$\displaystyle dx/{dt}=-ax$ where x=x(t)
$\displaystyle x(t)=Ce^{-t}$ where do the constant C come from?
$\displaystyle dy/dt=ax-by$ where y=y(t) How do i find y(t)
thanks for any help.
Dear charikaar,
$\displaystyle \frac{dx}/{dt}=-ax$
By seperation of variables,
$\displaystyle \frac{dx}{x}=-adt$
$\displaystyle \int{\frac{dx}{x}}=-a\int{dt}$
$\displaystyle lnx=-at+lnC$ where lnC is an arbirary constant.
$\displaystyle \frac{x}{C}=e^{-at}$
$\displaystyle x=x(t)=Ce^{-at}$ ; there must be a typo, are you sure that the answer is $\displaystyle x(t)=Ce^{-t}$????
Still thinking the second part......
Dear charikaar,
$\displaystyle \frac{dy}{dx}=ax-by$
$\displaystyle \frac{dy}{dx}=aCe^{-at}-by$ by substituting the result of the first part....
$\displaystyle \frac{dy}{dx}+by=aCe^{-at}$
This is a linear differential equation. i.e: $\displaystyle \frac{dy}{dx}+p(x)y=q(x)$ Therefore it could be solved by using an integrating factor.
Can you do it from here??