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Thread: How to find the minimum value of a solution

  1. #1
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    How to find the minimum value of a solution

    Solve the initial value problem

    $\displaystyle dy/dx=2y^2+xy^2$

    I found the explicit solution to be:

    $\displaystyle y=-1/(2x+x^2/2-1)$

    and now the question is how do I attain its minimum. I know I'm supposed to take a derivative somewhere but I'm not sure of what. The answer is $\displaystyle x=-2$

    I hate how I keep getting parts of every question but not a single entire question.
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  2. #2
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    Quote Originally Posted by crabchef View Post
    and now the question is how do I attain its minimum.
    solve $\displaystyle \frac{dy}{dx}= 0$

    $\displaystyle 2y^2+xy^2=0$

    $\displaystyle y^2(2+x)=0$

    $\displaystyle 2+x=0\implies x=-2$
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  3. #3
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    Quote Originally Posted by crabchef View Post
    Solve the initial value problem

    $\displaystyle dy/dx=2y^2+xy^2$

    I found the explicit solution to be:

    $\displaystyle y=-1/(2x+x^2/2-1)$

    and now the question is how do I attain its minimum. I know I'm supposed to take a derivative somewhere but I'm not sure of what. The answer is $\displaystyle x=-2$

    I hate how I keep getting parts of every question but not a single entire question.

    $\displaystyle y = -\frac{2}{x^2+4x-2}$

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

    note that $\displaystyle y \ne 0$ for all $\displaystyle x$ ... therefore, $\displaystyle 2+x = 0$, indicating that $\displaystyle x = -2$.

    also note that $\displaystyle \frac{dy}{dx}$ changes sign from negative to positive for values of $\displaystyle x$ on either side of $\displaystyle x = -2$, indicating a minimum for $\displaystyle y$.
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  4. #4
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    thanks for the detailed explanations guys

    I don't know where I'd be without you guys...

    I wonder how people can stand to do math for so long. I did MAYBE 5-6 hours of math today and I want to drive out south to get murdered cuz I'm too tired to do it myself.
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