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Math Help - Homogeneous ODE

  1. #1
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    Homogeneous ODE

    Equation is xy' = y \ln \frac{y}{x}.
    I have tried to substitute v = \frac{y}{x} and I am stuck at \frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}

    To be exact I do not know where to go from there in calculus sense.
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  2. #2
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    Re: Homogeneous ODE

    So you have (1/x)dx = -(1/(vlnv-v) )dv= -1/(v(lnv-1)

    Now the integral of -1/(v(lnv-1) is -ln(lnv - 1)

    Then proceed accordingly, hope I helped
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  3. #3
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    Re: Homogeneous ODE

    Quote Originally Posted by losm1 View Post
    Equation is xy' = y \ln \frac{y}{x}.
    I have tried to substitute v = \frac{y}{x} and I am stuck at \frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}

    To be exact I do not know where to go from there in calculus sense.
    \frac{dv}{v(\ln v-1)}=\ln (\ln v-1)+C
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  4. #4
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    Re: Homogeneous ODE

    In my example differentials dx and dv are in denominator:

    \frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}

    I'm having trouble applying your formula in this case. Can you please clarify further?
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  5. #5
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    Re: Homogeneous ODE

    Quote Originally Posted by losm1 View Post
    Equation is xy' = y \ln \frac{y}{x}.
    I have tried to substitute v = \frac{y}{x} and I am stuck at \frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}

    To be exact I do not know where to go from there in calculus sense.
    Make the substitution \displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{dy}{dx} = \frac{dy}{dx} = v + x\,\frac{dv}{dx} \end{align*} and the DE becomes

    \displaystyle \begin{align*} x\,\frac{dy}{dx} &= y\ln{\left(\frac{y}{x}\right)} \\ x\left(v + x\,\frac{dv}{dx}\right) &= v\,x\ln{v} \\ v + x\,\frac{dv}{dx} &= v\ln{v} \\ x\,\frac{dv}{dx} &= v\ln{v} - v \\ x\,\frac{dv}{dx} &= v\left(\ln{v} - 1\right) \\ \frac{1}{v\left(\ln{v} - 1\right)}\,\frac{dv}{dx} &= \frac{1}{x} \\ \int{\frac{1}{v\left(\ln{v} - 1\right)}\,\frac{dv}{dx}\,dx} &= \int{\frac{1}{x}\,dx} \\ \int{\frac{1}{\ln{v} - 1}\,\frac{1}{v}\,dv} &= \ln{|x|} + C_1 \\ \int{\frac{1}{u}\,du} &= \ln{|x|} + C_1 \textrm{ after making the substitution }u = \ln{v} - 1 \implies du = \frac{1}{v}\,dv \\ \ln{|u|} + C_2 &= \ln{|x|} + C_1 \\ \ln{|u|} - \ln{|x|} &= C \textrm{ where }C = C_1 - C_2 \\ \ln{\left|\frac{u}{x}\right|} &= C \\ \ln{\left|\frac{\ln{v} - 1}{x}\right|} &= C \\ \ln{\left|\frac{\ln{\left(\frac{y}{x}\right)} - 1}{x}\right|} &= C \\ \frac{\ln{\left(\frac{y}{x}\right)} - 1}{x} &= A \textrm{ where } A = \pm e^C\end{align*}

    You could get y in terms of x if you wanted to.
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  6. #6
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    Re: Homogeneous ODE

    Quote Originally Posted by losm1 View Post
    In my example differentials dx and dv are in denominator:

    \frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}

    I'm having trouble applying your formula in this case. Can you please clarify further?

    So you have x/dx = v(lnv - 1)/dv implies xdv = v(lnv-1)dx implies (1/x)dx = (1/v(lnv-1)) dv
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  7. #7
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    Re: Homogeneous ODE

    Yes, that is the idea. However, just in case an overview helps...





    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).



    __________________________________________________ __________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

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