Differential equations

  • Thread starter wel
  • Start date
  • #1
wel
Gold Member
36
0
Simple harmonic motion: ##y'= -z,~z'= f(y)##


the modified explicit equation are


$$y'=-z+\frac {1}{2} hf(y)$$


$$y'=f(y)+\frac {1}{2} hf_y z$$
and deduce that the coresponding approximate solution lie on the family of curves
$$2F(y)-hf(y)y+z^2=\textrm{constant}$$


where ##f_y= f(y)##.
What are the curves when verify ##f(y)=y##


=>


for the solution of the system lie on the family of curves, i was thinking


$$\frac{d}{dt}[F(y)^2+z^2]= y \frac{dy}{dt} + z \frac{dz}{dt}$$


but i am not sure


if ##f(y)=y##, then the differential equation is ##y'' + y =0##, meaning that ##y=A \cos x +B \sin x## and ##z=-y'= - A \sin x +B \cos x##.
 
Last edited by a moderator:

Answers and Replies

  • #2
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,129
3,298
Dear wel,

This thread involves derivatives and diff eq. Thus it belongs in the Calculus & Beyond homework and not precalculus. Please post there in the future.
 

Related Threads on Differential equations

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
702
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
862
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
Top