# Thread: How to find the minimum value of a solution

1. ## 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.

2. Originally Posted by crabchef
and now the question is how do I attain its minimum.
solve $\displaystyle \frac{dy}{dx}= 0$

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

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

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

3. Originally Posted by crabchef
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 $\displaystyle y$.

4. 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.