How to find the minimum value of a solution

**crabchef** · June 29th 2010, 04:25 PM

Solve the initial value problem

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

I found the explicit solution to be:

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

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 $x=-2$

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

**pickslides** · June 29th 2010, 04:35 PM

Originally Posted by crabchef

and now the question is how do I attain its minimum.

solve $\frac{dy}{dx}= 0$

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

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

$2+x=0\implies x=-2$

**skeeter** · June 29th 2010, 04:38 PM

Originally Posted by crabchef

Solve the initial value problem

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

I found the explicit solution to be:

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

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 $x=-2$

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

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

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

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

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

**crabchef** · June 29th 2010, 08:22 PM

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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