anonymous
  • anonymous
Write and then solve the differential equation for the statement: "The rate of change of y with respect to x is inversely proportional to the square root of y."
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ dy }{ dx }=\sqrt{y}\]
anonymous
  • anonymous
@freckles
anonymous
  • anonymous
Have I written out the prompt properly?

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anonymous
  • anonymous
@SolomonZelman
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{dy }{dx} =k\times\sqrt{y} }\) that would be true, if you say "varies"
SolomonZelman
  • SolomonZelman
if you say "inversely varies" then \(\large\color{#000000 }{ \displaystyle \frac{dy }{dx} =k\times(1/\sqrt{y}) }\)
SolomonZelman
  • SolomonZelman
(I think that is how it is....)
anonymous
  • anonymous
Hmm I don't "varies" in the question so would I still include a constant, k?
SolomonZelman
  • SolomonZelman
proportional=varies
SolomonZelman
  • SolomonZelman
inversely proportional = varies inversely
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{dy }{dx} =k\times(1/\sqrt{y}) }\) \(\large\color{#000000 }{ \displaystyle \sqrt{y}\frac{dy }{dx} =k }\) \(\large\color{#000000 }{ \displaystyle \color{#ff0000}{\int}\sqrt{y}\frac{dy }{dx} \color{#ff0000}{~dx}=\color{#ff0000}{\int}k\color{#ff0000}{~dx} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \int\sqrt{y}~dy=\int k~dx }\)
SolomonZelman
  • SolomonZelman
go from there ...
SolomonZelman
  • SolomonZelman
(Note, you need to solve this equation for y, and don't forget about the integration constant C)
SolomonZelman
  • SolomonZelman
I got to go...
anonymous
  • anonymous
I'll try to figure it out. Thanks
anonymous
  • anonymous
I got \[y=(\frac{ 3 }{ 2}kx)^{2/3}+C\]
anonymous
  • anonymous
@ganeshie8 @freckles Can any of you confirm for me?
anonymous
  • anonymous
@SithsAndGiggles
anonymous
  • anonymous
@baru
baru
  • baru
i think you have gone wrong in the integration part. can you tell the rule for integrating \(x^n\)?
anonymous
  • anonymous
The main thing that has me confused is C, I am not sure if they cancel each other out when you have something like this: \[x+C=12x+C\]
anonymous
  • anonymous
@baru integration is correct, checked using wolfram
baru
  • baru
no..they do not cancel
anonymous
  • anonymous
So I would still have \[x=12x+C\] When moving C to the other side?
baru
  • baru
yes, well actually, if it confuses, you.. this is how it is actually supposed to be written \[x+C_1=12x+C_2\]
baru
  • baru
\[x=12x+(C_2-C_1)\]
baru
  • baru
since c2 and C1 are both arbitrary constants, we replace C2-C1 with just C
baru
  • baru
i seem to be getting\[\frac{2}{3}y^{2/3}=kx +c\\y^{2/3}=\frac{3}{2}kx +C\\y=(\frac{3}{2}kx +C)^{3/2}\]
anonymous
  • anonymous
should be y=(3/2kx + C)^2/3
whpalmer4
  • whpalmer4
@baru \[\int y^n\ dy = \frac{1}{n+1}y^{n+1} + C \]\[\int \sqrt{y}\ dy = \int y^{1/2}\ dy = \frac{1}{1+\frac{1}{2}} y^{1+\frac{1}{2}} + C = \frac{2}{3}y^{3/2} +C\] your exponent is incorrect, should be \[\frac{2}{3}y^{3/2} = kx + c\]\[y^{3/2} = \frac{3}{2}kx + C\]\[y = (\frac{3}{2}kx + C)^{2/3}\]
baru
  • baru
yes...my mistake :)

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