Question

1.求解方程 (dy)/(dx)=(4x-y+1)/(x-2y+3)

Answer 1

解:微分方程为dy/dx=(4x-y+1)/(x-2y+3)，化为(4x-y+1)dx-(x-2y+3)dy=0，(4x-y+1)dx+(2y-x-3)dy=0
∵∂(4x-y+1)/∂y=-1，∂(2y-x-3)/∂x=-1
∴有2x²-yx+x+f(y)=y²-xy-3y+g(x)，则
g(x)=2x²+x，f(y)=y²-3y，微分方程的通解为2x²-yx+x+y²-3y=c(c为任意常数)

Answer 2

联立解 4x-y+1 = 0， x-2y+3 = 0， 得交点 (1/7, 11/7)
将坐标原点平移至交点处，即令 x = X - 1/7， y = Y - 11/7
则原微分方程可化为齐次方程 dY/dX = (4X-Y)/(X-2Y)
令 Y = XU， 则 U+XdU/dX = (4-U)/(1-2U)
XdU/dX = (4-U)/(1-2U)-U = -2(U^2-U+2)/(2U-1)
(2U-1)dU/(U^2-U_2) = -2dX/X
ln(U^2-U+2) = -ln(X^2) + lnC
(U^2-U+2)(X^2) = C, Y^2-XY+2X^2 = C,
通解 (y+11/7)^2 - (x+1/7)(y+11/7) + 2(x+1/7)^2 = C