Question

$xy''=y'+x^2$

Answer 1

xy''=y'+x^2

xy''-y'=x^2

(1/x)*y''+(-1/x^2)*y'=1

d((1/x)*y')=1dx

(1/x)*y'=x+C1

y'=x^2+C1*x

y=(1/3)x^3+C2*x^2+C3