Question

Solve the given bernoulli equation

dy/dx=y(xy^3 -1)

Answer 1

dy/dx = y(xy³-1)=> dy/dx = xy⁴-y=> dy/dx + y = xy⁴ Dividing the above equation by y³ to get -

y^-4 (dy/dx) + y^-3 = x ---- A

Put y^-3 = t ,then -3y^-4 (dy/dx) =(dt/dx)
=> y^-4(dy/dx) = (-1/3) (dt/dx)
Substituting above in equation A to get -(-1/3) (dt/dx) + t = x Multiplying both sides by -3=> dt/dx +(-3)t = -3xThe above resides as a linear first order equation of the form-t^'+P(x)t=Q(x), where P(x)=-3 and Q(x)=-3xSo, the integrating factor for above bernoulli equation is - M(x)=exp( $\int$-3 dx) = e^-3x.Multiplying this by both sides of the equation -
e^-3x t^' + (-3)t e^-3x =-3x e^-3xAs left hand side represents d/dx(t e^-3x) :d/dx(t e^-3x) = -3xe^-3x Integrating both sides with respect to x :$\int$ d/dx(t e^-3x) dx = $\int$ -3xe^-3x dx=> t e^-3x = (-3) [ x$\int$ e^-3x dx - $\int$ {{(d/dx) x} $\int$ e^-3x dx} ]$\implies$t e^-3x = (-3) [ (-1/3) x e^-3x - $\int$ 1. (-1/3) e^-3x dx ]$\implies$t e^-3x = x e^-3x + (-1/3) e^-3x $\implies$t = x -(1/3) Put t = y^-3 to get - y^-3 = x - (1/3)