全微分方程的通解（B028）

问题

已知，$M(x, y)$ $\mathrm{d} x$ $+$ $N(x, y)$ $\mathrm{d} y$ $=$ $0$ 为全微分方程：$M(x, y)$ $\mathrm{d} x$ $+$ $N(x, y)$ $\mathrm{d} y$ $=$ $0$ $\Leftrightarrow$ $\frac{\partial M}{\partial y}$ $=$ $\frac{\partial N}{\partial x}$

则，该全微分方程的通解 $?$

选项

[A].   $u(x, y)$ $=$ $\int_{x_{0}}^{x}$ $M(x, y)$ $\mathrm{d} x$ $-$ $\int_{y_{0}}^{y}$ $N(x, y)$ $\mathrm{d} y$ $=$ $C$

[B].   $u(x, y)$ $=$ $\int_{x_{0}}^{x}$ $M(x, y)$ $\mathrm{d} x$ $+$ $\int_{y_{0}}^{y}$ $N(x, y)$ $\mathrm{d} y$ $=$ $C$

[C].   $u(x, y)$ $=$ $\int_{0}^{x}$ $M(x, y)$ $\mathrm{d} x$ $+$ $\int_{0}^{y}$ $N(x, y)$ $\mathrm{d} y$ $=$ $C$

[D].   $u(x, y)$ $=$ $\int_{x}^{x_{0}}$ $M(x, y)$ $\mathrm{d} x$ $+$ $\int_{y}^{y_{0}}$ $N(x, y)$ $\mathrm{d} y$ $=$ $C$

   答 案   

$u(x, y)$ $=$ $\int_{x_{0}}^{x}$ $M(x, y)$ $\mathrm{d} x$ $+$ $\int_{y_{0}}^{y}$ $N(x, y)$ $\mathrm{d} y$ $=$ $C$