Question

hey guys, how would I find the general solution to this?
(x+2)^2 dy/dx
= -4xy - 8y +5 ... pretty please?

Answer 1

@terenzreignz or @oldrin.bataku ... if you've got a sec?

Answer 2

I need a moment to digest this... if I can XD

Answer 3

i'm studying these at the moment but it looks too tricky for me

I think you would start by dividing through by (x+2)^2

to get it into the form dy/dx = Py + Q

where P and Q are functions of x

then you would use an integrating factor I

I = e^ (integral Pdx)

Answer 4

cant isolate to get integrating factor
i understand the general way to do it but this and one other q has me stumped

Answer 5

Allow me to do just that... First,
\[\Large \frac{dy}{dx}=\left[ \frac{-4x-8}{(x+2)^2}\right]y+\frac5{(x+2)^2}\]

Answer 6

hmmm... ok

Answer 7

so e^( -4 ln x+2) = integrating factor?

Answer 8

So we'd get...
\[\Large \frac{dy}{dx}+\left[ \frac{4x+8}{(x+2)^2}\right]y=\frac5{(x+2)^2}\]

IN FACT
Why don't we go ahead and factor out 4 here...
\[\Large \frac{dy}{dx}+\left[ \frac{\color{red}{4(x+2)}}{(x+2)^2}\right]y=\frac5{(x+2)^2}\]

I bet it looks so much better now XD

Answer 9

oh sorry, my internet, it plays games with me :D

Answer 10

no, integrating factor equals integral of e^( -4 ln x+2)
so = ( -1/ 3(x^2 + 2)^3) ...?

Answer 11

It must be a method I'm yet unfamiliar with, but it should go smoothly once you know your integrating factor, right?

(The way I do it has y' + P(x)y = Q(x) )

Answer 12

i got it now, thanks for clearing up the origioinal equation, that's where i was struggling

Answer 13

Okay great :)
I'll post my solutions here... maybe you could do the same, see if we get the same answers...

Answer 14

Getting this integral...\[\Large \int \frac4{x+2}dx\]
\[\Large = 4\ln (x+2)\]

So (my) integrating factor is
\[\Large e^{4\ln(x+2)}=(x+2)^4\]

Answer 15

Multiplying this with both sides of the differential equation yields
\[\Large (x+2)^4\frac{dy}{dx} +4(x+2)^3y=5(x+2)^2\]

Answer 16

The left side is now equal to...
\[\Large \frac{d}{dx}\left[(x+2)^4y\right]= 5(x+2)^2\]

Answer 17

so y = 5/3(x+2) + c / ( x+2)^4
...?

Answer 18

Now essentially a separable differential equation, we get
\[\Large d\left[(x+2)^4y\right]=5(x+2)^4dx\]

integrate both sides
\[\Large \int d\left[(x+2)^4y\right]=\int5(x+2)^4dx\]
We get...
\[\Large (x+2)^4y=(x+2)^5+C\]

Answer 19

I think we can stop here, only trimming the equation is necessary, I'll leave that to you :3

Answer 20

This is my problem with your way of using integrating factor, it gives you that annoying fraction :3

Answer 21

... i got a different answer to you tho...?

Answer 22

which is correct?

Answer 23

i got it down to d/dx ((x+2)^4) y = 5 (x+2)^2

Answer 24

then came up with y = 5/3(x+2) + c / ( x+2)^4

Answer 25

Better consult some experts :D
@cwrw238 ?

Answer 26

@cwrw238 ...? please?

Answer 27

or @ganeshie8 ...?

Answer 28

ah, sall good, im sure i can get part marks in exam if i show enough working

Answer 29

No luck so far? :D

Answer 30

nada yet

Answer 31

urgh

Answer 32

meh, s'cool, maybe i should go and re-watch the lecture on it again

Answer 33

cheers anyway terenz

Answer 34

Cheers Jack :D

Answer 35

Signing off...
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Terence out

Answer 36

$$(x+2)^2 \frac{dy}{dx}=-4y(x+2)+5\\\frac{dy}{dx}=-\frac{4y}{\underbrace{x+2}_u}+\frac5{(x+2)^2}\\\frac{dy}{du}=-\frac{4y}u+\frac5{u^2}\\\frac{dy}{du}+\frac4uy=\frac5{u^2}$$We want an integration factor \(\mu\) s.t. $$\mu\frac{dy}{du}+\frac4u\mu y=(\mu y)'=\mu\frac{dy}{dx}+\mu'y$$It's clear, then, that we want \(\dfrac4u\mu=\mu'\). This is separable:$$\frac4udu=\frac1\mu d\mu\\\int\frac4udu=\int\frac1\mu d\mu\\4\log u=\log\mu\\u^4=\mu$$Now multiply throughout by \(\mu=u^4\):$$u^4\frac{dy}{du}+4u^3y=5u^2\\(u^4y)'=5u^2\\u^4y=\int5u^2du=\frac53u^3+C\\y=\frac5{3u}+\frac{C}{u^4}=\frac5{3(x+2)}+\frac{C}{(x+2)^4}$$

Answer 37

@terenzreign you screwed up your *exponent* mate.

Answer 38

good work oldrin

Answer 39

Let's compare with WolframAlpha's output: http://www.wolframalpha.com/input/?i=%28x%2B2%29%5E2+y%27+%2B+4y%28x%2B2%29+%3D+5
$$y=\frac5{3(x+2)}+\frac{C}{(x+2)^4}\quad\text{vs}\quad y=\frac{c_1}{(x+2)^4}+\frac{5x^3}{3(x+2)^4}+\frac{10x^2}{(x+2)^4}+\frac{20x}{(x+2)^4}$$... huh? These look pretty different! In reality, though, they're not:$$\frac5{3(x+2)}+\frac{C}{(x+2)^4}=\frac{5(x+2)^3+3C}{3(x+2)^4}=\frac{5(x^3+6x^2+12x+27)+3C}{3(x+2)^4}$$Now split this into separate terms:$$\frac{5x^3}{3(x+2)^4}+\frac{10x^2}{(x+2)^4}+\frac{20x}{(x+2)^4}+\frac{27(5)+3C}{3(x+2)^4}$$Paying closer attention to our last term, we see:$$\frac{3(45+C)}{3(x+2)^4}=\frac{45+C}{(x+)^4}$$... so we see our solutions are equivalent where \(c_1=45+C\).

Answer 40

oops I meant for \((x+2)^4\) to be in that last denominator

Answer 41

So I did. Oh well...
Better pay closer attention to details next time~

Answer 42

cheers for that @oldrin.bataku and also @terenzreignz