How you solve this? #x^2y'+x^2y^2-3xy+3=0# with solution #y_1=3/x#

1 Answer
Apr 11, 2018

Answer:

#y = (C_1+3C_2x^2)/(C_1 x+ C_2 x^3)#

Explanation:

Making the substitution

#y = (xi')/xi# we have after substitution

#x^2xi''-3x xi'+3xi = 0#

this is linear differential equation with solution

#xi(x) = C_1 x+ C_2 x^3# then

#y = (C_1+3C_2x^2)/(C_1 x+ C_2 x^3)#