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Thread: Differential Equations - Intergrating by factors

  1. #1
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    Differential Equations - Intergrating by factors

    Hi

    Need help on the following question:

    Find the general solution of the differential equation

    $\displaystyle \frac{dy}{dx}-\frac{4y}{x}=e^{x^{-3}}$

    $\displaystyle p(x)=\frac{-4}{x}$

    $\displaystyle q(x)=e^{x^{-3}}$

    $\displaystyle I=e^{\int \frac{-4}{x} dx$
    $\displaystyle I=e^{ln|x^{-4}|}$
    $\displaystyle I=x^{-4}$

    $\displaystyle x^{-4}y=\int x^{-4}e^{x^{-3}} dx$

    ok, i tried to integrate by parts however i never get the correct answer.

    this is what i did

    $\displaystyle u = e^{x^{-3}}$
    $\displaystyle du = -3x^{x^{-3}}$

    $\displaystyle dv = x^{-5}$
    $\displaystyle v = \frac{x^{-4}}{-4}$

    i have also tried the other way but still not correct.

    P.S
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  2. #2
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    $\displaystyle \frac{dy}{dx} - \frac{4}{x}y = e^{x^{-3}}$.

    The integrating factor is $\displaystyle e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{x^{-4}}} = x^{-4}$.


    Multiplying through by the integrating factor gives

    $\displaystyle x^{-4}\frac{dy}{dx} - 4x^{-5}y = x^{-4}e^{x^{-3}}$

    $\displaystyle \frac{d}{dx}(x^{-4}y) = x^{-4}e^{x^{-3}}$

    $\displaystyle x^{-4}y = \int{x^{-4}e^{x^{-3}}\,dx}$

    $\displaystyle x^{-4}y = -\frac{1}{3}\int{e^{x^{-3}}(-3x^{-4})\,dx}$.


    Now let $\displaystyle u = x^{-3}$ so that $\displaystyle \frac{du}{dx} = -3x^{-4}$ and the DE becomes

    $\displaystyle x^{-4}y = -\frac{1}{3}\int{e^u\,\frac{du}{dx}\,dx}$

    $\displaystyle x^{-4}y = -\frac{1}{3}\int{e^u\,du}$

    $\displaystyle x^{-4}y = -\frac{1}{3}e^u + C$

    $\displaystyle x^{-4}y = -\frac{1}{3}e^{x^{-3}} + C$

    $\displaystyle y = -\frac{1}{3}x^4e^{x^{-3}} + Cx^4$.
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  3. #3
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    i don't understand how you got the following line?
    $\displaystyle x^{-4}y = -\frac{1}{3}\int{e^{x^{-3}}(-3x^{-4})\,dx}$
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  4. #4
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    I knew I needed to "create" the derivative of $\displaystyle x^{-3}$, which is $\displaystyle -3x^{-4}$.

    I already had the $\displaystyle x^{-4}$, so all I needed to do was multiply by $\displaystyle -3$. But to keep the equation balanced, that means I had to multiply by $\displaystyle -\frac{1}{3}$ as well.
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