austinL
  • austinL
Differential Equations Initial Value Problem I worked it out, but I arrived at a different answer than is given in my text book. \(ty\prime + (t+1)y=t\) \(y(\ln(2))=1,~t>0\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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austinL
  • austinL
I arrived at y = t + 1 - ln(2) My books answer.... \(y=\dfrac{(t-1+2e^{-t})}{t},~t\ne0\)
terenzreignz
  • terenzreignz
You have a new DE and you didn't invite me? T.T
austinL
  • austinL
@terenzreignz :P

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austinL
  • austinL
My way is clearly incorrect. I think I need to treat the t+1 term differently. Any thoughts?
terenzreignz
  • terenzreignz
I say go for that integrating factor again... After dividing everything by t, then we get the integrating factor ro be \[\Large \mu = \int \frac{t+1}{t}dt\]
austinL
  • austinL
Darn it.... stupid stupid stupid stupid stupid Austin.
terenzreignz
  • terenzreignz
Whoops... I meant...\[\LARGE e^\mu\] as the integrating factor...
austinL
  • austinL
\(\huge{\mu(t)=te^t}\)
austinL
  • austinL
Look ok?
terenzreignz
  • terenzreignz
That's the integrating factor itself, so... okay, great :D Multiply everything with it... you get \[\Large te^t y' + (t+1)e^ty = te^t\]
terenzreignz
  • terenzreignz
That's the integrating factor itself, so... okay, great :D Multiply everything with it... you get \[\Large te^t y' + (t+1)e^ty = te^t\]
austinL
  • austinL
\(\dfrac{(t+1)}{t}\) Right?
terenzreignz
  • terenzreignz
What do you mean? THIS term? \[\Large te^t y' +\color{red}{ (t+1)e^ty} = te^t\]
austinL
  • austinL
Yes.
anonymous
  • anonymous
the standard equation for a linear differential equation is y' +Py = Q the first step towards solving the differential equation lies in identifying weather the given equation is in standard form or not. This can be achieved by comparing the given equation with the standard form. By doing so, we find that the given eq. isn't in the standard form so we must do certain operations to bring it to the standard form in this case We divide the both sides with 't' to obtain y' + (t+1)/t = 1 here, P=(t+1)/t or P =t/t +1/t or P= 1+1/t IF = e^int(p dt) IF =e^[int(1) + int(1/t )] IF =e^[t+log t] IF = e^t + e^log t IF = e^t + t the solution is given by ye^int(p dx) = int(Q e^int(p dx))
terenzreignz
  • terenzreignz
Remember, we started with this: \[\Large t y' + (t+1)y = t\] divided everything by t \[\Large y' + \frac{(t+1)}ty =1\] Multiplied everything by \(\large te^t\), the integragint factor, arriving finally at: \[\Large te^t y' + (t+1)e^ty = te^t\] Questions? :P
austinL
  • austinL
Oh... duh Austin. I swear I don't function without caffeine. I feel like I just went full dipstick.
terenzreignz
  • terenzreignz
Okay, well, I don't need caffeine, but some mocha ice cream would be awsum ^_^ Anyway, left side is now \[\Large \frac{d}{dt}[te^ty]= te^t\] Let's have at it Austin :3
austinL
  • austinL
Yep, I have worked that out. And then I have arrived at C=2
austinL
  • austinL
\(te^ty=(t-1)e^t+2\)
terenzreignz
  • terenzreignz
Does that work out?
terenzreignz
  • terenzreignz
Oh, it IS right, then?
austinL
  • austinL
I believe so!
terenzreignz
  • terenzreignz
Then... ice cream time :D More ice cream, that is ^_^
austinL
  • austinL
http://www.simplyrecipes.com/wp-content/uploads/2007/04/coffee-ice-cream.jpg?ea6e46
terenzreignz
  • terenzreignz
Lovely :D Well, I'm off ^_^ Have fun XD ------------------------------ Terence out
austinL
  • austinL
Thanks for all of your help Terence!

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