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Thread: Problem finding integrating factor of simple first order ODE

  1. #1
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    Problem finding integrating factor of simple first order ODE

    Hi ! So I'm having trouble with a problem asking to find the integrating factor of this equation :

    t^{2}(\frac{dx}{dt})+t^{5}x = 4t

    So my approach was to set the right-hand side to 0 to get a homogeneous DE and then use separation of variables.

    This got me :

    \frac{dx}{dt} = -t^{5}x/t^{2} =  -t^{3}x

    \frac{dx}{x} =   -t^{3}dt

    x =   ce^{\frac{-t^{4}}{4}}

    So the general solution to the inhomogeneous one with u as integrating factor is : x= ue^{\frac{-t^{4}}{4}}
    But plugging this into the original equation I arrive at :

    u'=\frac{4e^{\frac{-t^{4}}{4}}}{t}


    Which I don't know how to integrate and doesn't look like it can be any of the answers I was offered (multiple-choice test).

    Where did I go wrong ?

    Thank you for your help !
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  2. #2
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    Re: Problem finding integrating factor of simple first order ODE

    You lost the 4t term when you moved everything to one side.

    $\displaystyle \begin{align*} t^2\,\frac{\mathrm{d}x}{\mathrm{d}t} + t^5\,x &= 4\,t \\ \frac{\mathrm{d}x}{\mathrm{d}t} + t^3\,x &= \frac{4}{t} \end{align*}$

    The integrating factor is $\displaystyle \begin{align*} \mathrm{e}^{\int{t^3\,\mathrm{d}t}} = \mathrm{e}^{\frac{t^4}{4}} \end{align*}$ so we get

    $\displaystyle \begin{align*} \mathrm{e}^{\frac{t^4}{4}}\,\frac{\mathrm{d}x}{ \mathrm{d} t} + t^3\,\mathrm{e}^{\frac{t^4}{4}} \,x &= \frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t} \\ \frac{\mathrm{d}}{\mathrm{d}t} \left( \mathrm{e}^{\frac{t^4}{4}} \, x \right) &= \frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t} \\ \mathrm{e}^{\frac{t^4}{4}}\,x &= \int{\frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t}\, \mathrm{d} t } \\ x &= \mathrm{e}^{-\frac{t^4}{4}} \int{ \frac{4\,\mathrm{e}^{\frac{t^4}{4}}}{t}\,\mathrm{d }t } \end{align*}$

    This is the exact solution, although it can't be written explicitly in terms of elementary functions.
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    Re: Problem finding integrating factor of simple first order ODE

    Thank you very much for your help !
    I can't see where I missed the 4/t term ? Except setting it to 0 in the original equation at first to make it homogeneous.
    Also is it not e^( -t^4/4) instead of t^4/4 ? Or did I make a mistake.
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    Re: Problem finding integrating factor of simple first order ODE

    Another way of looking at it:
    First, divide by t^2 to get \frac{dx}{dt}+ t^3x= \frac{4}{t}.

    Now, by the definition of "integrating factor" we are looking for a function, u(t), such that \frac{d(ux)}{dt}= u\frac{dx}{dt}+ \left(\frac{du}{dt}\right)x= u\frac{dx}{dt}+ t^3ux. That is, u must satisfy the equation \frac{du}{dt}= t^3u. We can separate that into \frac{du}{u}= t^3dt. Integrating, ln(u)= \frac{1}{4}t^4+ C. Solving for u, u= C'e^{(1/4)t^4} where C'= e^C. We only need one function so we can take C'= 1. Our integrating factor is e^{(1/4)t^4}.

    Multiplying by that, the differential equation becomes e^{(1/4)t^4}\frac{dx}{dt}+ t^3e^{(1/4)t^4}x= \frac{d}{dt}\left(e^{(1/4)t^4}x(t)\right)= \frac{e^{(1/4)t^4}}{t}.
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    Re: Problem finding integrating factor of simple first order ODE

    Quote Originally Posted by Jo37 View Post
    Thank you very much for your help !
    I can't see where I missed the 4/t term ? Except setting it to 0 in the original equation at first to make it homogeneous.
    Also is it not e^( -t^4/4) instead of t^4/4 ? Or did I make a mistake.
    You made several mistakes. Instead of bleating about your own incorrect solution you should look through the correct solution given and try to determine where your mistakes are.
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