![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
常微分方程组求解 255
y^3=[(B/A)x^3]+[(C/D)z^3]+Ex=y=z=x0时,能求出E值。x=0时,y与z不等于0...
y^3=[(B/A)x^3]+[(C/D)z^3]+Ex=y=z=x0时,能求出E值。x=0时,y与z不等于0
展开
展开全部
①x'=A*[1/(y-x)]*(y/x)
2xx'=2Ay/(y-x)
(x^2)'=2Ay/(y-x)
(B/A)*(x^2)'=2By/(y-x)
②z'=D*[1/(z-y)]*(y/z)
2zz'=2Dy/(z-y)
(z^2)'=2Dy/(z-y)
(C/D)*(z^2)'=2Cy/(z-y)
③y'=B*[1/(y-x)]*(x/y)+C*[1/(z-y)]*(z/y)
2yy'=2Bx/(y-x)+2Cz/(z-y)
(y^2)'=2By/(y-x)-2B+2Cy/(z-y)+2C
③-①-②,(y^2)'-(B/A)*(x^2)'-(C/D)*(z^2)'=2C-2B
[y^2-(B/A)*x^2-(C/D)*z^2]'=2C-2B
y^2-(B/A)*x^2-(C/D)*z^2=(2C-2B)*t+E,其中E是任意常数
当t=0时,x=y=z=x0,则E=(1-B/A-C/D)*x0^2
所以y^2-(B/A)*x^2-(C/D)*z^2=(2C-2B)*t+(1-B/A-C/D)*x0^2
两边同乘以AD
ADy^2-BDx^2-ACz^2=(2ACD-2ABD)*t+(AD-BD-AC)*x0^2
2xx'=2Ay/(y-x)
(x^2)'=2Ay/(y-x)
(B/A)*(x^2)'=2By/(y-x)
②z'=D*[1/(z-y)]*(y/z)
2zz'=2Dy/(z-y)
(z^2)'=2Dy/(z-y)
(C/D)*(z^2)'=2Cy/(z-y)
③y'=B*[1/(y-x)]*(x/y)+C*[1/(z-y)]*(z/y)
2yy'=2Bx/(y-x)+2Cz/(z-y)
(y^2)'=2By/(y-x)-2B+2Cy/(z-y)+2C
③-①-②,(y^2)'-(B/A)*(x^2)'-(C/D)*(z^2)'=2C-2B
[y^2-(B/A)*x^2-(C/D)*z^2]'=2C-2B
y^2-(B/A)*x^2-(C/D)*z^2=(2C-2B)*t+E,其中E是任意常数
当t=0时,x=y=z=x0,则E=(1-B/A-C/D)*x0^2
所以y^2-(B/A)*x^2-(C/D)*z^2=(2C-2B)*t+(1-B/A-C/D)*x0^2
两边同乘以AD
ADy^2-BDx^2-ACz^2=(2ACD-2ABD)*t+(AD-BD-AC)*x0^2
追问
能求出x,y,z各自关于t的表达式吗
追答
求不出的,因为你这个方程组不是线性的
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询