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Math Help - Differential Equation (ii)

  1. #1
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    Differential Equation (ii)

    I'm stuck at this question.

    Express G in terms of t and k.

    \frac{dG}{dt}=\frac{25k-G}{25}

    ANSWER: G=25k(1-e^{\frac{-t}{25}})
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  2. #2
    Eater of Worlds
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    We can use an integrating factor.

    \frac{dG}{dt}=k-\frac{G}{25}

    \frac{dG}{dt}+\frac{G}{25}=k

    The integrating factor is

    e^{\int\frac{1}{25}dt}=e^{\frac{t}{25}}

    \frac{d}{dt}[e^{\frac{t}{25}}G]=ke^{\frac{t}{25}}

    Integrate:

    Ge^{\frac{t}{25}}=25ke^{\frac{t}{25}}+C

    \boxed{G=25k+C_{1}e^{\frac{-t}{25}}}
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by r_maths View Post
    I'm stuck at this question.

    Express G in terms of t and k.

    \frac{dG}{dt}=\frac{25k-G}{25}

    ANSWER: G=25k(1-e^{\frac{-t}{25}})
    Separate the variables:

    \frac{dG}{25k-G}=\frac{dt}{25}

    Integrate:

    -ln{\left|25k-G\right|}=\frac{t}{25}+C.

    Solve for G:

    ln{\left|25k-G\right|}=-\frac{t}{25}+C
    {\left|25k-G\right|}=e^{-\frac{t}{25}+C}
    {\left|25k-G\right|}=e^{C}e^{-\frac{t}{25}}
    25k-G=\pm e^{C}e^{-\frac{t}{25}}
    G=25k-Ce^{-\frac{t}{25}}

    Was there an initial condition? G(0)=0?

    0=25k-Ce^{-\frac{0}{25}}\rightarrow C=25k

    \therefore G=25k-25ke^{-\frac{t}{25}}
    \color{red}\boxed{G=25k{\left(1-e^{-\frac{t}{25}}\right)}}

    Hope that helped you out!
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