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Math Help - Trying to solve by substitution...getting stuck near the end

  1. #1
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    Trying to solve by substitution...getting stuck near the end

    Hi, I have a differential equation that I'm trying to solve my substitution. I've got most of it, but I can figure out how to end it properly.

    The equation: x y' - y = y [1 - ln2(y/x)]

    I divide by x
    let v = y/x
    y' = v + xv'

    sub v and y' in the equation to get:

    dx/x = dv/[v(1-ln2v)]

    let u = ln v
    du = dv/v

    sub these to get:
    dx/x = du/(1-u2)

    Integrating this I get:
    ln |x| = 1/2 * ln | (1+u)/(1-u) |

    reverse substituting v:
    ln |x| = 1/2 * ln | (1+ ln v) / (1 - ln v ) |

    reverse substituting y:
    ln |x| = 1/2 * ln | [1+ ln (y/x)] / [1- ln (y/x)] |

    How do I proceed after this. Is there any way to get the solution in terms of y?

    Or should I use a different approach from the beginning itself?

    Any help is appreciated, thanks
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  2. #2
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    Re: Trying to solve by substitution...getting stuck near the end

    Hello, fractal5!

    Hi, I have a differential equation that I'm trying to solve my substitution.
    I've got most of it, but I can't figure out how to end it properly.

    The equation: . x\frac{dy}{dx} - y \:=\: y \left[1 - \ln^2\!\left(\tfrac{y}{x}\right)\right]

    I divide by x\!:\;\;\frac{dy}{dx} -\frac{y}{x} \:=\:\frac{y}{x}\left[1 - \ln^2\!\left(\tfrac{y}{x}\right)\right]


    Let v \,=\, \frac{y}{x} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}


    Substitute: .  \frac{dv}{1-\ln^2v} \:=\:\frac{dx}{x}

    Let u \,=\,\ln v \quad\Rightarrow\quad du \,=\, \frac{dv}{v}

    Substitute: . \frac{du}{1-u^2} \:=\:\frac{dx}{x}

    Integrating: . \tfrac{1}{2}\ln\left|\frac{1-u}{1+u}\right| \:=\:\ln|x| + c_1

    Back-substitute: . \tfrac{1}{2}\ln\left|\frac{1-\ln v}{1 + \ln v }\right| \;=\;\ln|x| + \ln c_2

    Back-substitute: . \tfrac{1}{2}\ln\left|\frac{1-\ln\frac{y}{x}}{1+\ln\frac{y}{x}}\right| \;=\;\ln|c_2x|

    How do I proceed after this? . You could stop here.
    Is there any way to get the solution in terms of y? . Yes, if you dare.

    We have: . \tfrac{1}{2}\ln\left|\frac{1-\ln\frac{y}{x}}{1+\ln\frac{y}{x}}\right| \;=\;\ln|c_2x|

    . . . . . . . . . \ln\left|\frac{1-\ln\frac{y}{x}}{1 + \ln\frac{y}{x}}\right| \;=\;2\ln|c_2x| \;=\;\ln|c_2x|^2 \;=\;\ln|c_2^2x^2| \;=\;\ln|Cx^2|

    . . . . . . . . . . . . \frac{1-\ln\frac{y}{x}}{1+\ln\frac{y}{x}} \;=\;Cx^2

    . . . . . . . . . . . . 1 - \ln\tfrac{y}{x} \;=\;Cx^2\left(1 + \ln\tfrac{y}{x}\right)

    . . . . . . . . . . . . 1 - \ln\tfrac{y}{x} \;=\;Cx^2 + Cx^2\ln\tfrac{y}{x}

    . . . . . . . Cx^2\ln\tfrac{y}{x} + \ln\tfrac{y}{x} \;=\;1 - Cx^2

    .n. . . . . . \left(Cx^2 + 1\right)\ln\tfrac{y}{x} \;=\;1 - Cx^2

    n . . . . . . . . . . . . . \ln\tfrac{y}{x} \;=\;\frac{1-Cx^2}{1+Cx^2}

    n . . . . . . . . . . . . . . \frac{y}{x} \;=\;e^{\frac{1-Cx^2}{1+Cx^2}}

    n . . . . . . . . . . . . . . y \;=\;x\cdot e^{\frac{1-Cx^2}{1+Cx^2}}


    This time we were lucky (if you can call it that) . . . we could solve for y.

    But 99% of the solutions cannot be solved for y.
    . . And we are not expected to.
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  3. #3
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    Re: Trying to solve by substitution...getting stuck near the end

    Thanks Soroban, seems easy once I've seen the solution. I'd just like to point out for anyone using this in the future, after integrating it comes to ln|(1+u)/(1-u)|. I believe you reciprocated it by mistake, but no worries I got my answer. Thanks again.
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