#xy^2*(dy/dx)=y^3-x^3, y(1)=2# (its a initial value problem) Any kind soul here who can give the solution of this eqn?

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Answer 1 · 2017-12-16T22:36:05

#y = x root(3)(1/3(C_0-log_ex))#

Making the variable change

#y = lambda x rArr y' = x lambda'+lambda# we have

#x^3lambda^2(x lambda'+lambda)=x^3lambda^3-x^3# or

#lambda' lambda^2x+1=0#

This is a separable differential equation. Grouping variables

#lambda^2d lambda=-dx/x# and integrating

#1/3lambda^3=-log_e x + C_0# or

#lambda = root(3)(1/3(C_0-log_ex))# or

#y = x root(3)(1/3(C_0-log_ex))#

but #y(1)=2=1 xx root(3)(C_0/3) rArr C_0 = 24#