UsukiDoll
  • UsukiDoll
Construct a tangent field for y' = 2sqroot(y). Note that necessarily y>0 y' > 0 Find all solutions of y' = 2sqroot(y) as in the treatment of eq. 4 in the text How many solutions are there through each point on the x-axis? Each point off the x-axis?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
And here's the book that I'm using...section 1.3 page 12 http://scholarspace.manoa.hawaii.edu/handle/10125/21735 Also, I'll be putting up the tangent field and what I got in paint shortly.
UsukiDoll
  • UsukiDoll
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UsukiDoll
  • UsukiDoll

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UsukiDoll
  • UsukiDoll
There are many solutions on the x-axis, but it's only on the top part of the plane in the tangent field.
UsukiDoll
  • UsukiDoll
done...so yeah the book is cray cray
UsukiDoll
  • UsukiDoll
it goes according to separating the values like one side for dy and one side for dx
UsukiDoll
  • UsukiDoll
not separating the values. variables.
anonymous
  • anonymous
So: \[ \frac {y'}{\sqrt y} =2 \]?
UsukiDoll
  • UsukiDoll
|dw:1378452398226:dw|
UsukiDoll
  • UsukiDoll
I did the tangent field of it which was no problem. the issue comes in when I have to use separate variables to get to the answer.
UsukiDoll
  • UsukiDoll
you know how y' becomes dy/dx
anonymous
  • anonymous
Okay so separation of variables.
UsukiDoll
  • UsukiDoll
|dw:1378452469533:dw|
UsukiDoll
  • UsukiDoll
eeyup
anonymous
  • anonymous
\[ \frac{y'}{\sqrt{2y}} = 1 \]Then do \[ \int \frac{y'dx}{\sqrt{2y}} =\int dx \]
UsukiDoll
  • UsukiDoll
this is what I got http://assets.openstudy.com/updates/attachments/52297c77e4b0a8663d551dbe-usukidoll-1378451431318-slopefieldequationfor2sqrooty.png
anonymous
  • anonymous
Remember that \(y'dx = dy\) so \[ \int \frac{dy}{\sqrt{2y}} = \int dx \]
UsukiDoll
  • UsukiDoll
so do I take the antiderivative of both sides?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
I really don't like that notation. It irks me.
anonymous
  • anonymous
The way I did it does not abuse notation
UsukiDoll
  • UsukiDoll
wait is the part for y correct? it's 2 sqroot y
anonymous
  • anonymous
Well you wrote \(y' = \sqrt{2y}\) before. So it depends on which is correct.
UsukiDoll
  • UsukiDoll
let me recheck
UsukiDoll
  • UsukiDoll
oh crap it's 2sqrooty
anonymous
  • anonymous
But with your original equation it would be \[ \int \frac{dy}{\sqrt{y}} = \int 2 dx \]
anonymous
  • anonymous
\[ \int y^{-1/2}dy = \int 2dx \]
UsukiDoll
  • UsukiDoll
|dw:1378452748922:dw|
anonymous
  • anonymous
\[ 2y^{1/2} +C_1= 2x+C_2 \]
anonymous
  • anonymous
Yeah, what you have is correct.
anonymous
  • anonymous
\[ y = (x+C)^2 = x^2+Cx + C^2 \]
UsukiDoll
  • UsukiDoll
woo hoo, but how do I get it to that y=(x-c)^2 stuff ? multiply everything by one half?
anonymous
  • anonymous
Divide by 2. You can ignore what happens to \(C\)
UsukiDoll
  • UsukiDoll
ok. can you check on one more problem for me?
anonymous
  • anonymous
You wanna make another question for it?
UsukiDoll
  • UsukiDoll
sure
anonymous
  • anonymous
Also I am getting totally robbed on medals.

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