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Math Help - first order differential question

  1. #1
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    first order differential question

    hi i am studying some past papers for my exam at uni and would like help with this question if it is not a problem....
    y= Ae^-ax show that dy/dx + ay = 0

    this would be very much appreciated!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Ep!taph View Post
    hi i am studying some past papers for my exam at uni and would like help with this question if it is not a problem....
    y= Ae^-ax show that dy/dx + ay = 0

    this would be very much appreciated!
    All you need to do is verify that the differential equation is true.

    Since y=Ae^{-ax}, it follows that \frac{\,dy}{\,dx}=-aAe^{-ax}.

    Now,

    \frac{\,dy}{\,dx}+ay=-aAe^{-ax}+a\left(Ae^{-ax}\right)=aAe^{-ax}-aAe^{-ax}=0.

    Therefore, we showed that y=Ae^{-ax} satisfied the differential equation \frac{\,dy}{\,dx}+ay=0.

    Does this make sense?
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  3. #3
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    yes that is actually very simple.... what about this one?

    dy/dx + 3y = 6
    By the method of separating the variables determine the equation that describes 'y' given that y = 0 when x =0
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    Quote Originally Posted by Ep!taph View Post
    yes that is actually very simple.... what about this one?

    dy/dx + 3y = 6
    By the method of separating the variables determine the equation that describes 'y' given that y = 0 when x =0
    dy/dx= 6- 3y= 3(2-y)
    "Separating the variables" means writing it as
    \frac{dy}{2- y}= 3 dx

    Can you integrate those?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Ep!taph View Post
    yes that is actually very simple.... what about this one?

    dy/dx + 3y = 6
    By the method of separating the variables determine the equation that describes 'y' given that y = 0 when x =0
    Note that \frac{\,dy}{\,dx}+3y=6\implies\frac{\,dy}{\,dx}=6-3y

    Now separate the variables:

    \frac{\,dy}{3\left(2-y\right)}=\,dx

    Then integrate and solve for y:

    \int\frac{\,dy}{3\left(2-y\right)}=\int\,dx\implies-\tfrac{1}{3}\ln\left|2-y\right|=x+C \implies\left|2-y\right|=Ce^{-3x}\implies y=2-Ke^{-3x}

    Now apply the initial condition y\left(0\right)=0 to find K.

    Can you take it from here?

    P.S.: From now on, please post a new question in a new thread.
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