Question

Solve the given differential equation.

\[\cfrac{dy}{dx} = \cfrac{(x+y)^2}{(x+2)(y-2)}\]

Answer 1

@ganeshie8 @Australopithecus @robtobey

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

What next?

Answer 6

That is in incorrect form

Answer 7

You want the form:

\[M(x,y)dx + N(x,y)dy = 0\]

Answer 8

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

Answer 9

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

Answer 10

For setting it in the right form do you understand?

Answer 11

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

Answer 12

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

Answer 13

I understand that point. But, am not sure where to put that \((x+y)^2\) ...

\((y-2)dy = \cfrac{dx}{x+2} \times (x+y)^2\)

Answer 14

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)

Answer 15

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

Answer 16

Okay, so, I have :

\((y-2)(x+2) dy - (x+y)^2 dx = 0 \)

This is the standard form, right?

Answer 17

Yup :D

Answer 18

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

Answer 19

Or substitution sorry

Answer 20

This is clearly not seperable

Answer 21

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?

Answer 22

Expand (y-2)(x+2)

Answer 23

Also expand (x+y)^2

Answer 24

MY browser is lagging like crazy one second

Answer 25

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

Answer 26

Okay, it will give me :

\[(y-2)(x+2) dy - (x+y)^2 dx = 0 \\
xydy + 2(y-x) dy - 4dy - x^2 dx -y^2 dx -2xy dx = 0 \\
xydy - 2y dy - 2xdy - x^2 dx - y^2 dx - 2xydx = 0 \]

Answer 27

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

Answer 28

Now factor out dy and dx

Answer 29

\[(xy - 2y - 2x)dy - (x^2 + y^2 + 2xy)dx = 0 \\
(xy - 2y - 2x)dy = (x^2 + y^2 + 2xy)dx \]

Seems like I'm reverting everything back to the original equation :/ :(

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

hence p = q

Answer 34

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

Answer 35

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.

Answer 36

Wait lets at least check to see

Answer 37

because you might have to use another method

Answer 38

Yeah sure!

Answer 39

Is M(x,y) homogenous?

Answer 40

and to what degree

Answer 41

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

Answer 42

If you are lost I dont mind giving you another example

Answer 43

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

Answer 44

So you have to use a different method

Answer 45

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

Answer 46

Yup

Answer 47

You can check to see if it is exact

Answer 48

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

Answer 49

its actually a pretty long method

Answer 50

kind of

Answer 51

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

Answer 52

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

Answer 53

A DE is said to be exact when,

\[M_y(x,y) = N_x(x,y)\]

You know how to take partial derivatives right?

Answer 54

Yeah!

Answer 55

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

Answer 56

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

Answer 57

if they are not this question is even longer ha

Answer 58

\(\delta\)

You mean this?

Answer 59

that is lower case delta

Answer 60

\(\Delta\) .. ?