Question

Solve the differential equation by using substitution for homogeneous equations. see attachment...

Answer 1

It's like everything goes great until i have to integrate the left side...always crashing at that

Answer 2

What was the original problem?

Answer 3

This?
\[\Large y' = \frac{y}{x+y}\]

Answer 4

yeah

Answer 5

\[\Large y\color{red}{dx}-(x+y)\color{blue}{dy}=0\]

Answer 6

divide by 1/x to get y/x and then subsitute it with the u, but I've learned a shortcut that cuts steps off...ummmm I have to use subsitution....

Answer 7

Letting \[\Large x = vy\]

Answer 8

like.... du/F(1,u)-u = dx/x that formula

Answer 9

\[\Large \color{red}{dx}=v\color{blue}{dy}+y\color{green}{dv}\]

Answer 10

\[\Large y(v\color{blue}{dy}+ y\color{green}{dv})- (vy + y)\color{blue}{dy}=0\]

Answer 11

Look good so far?

Answer 12

what the heck???!?!?!?!?!?!!?!? the methods I learned from my class seemed less letterish than this.

Answer 13

Want me to run them by you? :)

Answer 14

and I'm more concerned about the ending part like when you turn it into separate variables and antiderivative... the attachment has the answer but the crazy book has it at x=ylny+C which I don't get it because it involves some manipulation

Answer 15

\[\Large y(v\color{blue}{dy}+ y\color{green}{dv})- (vy + y)\color{blue}{dy}=0\]

Answer 16

Let's see if we can get that from here :D
Simplifying, we get
\[\Large y^2\color{green}{dv}- y\color{blue}{dy}=0\]

Answer 17

\[\Large y^2\color{green}{dv} = y \color{blue}{dy}\]

Answer 18

\[\Large \color{green}{dv} = \frac{\color{blue}{dy}}{y}\]

Answer 19

\[\Large \int\color{green}{dv} =\int \frac{\color{blue}{dy}}{y}\]

Answer 20

yeah which leaves a v = lny and subsitute v back... heyhow did you get it so fast?

Answer 21

Because... I used that 'letterish' method? :D

Answer 22

So, everything cleared up @UsukiDoll ?

Answer 23

that was...fast.....the du/F(1,u)-u = 1/x dx had some additional stuff.. what the heckkkkkkk

Answer 24

eating at the moment...sorry I feel peckish

Answer 25

math makes me hungry :S

Answer 26

mhmm well, if you're not fast, you're not doing it right XD

LOL
jk

Answer 27

alright I need the run down on the letterish method. seems like you got it fast like a boss

Answer 28

oh I see it now... x=vy and then product rule it...put it back in simplify if possible and then use separate variables. I should do that on number 5 and see if I could get it lightning fast like a boss.

Answer 29

ok nevermind....too many letters going on...just what the heck???????

Answer 30

@UsukiDoll sorry was preoccupied... still up for it?

Answer 31

nuh uh...what is the first step ? did you rearranged the problem?

Answer 32

\[\Large y' = \frac{y}{x+y}\]

Answer 33

\[\Large \frac{\color{blue}{dy}}{\color{red}{dx}}= \frac{y }{x+y}\]
Got it up to here?

Answer 34

yeah....

Answer 35

Okay, so that rearranges into
\[\Large (x+y)\color{blue}{dy}= y\color{red}{dx}\]

Answer 36

A little algebraic manipulation gives us
\[\Large y\color{red}{dx}-(x+y)\color{blue}{dy}=0\]

Answer 37

So far so good?

Answer 38

ya

Answer 39

So... now we have it of the form
\[\Large M(x,y) \color{red}{dx} +N(x,y) \color{blue}{dy}=0\]

Answer 40

Ever heard of homogenous functions?

Answer 41

just leaarned about it the other day...that's how I got the formula du/F(1,u)-u = 1/x dx

Answer 42

Well, that's just missing the essence of it :3
In our case, a function f of two variables f(x,y) is homogeneous with degree k if for any number \(\lambda\),
\[\Large f(\lambda x,\lambda y) = \lambda^kf(x,y)\] understood?

Answer 43

oh yeah to test out if the equation is homogeneous you replace the x and y's with the landa x landa y and then the landas cancell

Answer 44

more like it can be factored out.
Now, when written in this form
\[\Large M(x,y) \color{red}{dx} +N(x,y) \color{blue}{dy}=0\]
and M and N are both homogeneous *of the same degree* then you can work a little magic to turn it into a separable equation :)

Answer 45

So... in our particular case
\[\Large y\color{red}{dx}-(x+y)\color{blue}{dy}=0\]
y and (x+y) are both homogeneous of degree 1, aren't they? :)

Answer 46

why ia there a negative at -(x+y) dy?

Answer 47

Do I really have to answer that? -.-

Answer 48

no it was moved over...

Answer 49

or switch places

Answer 50

Okay, back to reality... since we know both terms here are homogeneous with degree 1,
\[\Large y\color{red}{dx}-(x+y)\color{blue}{dy}=0\]

We can either let x = vy
or y = vx

Answer 51

I pick x, well, because the dx terms is simpler, being just a monomial and all.

Answer 52

So we let x = vy.
Note that this is the same as letting
\[\Large v = \frac{x}y\]

Answer 53

You copy? :)

Answer 54

yup... the v variable is by itself also. x = yv x' yv'+v

Answer 55

\[\Large x = vy\]\[\Large \frac{\color{red}{dx}}{\color{blue}{dy}}=v + y\frac{\color{green}{dv}}{\color{blue}{dy}}\]Yup ^_^

Answer 56

Multiplying both sides by \(\color{blue}{dy}\) gives
\[\Large \color{red}{dx}= v\color{blue}{dy} + y\color{green}{dv}\]

Answer 57

and then subsitute that back into the equation

Answer 58

Just do the necessary substitutions here:
\[\Large y\color{red}{dx}-(x+y)\color{blue}{dy}=0\]
Replacing dx with that new differential expression and x with vy

Answer 59

And everything follows ^_^

Answer 60

what about for y' = x divided by x+y ?

Answer 61

this is what I got so far.

dy/dx = x/x+y
(x+y)dy = xdx
xdy+ydy-xdx=0
xdx-(xdy+ydy)=0
x(v+ydv)-xdy-ydy=0
xv+xydv-xdy-ydy=0
v^2y+y^2vdv-vydy-ydy = o
I don't see any cancellations whatsoever..

Answer 62

Sorry again... let me see...

Answer 63

We start with this one...
\[\Large x\color{red}{dx}-(x+y)\color{blue}{dy}=0\]

Answer 64

And this still holds:
\[\Large \color{red}{dx}= v\color{blue}{dy} + y\color{green}{dv}\]

Answer 65

Ready?

Answer 66

@UsukiDoll talk to me ^_^

Answer 67

k k hang on ... let me write this down...

Answer 68

k got it

Answer 69

Okay, we let x = vy
and dx = vdy + ydv
\[\Large vy(v\color{blue}{dy}+y\color{green}{dv})-(vy+y)\color{blue}{dy}=0\]

Answer 70

Actually, the whole point of this substitution wasn't really to make things cancel out, it's just to make the entire differential equation separable! Okay? :)

We were just lucky the first time around, when things cancelled out.

Answer 71

yeah but this time it's what the heckk....

Answer 72

So, don't be afraid and let's just simplify :3
\[\Large v^2y\color{blue}{dy}+vy^2\color{green}{dv}-y(v+1)\color{blue}{dy }=0\]

Answer 73

Got that?

Answer 74

for the -y(v+1)dy can you distribute the dy?

Answer 75

Yes, but that only serves to complicate things.
In fact, as much as possible, we'd only like one dy and one dv. So instead of distributing the dy, we should just factor it out... but later. First, let's bring all the dy's to the right-side.

Answer 76

\[\Large vy^2\color{green}{dv}=y(v+1)\color{blue}{dy }-v^2y\color{blue}{dy}\]
Catch me so far?

Answer 77

ya

Answer 78

And THEN factor out the dy.
\[\Large vy^2\color{green}{dv}=[y(v+1)-v^2y]\color{blue}{dy}\]

Answer 79

got it

Answer 80

Actually, we can factor out the y on the right-hand side.
\[\Large vy^2\color{green}{dv}=y[v+1-v^2]\color{blue}{dy}\]

Answer 81

And the entire thing is now separable...
\[\Large \frac{v}{v+1-v^2}\color{green}{dv}= \frac{\color{blue}{dy}}{y}\]

Answer 82

ok then multiply the dsafsdfsdadfafsd more like divide the v's to the left and the y^2 goes to the right

Answer 83

Eventually you end up with that ^
Good luck integrating that, though :/
Looks terrible

Answer 84

b^2-4ac will tell me whether it's factorable.

Answer 85

It kind of is, it's just not pretty D:

Answer 86

and it's not..... gotta use quadratics... I know what it is it's 1/2+ - sqrt 5

Answer 87

You do that dirty bit... :3

Answer 88

but how to put it in place

Answer 89

partial fractions?

Answer 90

ha ha tried to fdo that yesterday

Answer 91

I'm not the best guy to help out with that... I get dizzy with equations :/

Answer 92

how is that going to work when it's not factorable to begin with unless you mean Ax+ nB all over that v^2-v-1

Answer 93

It is factorable... it's just that its roots aren't pretty... what ARE its roots anyway?

Answer 94

been there it's -1/2-sqroot5/2

Answer 95

and -1/2 +sqroot(5)/2

Answer 96

I believe it's 1/2 and not -1/2