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Thread: Seperable Differential Equation

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by charikaar View Post
    $\displaystyle dx/{dt}=-ax$ 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......
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  3. #3
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    Quote Originally Posted by charikaar View Post
    $\displaystyle dx/{dt}=-ax$ 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??
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  4. #4
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    Nicely done.

    Thank you very much.
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