Question

\[2(2xy + 4y - 3)dx + (x+2)^2 dy = 0\]

how to find the solution to this using linear DE?

Answer 1

for an equation in the form
\[ \frac{dy}{dx}+P(x)y= Q(x) \]
the integrating factor f(x) is
\[ f(x)= e^{\int{P(x) dx} }\]
and the solution is
\[ f(x)y= \int{f(x)Q(x)dx} +c \]

It looks like you can put this problem into this form.

Answer 2

how? divide by (x+2)^2 dx?

Answer 3

Divide both sides by dx:
\[2(2xy+4y-3)+(x+2)^2\frac{dy}{dx}=0\]
\[(x+2)^2 \frac{dy}{dx}+4xy+8y-6=0\]
Collect y terms together
leave y' and y terms on left hand side
everything else goes right hand side
\[(x+2)^2y'+(4x+8)y=6\]
Now to put this equation in the form @phi was talking about yes you need to divide both sides by (x+2)^2

Answer 4

OHHHH!!!! i see it now...thanks so much!!!

Answer 5

so it's linear in x right?

Answer 6

i mean no need for bernoulli's equation yada yada?

Answer 7

It is a first order linear differential equation
There has actually been a formula derived for these types

Answer 8

really? there is??

Answer 9

@phi actually mentions the formula above

Answer 10

the part where he says
"the solution is ..."

Answer 11

oh that..i thought you had a shortcut or something :p

Answer 12

You need to know what P(x) and Q(x) is
and find f(x) and just plug in

Answer 13

That is the shortcut lol

Answer 14

i thought that was the only way?

Answer 15

I can do these long way if I wanted

Answer 16

that means there's a bloodier way o.O

Answer 17

ok okay lets talk about the quadratic formula and completing the square
what phi gave you was like the "quadratic formula" but we could always "complete the square" instead of using the easy shortcut which is the "quadratic formula"

Answer 18

uhmm could we not :(

Answer 19

What?

Answer 20

Oh.

Answer 21

just let me believe this is the only way :P lol

Answer 22

lol. We don't have to "complete the square"
And I hope you know what I mean
I'm just trying to give you like analogy

Answer 23

hmm okay..maybe seeing it can help..

Answer 24

Ok hey I don't feel like typing it
I'm gonna draw it out

Answer 25

Is that okay?

Answer 26

sure

Answer 27

I mean scan is and junk
few moments please

Answer 28

okay :D

Answer 29

why did i uninstall that adobe reader :/ lol

Answer 30

Well I think what I just wrote can be found on a lot of different sites anyways

Answer 31

Let me see if I can find a good site

Answer 32

okay thanks :DDD

Answer 33

http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Answer 34

ohh paul's online notes..ive heard rumours bout him

Answer 35

So...what i was saying you can go through the steps that actually prove that formula or you can use the formula
You know just like if you wanted to use the steps that prove the quadratic formula (completing the square (and there is also my method of course ;) ) )
instead of using the quadratic formula

I actually like going the only way on these first order linear differential equations instead of using the formula

Answer 36

long*
not only sorry

Answer 37

why do you prefer the long way??

Answer 38

Because I like the extra work
I mean it is like poetry to me when I do it
lol
I'm weird okay?
I prefer the long way on a lot of things
like trig substitution and there are other things
I just like it
Don't know exactly why

Answer 39

lol =))) well i like trig sub too :p i usually prefer step0by-step rather than going by a formula because i dont want memorizing..but this lesson is just something that i should not do step by step lol :))))

Answer 40

It is not that bad. lol.

Answer 41

Hey one second I think I explained the long way on here before
Let me see if I can find it

Answer 42

hmm okay :) i hope it's not that hard...

Answer 43

wow I did it the long way here but this was a real confusing one for you at your stage I think
i didn't even remember doing this: http://openstudy.com/study#/updates/4e6d116f0b8b4a2b95e1bec4
I think I done an easier one before
let me see if I can find it

Answer 44

lol wow you werent kidding @_@

Answer 45

i cant even understand the first step :( lol

Answer 46

http://openstudy.com/study#/updates/4dd82fd8d95c8b0b735162c4

Answer 47

that might be a little easier because that actually was a first order linear differential equation

Answer 48

i think it resembles bernoulli's equation where you sub v = y^1-n or something like that

Answer 49

i dont think i want this method :( lol

Answer 50

Honestly I think I know less math now looking back at this.
That is crazy that first link I gave you lol