- #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##.

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: