# Solving 2 differential equations

• Mar 31st 2012, 04:51 AM
oxxiissiixxo
Solving 2 differential equations
dx/dt=ay and dy/dt=bx where x and y are function of t [x(t) and y(t)] and a and b are constant.

1) show what x and y satisfy the equation for a hyperbola: y^2-(b/a)*x^2=(y_0)^2-(b/a)*(x_0)^2

2) suppose at some time t_s, the point (x(t_s),y(t_s)) lies on the upper branch of hyperbola, show that: y(t_s)>sqrt(b/a)*x(t_s)

I dun know whether i am doing it right.

First, in integrate both equations,

dx/dt=ay >>> x/y+C_1=at+C_2 >>> x/y+C_5=at

dy/dt=bx >>> y/x+C_3=bt+C_4 >>> y/x+C_6=bt

then I say t = 0 and so

x/y+C_5=at >>> C_5=-x_0/y_0

y/x+C_6=bt >>> C_6=-y_0/x_0

then i say this happens only when C_5 and C_6 are 0

then going back to

x/y+C_5=at >>> x/y=at

y/x+C_6=bt >>> y/x=bt and isolating t to yield

y^2-(b/a)*x^2=0

and when t=0

y_0^2-(b/a)*x_0^2=0

so y^2-(b/a)*x^2=y_0^2-(b/a)*x_0^2

and can somebody give me some hints to deal with the second problem? thank you.
• Mar 31st 2012, 08:37 AM
HallsofIvy
Re: Solving 2 differential equations
Quote:

Originally Posted by oxxiissiixxo
dx/dt=ay and dy/dt=bx where x and y are function of t [x(t) and y(t)] and a and b are constant.

1) show what x and y satisfy the equation for a hyperbola: y^2-(b/a)*x^2=(y_0)^2-(b/a)*(x_0)^2

2) suppose at some time t_s, the point (x(t_s),y(t_s)) lies on the upper branch of hyperbola, show that: y(t_s)>sqrt(b/a)*x(t_s)

I dun know whether i am doing it right.

First, in integrate both equations,

dx/dt=ay >>> x/y+C_1=at+C_2 >>> x/y+C_5=at

dy/dt=bx >>> y/x+C_3=bt+C_4 >>> y/x+C_6=bt

No, this is not at all correct. You have integrated the first equation assuming that y is a constant and integrating the second equation assuming that x is constant and those assumptions are not true.

There are two standard ways to handling pairs of equations.
Simplest is to [b]differentiate[b] the first equation as second time with respect to t: d^2x/dt^2= a(dy/dt) and, from the second equation that becomes
d^2x/dt^2= a(bx) so you have the second degree equation d^2x/dt^2- abx= 0. The general solution to that will involve two undetermined constants. Once you have x, solve for y by using y= (dx/dt)/a.

Quote:

\then I say t = 0 and so

x/y+C_5=at >>> C_5=-x_0/y_0

y/x+C_6=bt >>> C_6=-y_0/x_0

then i say this happens only when C_5 and C_6 are 0

then going back to

x/y+C_5=at >>> x/y=at

y/x+C_6=bt >>> y/x=bt and isolating t to yield

y^2-(b/a)*x^2=0

and when t=0

y_0^2-(b/a)*x_0^2=0

so y^2-(b/a)*x^2=y_0^2-(b/a)*x_0^2