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Thread: clarification on differential equation workings

  1. #1
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    Question clarification on differential equation workings

    the question is:
    Q) Find the general solution of the differential equation dx/dt - x = t+1
    A) Here are my workings. Does this look correct?

    dx/dt = x + t +1
    dx/dt = t(x/t 1 + 1/t)

    Let u = x/t 1 + 1/t so du/dx = 1/t so dx = du.t

    Equation become: dut/dt = tu
    du/u = dt
    ln(u) = t + C
    u = exp(t + C)
    u = Aexp(t)

    As u = x/t 1 + 1/t rearranging gives x = ut - t - 1

    Putting u into equation gives final answer of x = Aexp(t)t - t - 1
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  2. #2
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    the question is:
    Q) Find the general solution of the differential equation dx/dt - x = t+1
    A) Here are my workings. Does this look correct?

    dx/dt = x + t +1
    dx/dt = t(x/t 1 + 1/t)

    Let u = x/t 1 + 1/t so du/dx = 1/t so dx = du.t

    Equation become: dut/dt = tu
    du/u = dt
    ln(u) = t + C
    u = exp(t + C)
    u = Aexp(t)

    As u = x/t 1 + 1/t rearranging gives x = ut - t - 1

    Putting u into equation gives final answer of x = Aexp(t)t - t - 1
    Your working looks quite odd...

    It is a first order linear DE, so you can solve it using an integrating factor.


    $\displaystyle \frac{dx}{dt} - x = t + 1$.


    The integrating factor is $\displaystyle e^{\int{-1\,dt}} = e^{-t}$.

    Multiply both sides of the equation by the integrating factor.


    $\displaystyle e^{-t}\frac{dx}{dt} - xe^{-t} = e^{-t}(t + 1)$


    Notice that the left hand side is the product rule expansion of $\displaystyle \frac{d}{dt}(xe^{-t})$.


    So $\displaystyle \frac{d}{dt}(xe^{-t}) = e^{-t}(t + 1)$

    $\displaystyle xe^{-t} = \int{e^{-t}(t + 1)\,dt}$


    Solve this integral using integration by parts.

    $\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

    Let $\displaystyle u = t + 1$ so that $\displaystyle du = 1$

    Let $\displaystyle dv = e^{-t}$ so that $\displaystyle v = -e^{-t}$.


    Therefore

    $\displaystyle xe^{-t} = -e^{-t}(t + 1) - \int{-e^{-t}\,dt}$

    $\displaystyle xe^{-t} = -e^{-t}(t + 1) - e^{-t} + C$

    $\displaystyle xe^{-t} = -e^{-t}(t + 2) + C$

    $\displaystyle x = Ce^t - (t + 2)$

    $\displaystyle x = Ce^t - t - 2$.
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  3. #3
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    the question is:
    Q) Find the general solution of the differential equation dx/dt - x = t+1
    A) Here are my workings. Does this look correct?

    dx/dt = x + t +1
    dx/dt = t(x/t 1 + 1/t)

    Let u = x/t 1 + 1/t so du/dx = 1/t so dx = du.t

    Equation become: dut/dt = tu
    du/u = dt
    ln(u) = t + C
    u = exp(t + C)
    u = Aexp(t)

    As u = x/t 1 + 1/t rearranging gives x = ut - t - 1

    Putting u into equation gives final answer of x = Aexp(t)t - t - 1
    Check that your solution satisfies the original differential equation, now tell us if it is correct.

    CB
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