Solve the given differential equation.
\[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y-2)}\]

- mathslover

Solve the given differential equation.
\[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y-2)}\]

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- mathslover

@ganeshie8 @Australopithecus @robtobey

- Australopithecus

What methods have you learned? I don't think this is separable

- mathslover

I tried putting y = vx and x + y = v but none of them seem to work.

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## More answers

- Australopithecus

First bring everything to one side and make it equal to zero

- mathslover

\(\cfrac{(x+2)(y-2) dy}{(x+y)^2 dx} - 1 = 0 \)
What next?

- Australopithecus

That is in incorrect form

- Australopithecus

You want the form:
\[M(x,y)dx + N(x,y)dy = 0\]

- Australopithecus

You can use substitution or you can check to see if it is exact and use that method

- Australopithecus

Substitution is as follows:
You have two choices for substitution:
u = x/y
then,
dx = udy + ydu
Or
u = y/x
then,
dy = xdu + udx

- Australopithecus

For setting it in the right form do you understand?

- mathslover

Yeah, I've studied the method of substitution. In fact, I did try substituting y = vx but that didn't seem to work.

- Australopithecus

You wrote it in the wrong form so that could be your problem

- mathslover

I understand that point. But, am not sure where to put that \((x+y)^2\) ...
\((y-2)dy = \cfrac{dx}{x+2} \times (x+y)^2\)

- Australopithecus

So, for example:
\[\frac{dx}{dy} = \frac{(y + x)}{(x + 2y)^2}\]
in standard form would be:
\[dx(x+2y)^2 - (y+x)dy = 0\]
M(x,y) = (x + 2y)^2
N(x,y) = -(y+x)

- Australopithecus

Sorry this would be standard form
\[(x+2y)^2dx - (y+x)dy = 0\]

- mathslover

Okay, so, I have :
\((y-2)(x+2) dy - (x+y)^2 dx = 0 \)
This is the standard form, right?

- Australopithecus

Yup :D

- Australopithecus

Ok now you have to test to see if you can even use the separation method

- Australopithecus

Or substitution sorry

- Australopithecus

This is clearly not seperable

- mathslover

Now, I will simply use substitution : y = vx
\(dy = vdx + xdv \)
The equation becomes:
\((vx - 2)(x+2)(vdx + xdv) -(x+vx)^2 dx = 0 \)
\((vx-2)(x+2)(vdx + xdv) = x^2 (v+1)^2 dx \)
Should I proceed further or am I doing something wrong?

- Australopithecus

Expand (y-2)(x+2)

- Australopithecus

Also expand (x+y)^2

- Australopithecus

MY browser is lagging like crazy one second

- Australopithecus

there is a test you have to do to make sure you can even use substitution method

- mathslover

Okay, it will give me :
\[(y-2)(x+2) dy - (x+y)^2 dx = 0 \\
xydy + 2(y-x) dy - 4dy - x^2 dx -y^2 dx -2xy dx = 0 \\
xydy - 2y dy - 2xdy - x^2 dx - y^2 dx - 2xydx = 0 \]

- mathslover

Test? Never heard of it. Can you please explain?

- Australopithecus

Now factor out dy and dx

- mathslover

\[(xy - 2y - 2x)dy - (x^2 + y^2 + 2xy)dx = 0 \\
(xy - 2y - 2x)dy = (x^2 + y^2 + 2xy)dx \]
Seems like I'm reverting everything back to the original equation :/ :(

- Australopithecus

You can only do substitution method if
\[M(\lambda x, \lambda y) = \lambda^pM(x,y)\ for\ all\ \lambda,x,y\]
and
\[N(\lambda x, \lambda y) = \lambda^qN(x,y)\ for\ all\ \lambda,x,y\]
to use separation p must equal q

- Australopithecus

First to check if M(x,y) or N(x,y) is homogenous here are some examples:

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- Australopithecus

The degree of homogeneity must be equal for both terms N(x,y) and M(x,y) to use the substitution method

- Australopithecus

hence p = q

- Australopithecus

its a pretty simple check you can do in your mind almost

- mathslover

I seem to get your point! Will give it a try next morning. Gotta sleep now! :)
Thanks a lot for your help! I'll let you know if I get stuck anywhere.

- Australopithecus

Wait lets at least check to see

- Australopithecus

because you might have to use another method

- mathslover

Yeah sure!

- Australopithecus

Is M(x,y) homogenous?

- Australopithecus

and to what degree

- Australopithecus

Sorry I know you are probably tired just I want to save you time

- Australopithecus

If you are lost I dont mind giving you another example

- mathslover

It doesn't seem to be homogeneous :( because of that constant there

- Australopithecus

So you have to use a different method

- mathslover

Oh! :( Any ideas for which method to use?

- Australopithecus

Yup

- Australopithecus

You can check to see if it is exact

- Australopithecus

Do you want to sleep we can work on this later if you want?

- Australopithecus

its actually a pretty long method

- Australopithecus

kind of

- mathslover

Oh! Well, I will love to give it a try now. I can control my sleep for studies!

- Australopithecus

Alright, first step you need to test to see if the DE is exact or not

- Australopithecus

A DE is said to be exact when,
\[M_y(x,y) = N_x(x,y)\]
You know how to take partial derivatives right?

- mathslover

Yeah!

- Australopithecus

Also I would put it in another form but I am not sure how to use write daran in latex

- Australopithecus

So first step take both partial derivatives and check to see if they are equal

- Australopithecus

if they are not this question is even longer ha

- mathslover

\(\delta\)
You mean this?

- Australopithecus

that is lower case delta

- mathslover

\(\Delta\) .. ?

- Australopithecus

|dw:1432921439849:dw|

- Australopithecus

anyways we are getting off topic

- Australopithecus

Have you taken the partial derivatives? You should always take both of them because you will always need them to solve this kind of problem

- mathslover

I will do that part. What I'm thinking right now is that this question appeared in my book before the topic of "Exact DE" ... and it wil surely involve the method of solving H.D.E

- Australopithecus

I Dont know the HDE method

- mathslover

I meant, solving Homogeneous Differential Equations

- Australopithecus

oh im out of my mind lol

- Australopithecus

anyways this is the I would use to solve this problem

- Australopithecus

it should work

- Australopithecus

I mean unless you make a mistake with algebra substitution method wont work with this

- mathslover

Well, I guess, I might have made a mistake then. I will see it tomorrow morning and will let you know about the progress, if you don't mind. As I'm getting a bit out of mind too :(

- Australopithecus

That's fine just message me or tag me or something I can keep going over this with you,
This is essentially a really brief note on how to solve this problem IF it is exact

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- Australopithecus

If you need help later have a nice sleep

- mathslover

:) Thanks a lot for that, @Australopithecus . And same to you (whenever you schedule to sleep :D )

- Australopithecus

It is never

- mathslover

Sleep is important ! Have one now, please! :D

- mathslover

Well, I got it! We had to substitute x + 2 = u and y -2 = v and then work out with y = kx

- anonymous

The ODE as given isn't homogeneous. For \(\dfrac{dy}{dx}=f(x,y)\) to be considered homogeneous, \(f\) has to satisfy \(f(tx,ty)=t^kf(x,y)\) for some real \(k\).
\[\begin{align*}f(x,y)&= \frac{(x+y)^2}{(x+2)(y-2)}\\\\
f(tx,ty)&= \frac{(tx+ty)^2}{(tx+2)(ty-2)}\\
&= t^2\frac{(x+y)^2}{(tx+2)(ty-2)}\\
&\neq t^kf(x,y)
\end{align*}\]
which explains why you were running into trouble with the substitution \(y=vx\).
As for your substitution, I was actually in the process of suggesting that! Nice work.

- mathslover

Yeah! :) I don't know how did it click just when I woke up ! :D
Indeed, I solved it in 5 minutes :P As compared to hour of work last night.
And thanks for that explanation, @SithsAndGiggles ! :)

- anonymous

Sometimes the hardest thing is remembering that \(2-2=0\) :P

- mathslover

Haha! Agreed. This is what "Maths" is...! Happens in programming too, when the error is ";" .. and we keep finding it throughout the code :D It's fun...

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