Question

Help finding the particular solution to a differential equation

Answer 1

\[\frac{ x ^{2}}{y ^{2}-5} \frac{dy}{dx}=\frac{1}{2y} \]
Satisfy the initial condition y(1)=sqrt(6).

Answer 2

So what part are you stuck on?
This appears to be separable yes?

Answer 3

Hey Zepdrix, honestly I don't even understand how to start it off. Do I get the dy/dx alone on one side and then just take the integral?

Answer 4

nice question :D

Answer 5

You want to get all of the x stuff on one side, and all of the y stuff on the other.
And you need to make sure your `differential` is in the numerator.
So that's telling us that the y's should be on the left, yes?
Since the dy is already in the numerator.

Answer 6

So let's "multiply" the dx to the other side,\[\Large\rm \frac{x^2}{y^2-5}dy=\frac{1}{2y}dx\]

Answer 7

just remember this hint
when x is alone and y is alone lolz
and x is not related to y somehow hehe like xy
just need to make y in a side and x on the other see simple :3

Answer 8

Then we'll further separate by dividing each side by x^2, and multiplying each side by 2y,\[\Large\rm \frac{2y}{y^2-5}dy=\frac{1}{x^2}dx\]

Answer 9

From there you can integrate^
Did those steps make a lil bit of sense? :O

Answer 10

Yes that makes a lot of sense. Now I should just integrate to get y equals something?

Answer 11

\[\Large\rm \int\limits\frac{2y}{y^2-5}dy=\int\limits \frac{1}{x^2}dx\]Yah you'll get some y stuff and some x stuff.
In this particular case we `will` be able to get an expliciti solution in term of x which is nice.
There are a few more tricky steps to be careful on.

Do you understand how to integrate the left side?

Answer 12

If I integrate the left side for dy. I get ln(-5+y^2).. is that correct?

Answer 13

\[\Large\rm \ln|y^2-5|=\int\limits\frac{1}{x^2}dx\]k looks good.

Answer 14

now differentiate for dx?

Answer 15

mhm

Answer 16

So that would be -1/x I think. And now this is the part where I am once again completely lost. Do I just solve for y or x?

Answer 17

We should have a +C somewhere also yes?
Do you understand why we didn't need to write +C on each side?

Answer 18

We only need a plus C on one side. So it would be ln(-5+y^2)=-1/(x)+C

Answer 19

\[\Large\rm \ln|y^2-5|=-\frac{1}{x}+c\]Ok good.
So now let's "start" to solve for y.
To get rid of the log we need to "exponentiate" each side.
(Rewrite each side as a exponent with a base of e)

Answer 20

\[\LARGE\rm e^{\ln|y^2-5|}=e^{-1/x+c}\]

Answer 21

The exponential and log function are inverses of one another,
so they simply "undo" one another.
Left side will simplify,\[\LARGE\rm y^2-5=e^{-1/x+c}\]

Answer 22

So then +5. And then square root the entire right side and you have y= something. Correct?

Answer 23

Let's not do that quite yet.
See how we have an unknown C value?
It'll be easier to find that if we don't square root everything.

Answer 24

Let's first apply rules of exponents to move the c.\[\Large\rm y^2-5=e^{-1/x}\cdot e^c\]This e^c is just a new arbitrary constant that can only be positive,
let's call it something else,\[\Large\rm y^2-5=Ke^{-1/x}\]

Answer 25

So now we'll use our initial data to solve for our unknown constant K.

Answer 26

Make sure you understand function notation:
\[\Large\rm y(1)=\sqrt6\quad\text{ means }\quad y=\sqrt6\quad\text{when}\quad x=1\]

Answer 27

Yes. So we need to find a C or K that makes that true.

Answer 28

Plug the stuff in! :)
Solve for K

Answer 29

\[\Large\rm (\sqrt6)^2-5=Ke^{-1/(1)}\]Solve for K!! :D

Answer 30

So is it 1?

Answer 31

\[\Large\rm 1=Ke^{-1},\qquad\qquad\to\qquad\qquad K=e^1\]yes?

Answer 32

Oh no I just redid it. And I get 1=.367879K which makes k=e

Answer 33

Oh boy.. you're one of those people huh -_-
with your decimal values and all that lol

Answer 34

It turns out that for this particular problem it would have made more sense for us to leave it in this form:\[\LARGE\rm y^2-5=e^{-1/x+c}\]Because when we plug our K=e^1 back in, and then combine the exponentials using rules of exponents we end up with:\[\LARGE\rm y^2-5=e^{-1/x+1}\]

Answer 35

And NOW you can finally solve for y! :)
We're dealt with the constant, it's ready to be put into an explicit solution form.

Answer 36

So now I would add the 5 and then take the square root?

Answer 37

yes

Answer 38

So y=sqrt(e^(-1/x+1)+5 is the final answer?

Answer 39

One sec I'll check it on wolfram to make sure we didn't make any mistakes.

Answer 40

I guess the only thing you'd want to be careful about is uhh...
See how you took the square root of a squared y?
So we should end up with a +/- in front of our root I guess.

Answer 41

Sweet! That was pretty complicated but I feel like I am getting the hang of it. Thanks a lot for all the help!

Answer 42

np \c:/