Answer to Question #251767 in Differential Equations for Tunde

Question

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

Answer 1

$$M(x, y)=2y(x^2-y+x), M_y=2x^2-4y+2x$$

$$N(x, y)=x^2-2y, N_x=2x$$

$$\dfrac{M_y-N_x}{N}=\dfrac{2x^2-4y+2x-2x}{x^2-2y}=2$$

$$\mu(x)=e^{\int 2dx}=e^{2x}$$

$$2e^{2x}y(x^2 - y + x) dx +e^{2x} (x^2 - 2y) dy = 0$$

$$M_y=2e^{2x}(x^2-2y+x)$$

$$N_x=e^{2x}(2(x^2-2y)+2x)$$

$$M_y=2e^{2x}(x^2-2y+x)=N_y$$

$$u(x, y)=\int e^{2x} (x^2 - 2y)dy+g(x)$$

$$=e^{2x}x^2y-e^{2x}y^2+g(x)$$

$$\dfrac{\partial u}{\partial x}=2e^{2x}x^2y+2e^{2x}xy-2e^{2x}y^2+g'(x)$$

$$=2e^{2x}y(x^2 - y + x)$$

$$g'(x)=0=>g(x)=-C$$

$$e^{2x}x^2y-e^{2x}y^2-C=0$$

$$e^{2x}x^2y-e^{2x}y^2=C$$