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Thread: Can't solve this Differential equation

  1. #1
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    Can't solve this Differential equation

    Hey everyone, I've been trying to solve this differential equation for 2 days and I haven't gotten any closer to the answer. If someone could help that would be amazing, thank you. Here it is:
    $\displaystyle xy' + y = y^2$
    $\displaystyle y(1)=-11$

    I've tried manipulating it to get y and dy on one side, and x and dx on the other side, and then integrating but that didn't work. I was left with:
    $\displaystyle cx = y^2 - y$ where c is a constant. but i can't do anything from there.

    I also tried doing $\displaystyle xy = \int y^2 dy$ where x is the constant of integration, but again, that didn't work out.

    Can someone please help me?


    Thank you,
    CenturionMonkey
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    $\displaystyle xy'+y={{y}^{2}}\implies \frac{y'}{{{y}^{2}}}+\frac{1}{xy}=\frac{1}{x}.$ Now put $\displaystyle u=\frac1y.$
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  3. #3
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    $\displaystyle x^2(xy)'=x^2(xy'+y)=x^2y^2=(xy)^2.$ let $\displaystyle xy=u$ to get $\displaystyle \frac{u'}{u^2}=\frac{1}{x^2}.$ integrate both sides: $\displaystyle \frac{-1}{xy}=\frac{-1}{u}=\frac{-1}{x} + C.$ thus: $\displaystyle y=\frac{1}{1-Cx}.$ now use your initial condition to find $\displaystyle C.$
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  4. #4
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    that actually kind of makes sense. Thanks a lot you two.
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  5. #5
    MHF Contributor
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    Your equation is separable

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

    or

    $\displaystyle \frac{dy}{y^2-y} = \frac{dx}{x}$.
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