Solve the initial value problem y′=x^2−y^2/xy with y(2)=2.

Answer 1

Let y = vx, where v is v(x). Then y' v'x + v.

v'x + v = (x² - v²x²)/(vx²); v'x = (1 - v²/v - v; v'x = (1 - 2v²)/v; ∫vdv/(1 - 2v²) = ∫dx/x; - 1/4lnI1 - 2v²I = lnIxI - 1/4lnC; lnI1 - 2v²I = - 4lnIxI + lnC; I2v² - 1I = Cx^-4. v = y/x; I2y²/x² - 1I = Cx^-4; I2·4/2 - 1I = C·2^-4; C = 48;

I2y²/x² - 1I = 48x^-4