Question

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."

Answer 1

\[\frac{ dy }{ dx }=\sqrt{y}\]

Answer 2

@freckles

Answer 3

Have I written out the prompt properly?

Answer 4

@SolomonZelman

Answer 5

\(\large\color{#000000 }{ \displaystyle \frac{dy }{dx} =k\times\sqrt{y} }\)

that would be true, if you say "varies"

Answer 6

if you say "inversely varies" then

\(\large\color{#000000 }{ \displaystyle \frac{dy }{dx} =k\times(1/\sqrt{y}) }\)

Answer 7

(I think that is how it is....)

Answer 8

Hmm I don't "varies" in the question so would I still include a constant, k?

Answer 9

proportional=varies

Answer 10

inversely proportional = varies inversely

Answer 11

\(\large\color{#000000 }{ \displaystyle \frac{dy }{dx} =k\times(1/\sqrt{y}) }\)

\(\large\color{#000000 }{ \displaystyle \sqrt{y}\frac{dy }{dx} =k }\)

\(\large\color{#000000 }{ \displaystyle \color{#ff0000}{\int}\sqrt{y}\frac{dy }{dx} \color{#ff0000}{~dx}=\color{#ff0000}{\int}k\color{#ff0000}{~dx} }\)

Answer 12

\(\large\color{#000000 }{ \displaystyle \int\sqrt{y}~dy=\int k~dx }\)

Answer 13

go from there ...

Answer 14

(Note, you need to solve this equation for y, and don't forget about the integration constant C)

Answer 15

I got to go...

Answer 16

I'll try to figure it out. Thanks

Answer 17

I got \[y=(\frac{ 3 }{ 2}kx)^{2/3}+C\]

Answer 18

@ganeshie8
@freckles
Can any of you confirm for me?

Answer 19

@SithsAndGiggles

Answer 20

@baru

Answer 21

i think you have gone wrong in the integration part.
can you tell the rule for integrating \(x^n\)?

Answer 22

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:
\[x+C=12x+C\]

Answer 23

@baru integration is correct, checked using wolfram

Answer 24

no..they do not cancel

Answer 25

So I would still have \[x=12x+C\]
When moving C to the other side?

Answer 26

yes, well actually, if it confuses, you.. this is how it is actually supposed to be written
\[x+C_1=12x+C_2\]

Answer 27

\[x=12x+(C_2-C_1)\]

Answer 28

since c2 and C1 are both arbitrary constants, we replace C2-C1 with just C

Answer 29

i seem to be getting\[\frac{2}{3}y^{2/3}=kx +c\\y^{2/3}=\frac{3}{2}kx +C\\y=(\frac{3}{2}kx +C)^{3/2}\]

Answer 30

should be y=(3/2kx + C)^2/3

Answer 31

@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

\[\frac{2}{3}y^{3/2} = kx + c\]\[y^{3/2} = \frac{3}{2}kx + C\]\[y = (\frac{3}{2}kx + C)^{2/3}\]

Answer 32

yes...my mistake :)