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    I have:

    \frac {dy}{dt}-by = a

    that would go to:

    ye^{\int -b dt} = \int ae^{\int -b dt} dt + C

    Right? Not sure what to do next

    Initial conditions are y = 0 at t = 0 and the constant is meant to equal \frac {a}{b}

    The answer is meant to be:  y = \frac {a}{b} (e^{bt} - 1)

    How did they get to this result?
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    (Original post by G A B R I E L)
    I have:

    \frac {dy}{dt}-by = a

    that would go to:

    ye^{\int -b dt} = \int ae^{\int -b dt} dt + C

    Right? Not sure what to do next

    Initial conditions are y = 0 at t = 0 and the constant is meant to equal \frac {a}{b}

    The answer is meant to be:  y = \frac {a}{b} (e^{bt} - 1)

    How did they get to this result?
    I'm not sure what you mean by 'that would go to', as I have no idea what you've done there. The integrating factor for this is p(x)=e^{\int bdt} = e^{bt+C}.
    Now multiply through your differential equation by p(x) and spot the derivative of a product on the RHS. From there onwards, use integration.
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    (Original post by G A B R I E L)
    I have:

    \frac {dy}{dt}-by = a

    that would go to:

    ye^{\int -b dt} = \int ae^{\int -b dt} dt + C

    Right? Not sure what to do next

    Initial conditions are y = 0 at t = 0 and the constant is meant to equal \frac {a}{b}

    The answer is meant to be:  y = \frac {a}{b} (e^{bt} - 1)

    How did they get to this result?
    Hi

    You have an inhomogeneous linear first order DE
    First solve the homogeneous part
    \frac{dy}{dt}-by=0
    from this
    \frac{dy}{y}=bdt
    and itegrating
    lny=bt +c \rightarrow Y=Ce^{bt}
    similarly as you wrote as a result with difference of that -bt will be bt at the exponent because -(-b) =b
    But this is the homogeneous solution (Y).
    Second: You should to find a particular solution (yp)
    (Your final solution will be y=Y+yp)
    One method tofind particular solution is substituting k(t) function
    in the coefficient of C in the general solution
    So y_p=k(t)e^{bt}
    With differentiation substitute it the original equation and arrange to k'(t).
    WIth integration you will get k(t) and from it yp
    So y=Y+yp
    The value of C in the final solution you will get from the initial conditions
    and these give that
    \displaystyle y=\frac{a}{b}\left( e^{bt}-1 \right)
    as you wrote it
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    (Original post by Farhan.Hanif93)
    I'm not sure what you mean by 'that would go to', as I have no idea what you've done there.
    he's just done a few steps in one go, it's right

    ye^{-bt}=a\int e^{-bt}\ dt

    ye^{-bt}=-abe^{-bt}+c

    0=c-ab\Leftrightarrow c=ab

    ye^{-bt}=ab-abe^{-bt}

    y=abe^{bt}-ab

    y=ab(e^{bt}-1)

    where did i go wrong?
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    (Original post by Pheylan)
    he's just done a few steps in one go, it's right

    ye^{-bt}=a\int e^{-bt}\ dt

    ye^{-bt}=-abe^{-bt}+c

    0=c-ab\Leftrightarrow c=ab

    ye^{-bt}=ab-abe^{-bt}

    y=abe^{bt}-ab

    y=ab(e^{bt}-1)

    where did i go wrong?
    the bit where you integrated e^-bt
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    (Original post by anshul95)
    the bit where you integrated e^-bt
    lol ****
 
 
 
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