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Find the equilibrium solutions of the ordinary differential equation

y' = x^2 cos^3y

Which of these solutions are stable?

Note: The cosy is raised to the power of 3 (cubed), like the following: (cosy)^3

1 Answer

3 votes

Answer:


(1)/(2)(sin y)/(cos^2y)+(1)/(4)ln[(siny +1)/(siny-1)]=(x^3)/(3)+c

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\\(dy)/((cosy)^3)= x^2 dx\\(cos\ y)/(cos^4 y)\ dy = x^2 dx\\\int (cos\ y)/((1-sin^2 y)^2)\ dy = \int x^2dx\\\int (1)/((t^2-1)^2)\ dt = (x^3)/(3)+c

here sin y = t : cos y = dt


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


(1)/(2)(sin y)/(cos^2y)+(1)/(4)ln[(siny +1)/(siny-1)]=(x^3)/(3)+c

User RagAnt
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