大學程度..first- order differential equation..難...

Denote M=3x^2+y and N = 2yx^2-x
then M_y = dM/dy = 1, N_x = dN/dx = 4yx-1
(M_y - N_x) / N = - 2/x

Integrating factor = exp( int(-2/x) dx) = 1/x^2

New D.E.:

(y/x^2 + 3) dx + (2y- 1/x) dy = 0

f(x,y) = int( y/x^2 + 3) dx = - y/x +3x +C(y)
df/dy = 2y - 1/x
-1/x + C'(y) = 2y - 1/x
C'(y) = 2y
C(y) = y^2 +C
so f = -y/x +3x +y^2 +C = k
-y/x + 3x + y^2 = C #