Solve (2x^2+y)dx+(x^2y-x)dy=0

Answer 1

Let f(x) is integrated factor; then f(x)(2x² + y)dx + f(x)(x²y - x)dy = 0;

and ∂/∂y[f(x)(2x²+ y)] = ∂/∂x[f(x)(x²y - x)]; f(x) = f'(x)(x²y - x) +f(x)(2xy - 1); 0 = f'(x)x(xy - 1) + 2f(x)(xy - 1).

From here f'(x)x + 2f(x) = 0. f'(x)/f(x) = - 2/x; ∫d(f(x))/f(x)) = -∫2dx/x; ln(f(x)) = -2lnx; f(x) = x ^{- 2} = 1/x^·2.

We get exact equation: (2 + y/x²)dx + (y - 1/x)dy = 0.

Looking for solution F(x,y); ∂F/∂x = 2 + y/x²; F = ∫(2 + y/x²)dx = 2x - y/x + φ(y)

∂F/∂y = - 1/x + φ'(y) = y - 1/x; φ'(y) = y; φ(y) = ∫ydy = y²/2 + C

So, F(x,y) = 2x - y/x + y²/2 + C