Question

Solve the initial value problem and determine where the solution attains it's minimum value.

\(y\prime=2y^2+xy^2,~y(0)=1\)

Answer 1

isn't this seperable?

Answer 2

\[ \frac{dy}{dx} = 2y^{2} + xy^{2} \text{, }y(0)=1 \]
\[\frac{dy}{dx} = y^{2}(2 + x)\Rightarrow \frac{dy}{y^{2}} = (2 + x)\, dx\]

Answer 3

Yeah, separable. Looks like no issue from there.

Answer 4

Why do I always forget this??
Thanks for the help!

Answer 5

Because we still gotta get used to doing it xD

Answer 6

check with you subconscious...

Answer 7

I try to, but then it responds in russian.

Answer 8

do you speak russian?

Answer 9

Roshia-go hanasemasen.

(no IME on school computers -_-)

Answer 10

i'm lost...

Answer 11

No, I cant. I take Japanese, not Russian, haha

Answer 12

привет

Answer 13

darn subconscious... always making everything more difficult!

Answer 14

Yo quiero Taco Bell?

Answer 15

talking to yourself again, i see...

Answer 16

Ok, back on subject :P

Answer 17

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?

Answer 18

your solution is not right

Answer 19

\(\large{-\frac{1}{y}}\)

Answer 20

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

Answer 21

\(\large{y=-\dfrac{2}{x^2+4x-2}}\)

Answer 22

the denominator is an upward parabola, right? when it is bigger, y is smaller, right? so, it is bigger when x \(\rightarrow -\infty ~~ or +\infty, right?\)

Answer 23

and if x \(\pm \infty\) y \(\rightarrow 0\)

Answer 24

That is where the solution attains its minimum value? \(x\rightarrow \pm\infty\)?

Answer 25

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.

Answer 26

got me?

Answer 27

http://www.wolframalpha.com/input/?i=-2%2F%28x^2%2B4x-2%29

Answer 28

Looking at these graphs, it approaches zero as x approaches infinity....

Answer 29

but you have local minimum at x =-2 still.

Answer 30

So is that the minimum value that it attains? \(x = -2\)

Answer 31

I think so.

Answer 32

Thanks very much @Loser66

Answer 33

And thank you too @pgpilot326 :)

Answer 34

wait... if x->infinity, the whole thing goes to -infinity, right? that says no minimum

Answer 35

likewise if x goes to -infinity...

Answer 36

\(\lim_{x \to \infty}~\dfrac{-2}{x^2+4x-1}=0\)

Answer 37

sorry... brain fart

Answer 38

I dunno, I will just have to ask my professor.

Answer 39

yeah but the bottom had 2 real solutions...
\[2 \pm \sqrt{6}\]

Answer 40

so the min will occur there and will be \[-\infty\]

Answer 41

oops, \[-2 \pm \sqrt{6}\]