# Differential equations

1. Apr 11, 2014

### wel

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: Apr 11, 2014
2. Apr 11, 2014

### micromass

Staff Emeritus
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.