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Math Help - First order differentials

  1. #1
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    First order differentials

    \frac {dy}{dx} = \frac{2x}{x+y} using the substitution y = xu

    I started by stating that dy/dx = du/dx + u (product rule) and subbing so get the equation:

    \frac {du}{dx} + u = \frac {2x}{x+xu}

    Simplifying to get: \frac {du}{dx} + u = \frac {2}{1+u}

    Then found my integrating factor (e^(p(x))) which was e^x and multiplied, leaving me with:

    \frac {d(ue^x)}{dx} = \frac {2e^x}{1+u}

    Firstly, am I right so far?
    Secondly, if I am, I need help integrating \frac {2e^x}{1+u} with respect to x, having the 2 unknowns is giving me second thoughts about integration by parts so I'm not entirely sure what to do.
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  2. #2
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    Quote Originally Posted by alrightgeez View Post
    \frac {dy}{dx} = \frac{2x}{x+y} using the substitution y = xu

    I started by stating that dy/dx = du/dx + u (product rule) and subbing so get the equation:

    \frac {du}{dx} + u = \frac {2x}{x+xu}

    Simplifying to get: \frac {du}{dx} + u = \frac {2}{1+u}

    Then found my integrating factor (e^(p(x))) which was e^x and multiplied, leaving me with:

    \frac {d(ue^x)}{dx} = \frac {2e^x}{1+u}

    Firstly, am I right so far?
    Secondly, if I am, I need help integrating \frac {2e^x}{1+u} with respect to x, having the 2 unknowns is giving me second thoughts about integration by parts so I'm not entirely sure what to do.

    The I-factor only works on linear ODE's

    From here the equation is seperable

    \frac {du}{dx} + u = \frac {2}{1+u}

    \frac {du}{dx} = \frac {2-u(1+u)}{1+u}=-\frac{u^2+u-2}{u+1} =-\frac{(u+2)(u-1)}{u+1}

    Now seperate to get

    \frac{u+1}{(u+2)(u-1)}du=-dt

    From here use partial fractions to integrate.

    I hope this helps
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  3. #3
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    Ah thanks for that, turns out I differentiated wrongly at the beginning when I used the product rule so my final answer was wrong, but you helped me out anyway so thanks
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