(cosx+1/y)dx+(1/y-x/y^2)dy=0
(cosx+1/y)dx+(1/y-x/y^2)dy=0
设y=(1-x)³cosx,求dy/dx
求dy/dx=x/y+(cosx/y)^2通解
设y=1+cosx / 1-cosx,求dy / dx
dy/dx-y=cosx
dy/dx=y^2cosx
已知y=x^cosx/2,求dy/dx,
解, Dy/Dx + y = x , y(0) = 1
dy/dx=x(1+y^2)/y通解
dx/dy=x/y+[cos(x/y)]∧2,y(0)=1
y(x + y + 1) dx + (x + 2y) dy = 0:运用正合方程式求解
从(dx)/(dy)=1/y '导出:(d^2x)/(dy^2)=-y''/(y')^3