Jack1
  • Jack1
hey guys, how would I find the general solution to this? (x+2)^2 dy/dx = -4xy - 8y +5 ... pretty please?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Jack1
  • Jack1
@terenzreignz or @oldrin.bataku ... if you've got a sec?
terenzreignz
  • terenzreignz
I need a moment to digest this... if I can XD
cwrw238
  • cwrw238
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)

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Jack1
  • Jack1
cant isolate to get integrating factor i understand the general way to do it but this and one other q has me stumped
terenzreignz
  • terenzreignz
Allow me to do just that... First, \[\Large \frac{dy}{dx}=\left[ \frac{-4x-8}{(x+2)^2}\right]y+\frac5{(x+2)^2}\]
Jack1
  • Jack1
hmmm... ok
Jack1
  • Jack1
so e^( -4 ln x+2) = integrating factor?
terenzreignz
  • terenzreignz
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
terenzreignz
  • terenzreignz
oh sorry, my internet, it plays games with me :D
Jack1
  • Jack1
no, integrating factor equals integral of e^( -4 ln x+2) so = ( -1/ 3(x^2 + 2)^3) ...?
terenzreignz
  • terenzreignz
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) )
Jack1
  • Jack1
i got it now, thanks for clearing up the origioinal equation, that's where i was struggling
terenzreignz
  • terenzreignz
Okay great :) I'll post my solutions here... maybe you could do the same, see if we get the same answers...
terenzreignz
  • terenzreignz
Getting this integral...\[\Large \int \frac4{x+2}dx\] \[\Large = 4\ln (x+2)\] So (my) integrating factor is \[\Large e^{4\ln(x+2)}=(x+2)^4\]
terenzreignz
  • terenzreignz
Multiplying this with both sides of the differential equation yields \[\Large (x+2)^4\frac{dy}{dx} +4(x+2)^3y=5(x+2)^2\]
terenzreignz
  • terenzreignz
The left side is now equal to... \[\Large \frac{d}{dx}\left[(x+2)^4y\right]= 5(x+2)^2\]
Jack1
  • Jack1
so y = 5/3(x+2) + c / ( x+2)^4 ...?
terenzreignz
  • terenzreignz
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... \[\Large (x+2)^4y=(x+2)^5+C\]
terenzreignz
  • terenzreignz
I think we can stop here, only trimming the equation is necessary, I'll leave that to you :3
terenzreignz
  • terenzreignz
This is my problem with your way of using integrating factor, it gives you that annoying fraction :3
Jack1
  • Jack1
... i got a different answer to you tho...?
Jack1
  • Jack1
which is correct?
Jack1
  • Jack1
i got it down to d/dx ((x+2)^4) y = 5 (x+2)^2
Jack1
  • Jack1
then came up with y = 5/3(x+2) + c / ( x+2)^4
terenzreignz
  • terenzreignz
Better consult some experts :D @cwrw238 ?
Jack1
  • Jack1
@cwrw238 ...? please?
Jack1
  • Jack1
or @ganeshie8 ...?
Jack1
  • Jack1
ah, sall good, im sure i can get part marks in exam if i show enough working
terenzreignz
  • terenzreignz
No luck so far? :D
Jack1
  • Jack1
nada yet
terenzreignz
  • terenzreignz
urgh
Jack1
  • Jack1
meh, s'cool, maybe i should go and re-watch the lecture on it again
Jack1
  • Jack1
cheers anyway terenz
terenzreignz
  • terenzreignz
Cheers Jack :D
terenzreignz
  • terenzreignz
Signing off... ---------------------------------- Terence out
anonymous
  • anonymous
$$(x+2)^2 \frac{dy}{dx}=-4y(x+2)+5\\\frac{dy}{dx}=-\frac{4y}{\underbrace{x+2}_u}+\frac5{(x+2)^2}\\\frac{dy}{du}=-\frac{4y}u+\frac5{u^2}\\\frac{dy}{du}+\frac4uy=\frac5{u^2}$$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 \(\mu=u^4\):$$u^4\frac{dy}{du}+4u^3y=5u^2\\(u^4y)'=5u^2\\u^4y=\int5u^2du=\frac53u^3+C\\y=\frac5{3u}+\frac{C}{u^4}=\frac5{3(x+2)}+\frac{C}{(x+2)^4}$$
anonymous
  • anonymous
@terenzreign you screwed up your *exponent* mate.
cwrw238
  • cwrw238
good work oldrin
anonymous
  • anonymous
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 \(c_1=45+C\).
anonymous
  • anonymous
oops I meant for \((x+2)^4\) to be in that last denominator
terenzreignz
  • terenzreignz
So I did. Oh well... Better pay closer attention to details next time~
Jack1
  • Jack1
cheers for that @oldrin.bataku and also @terenzreignz

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