Seperable Differential Equation

**charikaar** · March 13th 2010, 05:47 AM

$dx/{dt}=-ax$ where x=x(t)

$x(t)=Ce^{-t}$ where do the constant C come from?

$dy/dt=ax-by$ where y=y(t) How do i find y(t)

thanks for any help.

**Sudharaka** · March 13th 2010, 06:24 AM

Originally Posted by charikaar

$dx/{dt}=-ax$ where x=x(t)

$x(t)=Ce^{-t}$ where do the constant C come from?

$dy/dt=ax-by$ where y=y(t) How do i find y(t)

thanks for any help.

Dear charikaar,

$\frac{dx}/{dt}=-ax$

By seperation of variables,

$\frac{dx}{x}=-adt$

$\int{\frac{dx}{x}}=-a\int{dt}$

$lnx=-at+lnC$ where lnC is an arbirary constant.

$\frac{x}{C}=e^{-at}$

$x=x(t)=Ce^{-at}$ ; there must be a typo, are you sure that the answer is $x(t)=Ce^{-t}$ ????

Still thinking the second part......

**Sudharaka** · March 13th 2010, 06:33 AM

Originally Posted by charikaar

$dx/{dt}=-ax$ where x=x(t)

$x(t)=Ce^{-t}$ where do the constant C come from?

$dy/dt=ax-by$ where y=y(t) How do i find y(t)

thanks for any help.

Dear charikaar,

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

$\frac{dy}{dx}=aCe^{-at}-by$ by substituting the result of the first part....

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

This is a linear differential equation. i.e: $\frac{dy}{dx}+p(x)y=q(x)$ Therefore it could be solved by using an integrating factor.

Can you do it from here??

**charikaar** · March 13th 2010, 06:41 AM

Nicely done.

Thank you very much.

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