Question

求解微分方程(y^2-1)dx+(y^2-y+2x)dy=0 具体过程～速度啊 急

Answer 1

求解微分方程(y²-1)dx+(y²-y+2x)dy=0
解：P=y²-1；Q=y²-y+2x；∂P/∂y=2y≠∂Q/∂x=2，故不是全微分方程；
下面求积分因子：由于(∂Q/∂x-∂P/∂y)/P=(2-2y)/(y²-1)=-2(y-1)/(y+1)(y-1)=-2/(y+1)=f(y)
故积分因子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
用积分因子1/(y+1)²乘原式两边得;
[(y-1)/(y+1)]dx+[(y²-y+2x)/(y+1)²]dy=0.............(1)
此时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，则(1)的解为：
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为y-3ln(y+1)+(xy-x-2)/(y+1)=C

Answer 2

此题前后两部分分别是对x求导和y求导然后相加，前半部分求导后为0，后半部分为2y-1,所以为0+2y-1=0 y=0.5