}{\mathrm{d} x} = x^2 cos^3y\\\dfrac{dy}{(cosy)^3}= x^2 dx\\\dfrac{cos\ y}{cos^4 y}\ dy = x^2 dx\\\int \dfrac{cos\ y}{(1-sin^2 y)^2}\ dy = \int x^2dx\\\int \dfrac{1}{(t^2-1)^2}\ dt = \dfrac{x^3}{3}+c$

here sin y  = t     :  cos y = dt

$\int(\dfrac{1}{2}{[\dfrac{1}{t-1}-\dfrac{1}{t+1}]}^2 = \dfrac{x^3}{3}+c\\\dfrac{1}{4}\int [\dfrac{1}{(t-1)^2}-\dfrac{1}{t-1}+\dfrac{1}{t-1}+\dfrac{1}{(t+1)^2}]= \dfrac{x^3}{3}+c$  

$\dfrac{1}{2}\dfrac{sin y}{cos^2y}+\dfrac{1}{4}ln[\dfrac{siny +1}{siny-1}]=\dfrac{x^3}{3}+c$