Question

Write and then solve for y = f(x) the differential equation for the statement: "The rate of change of y with respect to x is inversely proportional to y^4."

Answer 1

I beleive it to be dy/dx =1/y^4 ?

Answer 2

Yes, your equation is right.

Answer 3

but, I would say, it is
\(\color{#000000}{ \displaystyle \frac{dy}{dx} =\frac{k}{y^4} }\)

Answer 4

to simulate some constant

Answer 5

yes, because if you said that y is proportional to x, then y=kx.
In other words, there is some constant k such that x•k=y.

Answer 6

The simplest here is to separate the variables.

Answer 7

so then I would get that y^5/5 =kx

Answer 8

+c

Answer 9

integrate 1/y^4. so \[y^4dy=kdx , \int\limits_{}^{}y^4dy=\int\limits_{}^{}kdx\]
\[\frac{ y^5 }{ 5 }=kx+c\]

Answer 10

I think thats adequite, isolating y would be nice but i dont think its necessary, the equation is simple in this form

Answer 11

\(\color{#000000}{ \displaystyle \frac{dy}{dx} =\frac{k}{y^4} \quad \Longrightarrow \quad y^4\frac{dy}{dx} =k }\)

I like claiming that you don't just "integrate", rather you integrate with respect to a variable. (and then cancel differentials)

\(\color{#000000}{ \displaystyle \color{red}{\int} y^4\frac{dy}{dx} \color{red}{dx} =\color{red}{\int} k \color{red}{dx} }\)

\(\color{#000000}{ \displaystyle \int y^4\frac{dy}{\cancel{dx}} \cancel{dx}=\int k dx }\)

\(\color{#000000}{ \displaystyle \int y^4~dy=\int k~dx }\)

and then integrate both sides, BUT DON'T FORGET ABOUT INTEGRATION CONSTANT C.

Or, if you prefer, you can just go ahead and "integrate" without canceling differentials as I showed. That is,

\(\color{#000000}{ \displaystyle y^4~= k~ \quad \Longrightarrow \quad \int y^4~dy=\int k~dx}\)

and either way (when done correctly) you end up with the same result.

Answer 12

oh excuse me let me fix my last equation

Answer 13

\(\color{#000000}{ \displaystyle y^4\frac{dy}{dx}=k\quad\Longrightarrow\quad y^4{\tiny~}dy=k{\tiny~}dx \quad\Longrightarrow\quad \int y^4{\tiny~}dy=\int k{\tiny~}dx }\)

Answer 14

i usuallyt look at it as having to seperate dy/dx by multiplying dx out ect

Answer 15

Like I wrote just now?

Answer 16

Well i would conuse myself with thinking for example, that I have dy=45y^2/2x dx and when i went to divide the 45y^2 i would think i could not becuase i would effectively be making dy/45y^2. usually i made this error when tired but once i ingrained it into myself to disassociate the dx or dy with any other things i do in the equation it worked out

Answer 17

Ok)

Answer 18

But yeah It seemed dumb once i made sure but it was a common mistake for me for a while

Answer 19

So, basically, when separating variables, all you want to do is to rearrange your original equation (without preforming incorrect operations, and when possible of course), isuch that it is something like

\(\color{#000000}{ \displaystyle f(y)~dy~=~f(x)~dx }\)

Answer 20

and from there you can simply integrate both sides.

Answer 21

Yeah i understand all of that I was just sharing a personal struggle i had

Answer 22

Alright.

Answer 23

so, in the original problem we left off,

\(\color{#000000}{ \displaystyle \int y^4~dy=\int k~dx }\)

Answer 24

now, all you need to do is to integrate, but don't forget arbitrary constants.

Answer 25

oh yeah i alreasy finished the problem if you look back

Answer 26

yes, you say, y^5/5=kx

Answer 27

and that is incomplete

Answer 28

I made the +c adendam

Answer 29

Oh, ok.

Answer 30

Missed that; apologize.

Answer 31

NP I appreciate the help!

Answer 32

So, your final is,
\(\color{#000000}{ \displaystyle \frac{1}{5}y^5=kx+C}\)

Answer 33

you are welcome, if I helped:)

Answer 34

you did!

Answer 35

Just curious is this calculus I, II or a differential equatiosn class?

Answer 36

calc 1 one of the easier ones on the test just had not worked with a word problem in a while so i was unsure of myself

Answer 37

Yeah, I am not a great fan of them either. Especially if it is poorly worded, which it would usually be the case as it turns out....

Answer 38

yeah these practice tests can be brutal though sometimes I had a pretty bad run in with the area of a revolved object of which half of the instructions indicated a square and the other half a circle

spent like 90 min on openstudy nobody could figure it out XD

Answer 39

still have problem with those, or you solved them?

Answer 40

Oh no I am great at that those come easy to me as im cerified in 2 cad softwares so it was like second nature, but the wording was completely incorrect

Answer 41

yes, fortunatelly, it is readable mostly.

Answer 42

in any case, won't disturb.... good luck with calc!