積分因子を1変数函数と仮定したときの導出

導出

$ \exist\phi:\mathrm d\phi=\lambda(x)a_x(x,y)\mathrm dx+\lambda(x)a_y(x,y)\mathrm dy

$ \iff\frac{\partial\lambda(x)a_x(x,y)}{\partial y}=\frac{\partial\lambda(x)a_y(x,y)}{\partial x}

$ \iff\lambda(x)\frac{\partial a_x(x,y)}{\partial y}=\frac{\partial\lambda(x)}{\partial x}a_y(x,y)+\lambda(x)\frac{\partial a_y(x,y)}{\partial x}

$ \iff a_y(x,y)\lambda '(x)=\left(\frac{\partial a_x(x,y)}{\partial y}-\frac{\partial a_y(x,y)}{\partial x}\right)\lambda(x)

$ \underline{\iff\lambda(x)=\exp\left(\frac1{a_y(x,y)}\frac{\partial a_x(x,y)}{\partial y}-\frac1{a_y(x,y)}\frac{\partial a_y(x,y)}{\partial x}\right)\quad}_\blacksquare