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Math Help - linear differential!

  1. #1
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    linear differential!

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

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

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

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

     \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 \begin{align*} e^{\frac{y}{x}} \end{align*} you should actually get \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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  4. #4
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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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  5. #5
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    Re: linear differential!

     -\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 u= e^v.

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