求初值:dy/dx+y/x=x+1/x,当x=2,y=3
求初值:dy/dx+y/x=x+1/x,当x=2,y=3
求微分方程初值y^3d^2y/dx^2+1=0,y|x=1=1,dy/dx|x=1=0
求dy/dx 当y=2^(x^2)
设y=(x/1-x)^x,求dy/dx
求dy/dx=(x-y+5)/(x+y-2)
请问用matlab怎么编程这个微分方程啊dy/dx=y-2*x/y,初值是y(0)=1.求y(1) ,急
dy/dx-y/x=x^2
x^2+xy+y^3=1,求dy/dx
matlab 龙格库塔法求解微分方程dy/dx)=y^2+x,初值为x(0)=5 ,y(0)=2
dy/dx,y=(1+x+x^2)e^x
dy/dx-2y/(1+x)=(x+1)^3
dy/dx=2y/x+3x/2y