Verified answer

QA:

y'+p(x)y=q(x)y^n

Divide y^n to both sides, then y^(-n)y'+ p(x)y^(1-n)=q(x) -----(A)

Set u=y^(1-n), then du/dx=(1-n)y^(-n) (dy/dx)

thus y^(-n) (dy/dx)=(du/dx)/(1-n).

Putting into (A), then (du/dx)/(1-n)+ p(x) u= q(x)

or u'+(1-n)p(x) u= (1-n)q(x) is a linear diff. equation.

QB:

y'+y=xy^3 then y^(-3)y'+y^(-2)=x

Set u=y^(-2), then u'=-2y^(-3) y'

thus -0.5 u'+ u=x or u'-2u= -2x

(d/dx)[exp(-2x)u ]= -2x exp(-2x)

exp(-2x) u= x exp(-2x)+ 0.5exp(-2x) +C

u= x+ 0.5+ C exp(2x)

hence, y^(-2)=x+0.5+C exp(2x)