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Thread: linear differential!

  1. #1
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    linear differential!

    $\displaystyle (\exp^{\frac{y}{x}}-\frac{y}{x}\exp^{\frac{y}{x}}+\frac{1}{1+x^2})dx+\ exp^{\frac{y}{x}}dy=0 $

    have reduce $\displaystyle (1-\frac{y}{x}+\frac{\exp^{\frac{y}{x}}}{1+x^2})dx+dy =0 $

    $\displaystyle \frac{dy}{dx}=\frac{y}{x}-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1 $

    $\displaystyle \frac{dy}{dx}-\frac{y}{x}=-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1 $

    pls am stock here, whether am on the right part or not, help!

    thanks
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  2. #2
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    Re: linear differential!

    Quote Originally Posted by lawochekel View Post
    $\displaystyle (\exp^{\frac{y}{x}}-\frac{y}{x}\exp^{\frac{y}{x}}+\frac{1}{1+x^2})dx+\ exp^{\frac{y}{x}}dy=0 $

    have reduce $\displaystyle (1-\frac{y}{x}+\frac{\exp^{\frac{y}{x}}}{1+x^2})dx+dy =0 $

    $\displaystyle \frac{dy}{dx}=\frac{y}{x}-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1 $

    $\displaystyle \frac{dy}{dx}-\frac{y}{x}=-\frac{\exp^{\frac{y}{x}}}{1+x^2}-1 $

    pls am stock here, whether am on the right part or not, help!

    thanks
    When you divide by $\displaystyle \displaystyle \begin{align*} e^{\frac{y}{x}} \end{align*}$ you should actually get $\displaystyle \displaystyle \begin{align*} \left[ 1 - \frac{y}{x} + \frac{1}{\left( 1 + x^2 \right) e^{\frac{y}{x}}} \right] dx + dy = 0 \end{align*}$.
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  3. #3
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    Re: linear differential!

    You understand, do you not, that this is NOT a "linear" equation?
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    Re: linear differential!

    Quote Originally Posted by HallsofIvy View Post
    You understand, do you not, that this is NOT a "linear" equation?
    At least not yet. With the right substitution it may become one...
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    Re: linear differential!

    $\displaystyle -\frac{(1+x^2)\exp^v}{1+(1+x^2)\exp^v}dv=\frac{1}{x }dx $

    by substitution, i got that above. Do i need further substitution?
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  6. #6
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    Re: linear differential!

    I would recommend $\displaystyle u= e^v$.

    (Notation: either "$\displaystyle e^v$" or "exp(v)" but not "$\displaystyle exp^v$".)
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