1. ## Seperable Differential Equation

$\displaystyle dx/{dt}=-ax$ 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.

2. Originally Posted by charikaar
$\displaystyle dx/{dt}=-ax$ 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......

3. Originally Posted by charikaar
$\displaystyle dx/{dt}=-ax$ 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,

$\displaystyle \frac{dy}{dx}=ax-by$

$\displaystyle \frac{dy}{dx}=aCe^{-at}-by$ 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??

4. Nicely done.

Thank you very much.