x(y^2-1)dx+y(x^2-1)dy=0所有常数解是
x(y^2-1)dx+y(x^2-1)dy=0所有常数解是
方程x(y^2-1)dx+y(x^2-1)dy=0的常数解是y=±1,x=±1.是否正确?
dy/dx,y=(1+x+x^2)e^x
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
dy/dx-2y/(x+1)=(x+1)^2
dy/dx-2y/(1+x)=(x+1)^3
dy/dx=(e^x+x)(1+y^2)通解
dy/dx-y/x=x^2
从(dx)/(dy)=1/y '导出:(d^2x)/(dy^2)=-y''/(y')^3