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.
\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
and when t=0
am i right about it?
and can somebody give me some hints to deal with the second problem? thank you.