austinL
  • austinL
Solve the initial value problem and determine where the solution attains it's minimum value. \(y\prime=2y^2+xy^2,~y(0)=1\)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
isn't this seperable?
anonymous
  • anonymous
\[ \frac{dy}{dx} = 2y^{2} + xy^{2} \text{, }y(0)=1 \] \[\frac{dy}{dx} = y^{2}(2 + x)\Rightarrow \frac{dy}{y^{2}} = (2 + x)\, dx\]
Psymon
  • Psymon
Yeah, separable. Looks like no issue from there.

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More answers

austinL
  • austinL
Why do I always forget this?? Thanks for the help!
Psymon
  • Psymon
Because we still gotta get used to doing it xD
anonymous
  • anonymous
check with you subconscious...
Psymon
  • Psymon
I try to, but then it responds in russian.
anonymous
  • anonymous
do you speak russian?
Psymon
  • Psymon
Roshia-go hanasemasen. (no IME on school computers -_-)
anonymous
  • anonymous
i'm lost...
Psymon
  • Psymon
No, I cant. I take Japanese, not Russian, haha
austinL
  • austinL
привет
anonymous
  • anonymous
darn subconscious... always making everything more difficult!
Psymon
  • Psymon
Yo quiero Taco Bell?
anonymous
  • anonymous
talking to yourself again, i see...
austinL
  • austinL
Ok, back on subject :P
austinL
  • austinL
I have arrived at, \(\large{\frac{1}{y}=\frac{x^2}{2}+2x-1}\) How would I proceed from here in regards to the problem?
Loser66
  • Loser66
your solution is not right
austinL
  • austinL
\(\large{-\frac{1}{y}}\)
Loser66
  • Loser66
it should be \[-\frac{1}{y}=\frac{x^2}{2}+2x +C= \frac{x^2+4x+2C}{2}\\\frac{1}{y}=\frac{x^2+4x+2C}{-2} \]now, flip it up you have y = \(\dfrac{-2}{x^2+4x +2C}\) replace y(0) =1 to find out C you have C = -1 replace back to y , at that time, take limit of y you can see your solution
austinL
  • austinL
\(\large{y=-\dfrac{2}{x^2+4x-2}}\)
Loser66
  • Loser66
the denominator is an upward parabola, right? when it is bigger, y is smaller, right? so, it is bigger when x \(\rightarrow -\infty ~~ or +\infty, right?\)
Loser66
  • Loser66
and if x \(\pm \infty\) y \(\rightarrow 0\)
austinL
  • austinL
That is where the solution attains its minimum value? \(x\rightarrow \pm\infty\)?
Loser66
  • Loser66
oh, I may makes mistake, heehe, sorry, I forgot the minus sign in the front, I am so sorry, , so it turns your parabola is downward and its vertex is its maximum value . At that value, your y is minimum. You have to find out that vertex to plug it into your y.
Loser66
  • Loser66
got me?
austinL
  • austinL
http://www.wolframalpha.com/input/?i=-2%2F%28x^2%2B4x-2%29
austinL
  • austinL
Looking at these graphs, it approaches zero as x approaches infinity....
Loser66
  • Loser66
but you have local minimum at x =-2 still.
austinL
  • austinL
So is that the minimum value that it attains? \(x = -2\)
Loser66
  • Loser66
I think so.
austinL
  • austinL
Thanks very much @Loser66
austinL
  • austinL
And thank you too @pgpilot326 :)
anonymous
  • anonymous
wait... if x->infinity, the whole thing goes to -infinity, right? that says no minimum
anonymous
  • anonymous
likewise if x goes to -infinity...
austinL
  • austinL
\(\lim_{x \to \infty}~\dfrac{-2}{x^2+4x-1}=0\)
anonymous
  • anonymous
sorry... brain fart
austinL
  • austinL
I dunno, I will just have to ask my professor.
anonymous
  • anonymous
yeah but the bottom had 2 real solutions... \[2 \pm \sqrt{6}\]
anonymous
  • anonymous
so the min will occur there and will be \[-\infty\]
anonymous
  • anonymous
oops, \[-2 \pm \sqrt{6}\]
anonymous
  • anonymous
|dw:1378416827626:dw|
austinL
  • austinL
Would it be x=-2 because that is a solution that is achievable?
anonymous
  • anonymous
? are you looking for a relative min or an absolute min? also, do you have any sort of restricted domain (like x >= 0)?
austinL
  • austinL
The question is, "Solve the initial value problem and determine where the solution attains it's minimum value."
anonymous
  • anonymous
i understand... maybe ask your instructor what they are looking for there.
austinL
  • austinL
Vague I know. I will leave it there and ask about it. And, you already suggested that lol.
anonymous
  • anonymous
10-4

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