i need help in finding the solution of this ordinary differential equation 3(3x^2+y^2)dx-2xydy=0 using any method.

- anonymous

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- goformit100

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- anonymous

Lets answer :)

- anonymous

@E.ali ok :D

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## More answers

- Mimi_x3

\[ 3(3x^2+ y^2)dx - 2xy\frac{dy}{dx}=0\]
\[ 9x^2 + 3y^2dx - 2xy\frac{dy}{dx}=0\]
\[ -2xy\frac{dy}{dx} + 3y^2dx + 9x^2 =0 => -2xy\frac{dy}{dx} + 3y^2dx = -9x^2\]
\[ 2ydy - \frac{3y^2}{x} = 9x\]
Let \(u = y^2\)
\[ \frac{du}{dx} = 2y\frac{dy}{dx}\]
\[ => \frac{du}{dx} - \frac{3u}{x} =9x\]
so it's a linear ode
where the solution is
\[ y = e^{-h(x)} [ \int e^{h(x)rdx +c\]

- anonymous

hint, equation is exact

- Mimi_x3

i dont think so..

- anonymous

ups, you right. Didn't see the - sign

- anonymous

\[3(3x ^{2}+y ^{2})dx-2xydy=0\]
\[9x ^{2}+3y ^{2} -2xydy=0\]
\[9dx-\frac{ 2xydy-3y ^{2}dx }{ x ^{2} }=0\]
There is an integrable combination wherein:
\[d(\frac{ y ^{2} }{ x })=\frac{ 2xydy-y ^{2}dx }{ x ^{2} }\]
However I am having problems with the \[-3y ^{2}\] in order to apply that integrable combination
:)

- anonymous

this is homogeneous ode. Write it like this:
\(y'=\huge\frac{2xy}{3x^2+y^2}\)
now reduce it single variable by multiplaying up and down part by \(1/x^2\) . Later you can solve it by taking variable change \(u=y/x\)

- anonymous

which makes it separate variable, and simple integration, :)

- anonymous

@Mimi_x3 u'r solution is wrong

- Mimi_x3

how lol?

- anonymous

where that 3 come out from?

- Mimi_x3

the question lol?

- anonymous

@myko uhm its \[3(3x ^{2}+y ^{2})\]
so if you get its partial derivatives its not homogeneous and there is also a negative sign O.O

- anonymous

Hey guys !
I have a better way !:)

- Mimi_x3

is the ode exact?

- anonymous

equation is homogenious

- anonymous

and u solve the way i explaned

- anonymous

@: myko : Wait please !

- terenzreignz

Bernoulli?

- anonymous

We have :
9x^2+2y^2.dx -2(x-y^2-d)=0
ok ?!

- anonymous

\[f(x,y) = 3(3x ^{2}+y ^{2})dx - 2xydy=0\]
\[f(\lambda x,\lambda y) = 3(3(\lambda x)^{2}+(\lambda y)^{2})- 2(\lambda x)(\lambda y) =0\]
so it is homogenous lol =))))

- anonymous

And now we can have :
9x^2+2y^2.dx=2(-x+y^2+d)
Ok?!

- anonymous

And now :!
CAN you complate it ?!

- anonymous

@E.ali can you use the equation button I can't understand your equation...

- anonymous

the idea is to first separate the variables all y's on one side all x's on the other side
to accomplish that, let's start by distributing in the first parenthesis
go ahead and give that a try please.

- anonymous

@ ; Archie :
We dont have just x and y .!
We have d to !

- anonymous

distributing means multiplying the factor in front of the parentheses by each and every term inside the parenthesis
here is an example: a(x+y) becomes ax + ay that is called distributing the a to the two terms inside the parenthesis.

- anonymous

Sure we do, but we are trying to do this step by step. if you want, you can distribute both the 3 and the dx at the same time

- anonymous

but I suggest you do it step by step

- anonymous

@ⒶArchie☁✪
\[9x ^{2}dx+3y ^{2}dx=2xy dy\]
\[9x ^{2}dx=2xy dy-3y ^{2}dx\]

- anonymous

am i doing it right?

- anonymous

Perfect :)

- anonymous

but you don't need to move the 3y^2dx term over to the right
let's keep that on the left, together with the other term that has dx in it
have a look:

- anonymous

|dw:1377938975495:dw|

- anonymous

alright what do we need to do next, to only have y's on the right hand side?

- anonymous

It s not important !

- anonymous

if it is helping yeyenunez learn, then I definitely think it is important E.ali.

- anonymous

also yes it is important we want the variables separated before we can integrate each side

- Callisto

\[3(3x^2+y^2)dx-2xydy=0 \]\[3(3x^2+y^2)dx=2xydy\]\[\frac{dy}{dx}=\frac{9x^2+3y^2}{2xy}\]\[\frac{dy}{dx}=\frac{9x}{2y}+\frac{3y}{2x}\]Let y=wx
\[\frac{dy}{dx}=w+\frac{dw}{dx}x\]
Then, the DE becomes
\[w+w'x = \frac{9}{2w}+\frac{3}{2}w\]\[w'x=\frac{9}{2w}+\frac{w}{2}\]\[w'x=\frac{9+w^2}{2w}\]\[w'x=\frac{9}{2w}+\frac{w}{2}\]\[\frac{dw}{\frac{9+w^2}{2w}}=\frac{dx}{x} \]

- anonymous

@ Archie : We have :
9x^2 + 3y ^2=2xy^2
Ha?!

- anonymous

@ⒶArchie☁✪ |dw:1377939104234:dw|?

- anonymous

Exacly !

- anonymous

My way is such as your way !

- anonymous

right

- anonymous

Then use Radical !

- anonymous

@E.ali i think you are not helping me at all :( I'm having problems with your equations because i cannot understand it at all

- anonymous

It s
9x + 3y =2xy
!@yeyenunez:Ok?!

- anonymous

@yeyenunez Can we start fresh here? There is a better way to solve this equation.

- anonymous

@ⒶArchie☁✪ ok

- anonymous

|dw:1377939464495:dw|

- anonymous

I've written the original DE on the board. A quick bit of theory next! :)
Suppose we have a function F(x,y)
and look at its differential:

- anonymous

@E.ali you cannot use radicals on terms with a plus sign in between ~_~ so 9x^2+3y^2 is not equal to 9x +3y ~_~

- anonymous

LOOK !
WE can Reduse from all .
We can have Radical from all ... and ...

- anonymous

|dw:1377939552415:dw|

- anonymous

So the differential dF can be expressed as this sum that involves partials of F with respect to X and Y.
I trust your teacher has covered this type of differential expression.

- anonymous

Are you familiar with partial derivatives? @yeyenunez

- terenzreignz

@yeyenunez As you have already found and pointed out, the equation is homogeneous... so why don't you start with letting
y = vx
and working from there? ^_^

- anonymous

This expression for dF is the general formula for a function of 2 variables
Just want to make sure you are ok with this general formula for dF and know about partial derivatives.

- anonymous

@ⒶArchie☁✪ we are only allowed to use partial derivatives when proving exactness of ode's

- anonymous

This is what we have here with this DE.
oh okay then, well I suggest you to look at the previous posts given by other people and go on from there.

- anonymous

@ⒶArchie☁✪ thanks for the help :)

- anonymous

Your welcome, good luck with your studies. [:

- anonymous

@terenzreignz wait i'll try using the homogenoeus method

- anonymous

@Callisto what type of method did you use?

- Callisto

Homogeneous.

- anonymous

@Callisto our homogenous method is quite different from yours... i don't think our professor would accept that kind of solution but still thanks for you help :)

- Callisto

What's the difference?

- anonymous

@yeyenunez perhaps you would like to share your homogenous method with us? [:

- anonymous

we use the y=vx and our equation must be in the form Pdx+Qdy=0

- terenzreignz

@yeyenunez out of interest, who IS your professor?
(UPD kid here ^_^)

- Callisto

It is just the same. I've just rearranged the terms.

- anonymous

@terenzreignz sir sigua =)))))

- terenzreignz

Neeever heard of him.
Anyway, just let y = vx and the rest falls into place.
What Calli did isn't that much different, she just delayed the substitution until a bit later into the solution.

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