验证参数方程{x=e^t*sint y=e^t*cost 所确定的函数满足关系式(d^2y/dx^2)*(x+y)^2=

问题描述:

验证参数方程{x=e^t*sint y=e^t*cost 所确定的函数满足关系式(d^2y/dx^2)*(x+y)^2=2(x*dy/dx-y)
这两个式子我怎么算都不相等
1个回答 分类:数学 2014-10-27

问题解答:

我来补答
x=e^t*sint y=e^t*cost
所以dx/dt=e^t*(sint +cost) ,dy/dt=e^t*(cost-sint)
故dy/dx=(dy/dt) / (dx/dt)= (cost-sint) / (sint +cost)

d^2y/dx^2
=d(dy/dx) /dt * dt/dx
=d[(cost-sint) / (sint +cost)] /dt * dt/dx
= [(-sint-cost)*(sint+cost) -(cost-sint)*(cost-sint)] /(sint+cost)^2 * 1/[e^t*(sint +cost)]
= (-1-2sint*cost -1+2sint*cost)/[e^t*(sint+cost)^3]
= -2 / [e^t*(sint+cost)^3]
所以
(d^2y/dx^2)*(x+y)^2
= -2 / [e^t*(sint+cost)^3] * (e^t*sint +e^t*cost)^2
= -2e^t /(sint+cost)

2(x*dy/dx-y)
=2[e^t*sint * (cost-sint) / (sint +cost) - e^t*cost]
=2e^t *[sint * (cost-sint) -cost*(sint +cost)] /(sint +cost)
=2e^t *[sint*cost -(sint)^2 -cost*sint -(cost)^2] / (sint +cost)
= -2e^t /(sint+cost)

(d^2y/dx^2)*(x+y)^2 =2(x*dy/dx-y) = -2e^t /(sint+cost)
所以这两个式子是相等的
 
 
展开全文阅读
剩余:2000