katieb
  • katieb
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Astrophysics
  • Astrophysics
I need help understanding what's going on in this example, \[(4+t^2)\frac{ dy }{ dt }+2ty=4t\] solve the O.D.E then it shows the left side \[(4+t^2)\frac{ dy }{ dt }+2y=\frac{ d }{ dt }[(4+t^2)y]\] \[\frac{ d }{ dt } [(4+t^2)y]=4t\]
Astrophysics
  • Astrophysics
Oops it says \[(4+t^2)\frac{ dy }{ dt }+2ty=\frac{ d }{ dt }[(4+t^2)y]\] forgot the t
Empty
  • Empty
Calculate the derivative on the right hand side of that

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ganeshie8
  • ganeshie8
They are simply using the product rule in reverse : \(f'g+fg' = (fg)'\)
ganeshie8
  • ganeshie8
This might be easy to compare against : \(fy' + f'y = (fy)'\)
Astrophysics
  • Astrophysics
OOOOOH
Empty
  • Empty
Yeah, when in doubt, work it out.
Astrophysics
  • Astrophysics
Thanks haha, I think this leads up to the integrating factor, so it's just another of the techniques
ganeshie8
  • ganeshie8
Exactly, that integrating factor business is all about making up the left hand side so that it is ready for using product rule in reverse combined with the fundamental theorem of calc( integral is the inverse of derivative realized up to a constant factor.. )
Astrophysics
  • Astrophysics
I see I see, thanks!
Astrophysics
  • Astrophysics
I've never really used the product rule in reverse, this is nifty
ganeshie8
  • ganeshie8
** realized up to a constant term
Astrophysics
  • Astrophysics
This is so amazing, omg
Empty
  • Empty
yup it's definitely one of my favorite tricks I wish all ODEs could be solved by integrating factors haha
ganeshie8
  • ganeshie8
Basically we can solve "ANY" differential equation of below form using that integrating factor trick : \[f(t)y' + g(t)y = h(t)\] the solution is given by : \[y(t) = \dfrac{\int (e^{\int (g/f)\, dt }h/f)dt}{e^{\int (g/f)\, dt }}\]
ganeshie8
  • ganeshie8
It is my fav too, and it really becomes powerful in cases where you want to make a DE exact...
Astrophysics
  • Astrophysics
Sweet haha, ok thanks guys, I may have more questions as I read more about ODE xD, as sometimes the book seems to skip steps -.-...
Astrophysics
  • Astrophysics
So if I have \[\frac{ d }{ dt }[(4+t^2)y]=4t \implies d[(4+t^2)y]=4tdt \implies (4+t^2)y=2t^2+C\] so the d cancels out the integral
Astrophysics
  • Astrophysics
I feel I should've known this
Astrophysics
  • Astrophysics
\[(4+t^2)y=\int\limits 4t dt \implies (4+t^2)y=2t^2+C\] to be more clear
Astrophysics
  • Astrophysics
This is so Newtonian!!
ganeshie8
  • ganeshie8
Haha I think it was euler...

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