
Yefim S. answered 07/06/20
Math Tutor with Experience
This is Homojinius equation. Substitution y= xv, where v is new function of x. Then dy/dx = y' = v + xv'/
We put this in our equation: v + xv' = (3x2 + 2x2v2)/(x2v); v + xv' = (3 + 2v2)/v; xv' = 3/v + v;
xdv/dx = (3 + v2)/v; Now we can separate variables: v/(v2 + 3)dv = 1/xdx.
We takes integrals now: ∫v/(v2 + 3)dv = ∫1/xdx; 1/2ln(v2 + 3) = ln(absx) + 1/2lnC; ln(v2 + 3) = 2ln(absx) + LnC;
v2 + 3 = Cx2, v2 = Cx2 - 3; y2/x2 = Cx2 - 3, y2 = Cx4 - 3x2 and y = ± (Cx4 - 3x2)1/2.
Answer: general solution of this ODE: y = ± (Cx4 - 3x2)1/2.