求解微分方程(y²-1)dx+(y²-y+2x)dy=0
P=y²-1；Q=y²-y+2x；∂P/∂y=2y≠∂Q/∂x=2,故不是全微分方程；
下面求積分因子：由于(∂Q/∂x-∂P/∂y)/P=(2-2y)/(y²-1)=-2(y-1)/(y+1)(y-1)=-2/(y+1)=f(y)
故積分因子M(x,y)=e^[∫f(y)dy]=e^[-2∫dy/(y+1)]=e^[-2ln(y+1)]=e^[ln(y+1)ֿ²]=1/(y+1)²
用積分因子1/(y+1)²乘原式兩邊得;
[(y-1)/(y+1)]dx+[(y²-y+2x)/(y+1)²]dy=0.(1)
此時(shí)P=(y-1)/(y+1)；Q=(y²-y+2x)/(y+1)²；
由于∂P/∂y=[(y+1)-(y-1)]/(y+1)²=2/(y+1)²；∂Q/∂x=2/(y+1)²；即有∂P/∂y=∂Q/∂x=2/(y+1)²,
故(1)是全微分方程.取xo=0,yo=0,則(1)的解為：
u(x,y)=[0,x]∫[(y-1)/(y+1)]dx+[0,y]∫[(y²-y)/(y+1)²]dy
=(y-1)x/(y+1)+[0,y]∫[1-3/(y+1)+2/(y+1)²]dy
=(y-1)x/(y+1)+y-3ln(y+1)-2/(y+1)
=y-3ln(y+1)+(xy-x-2)/(y+1)
于是得原方程的通解為y-3ln(y+1)+(xy-x-2)/(y+1)=C