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Astrophysics
 one year ago
@ganeshie8
Astrophysics
 one year ago
@ganeshie8

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1Oops it says \[(4+t^2)\frac{ dy }{ dt }+2ty=\frac{ d }{ dt }[(4+t^2)y]\] forgot the t

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Calculate the derivative on the right hand side of that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4They are simply using the product rule in reverse : \(f'g+fg' = (fg)'\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4This might be easy to compare against : \(fy' + f'y = (fy)'\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, when in doubt, work it out.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Thanks haha, I think this leads up to the integrating factor, so it's just another of the techniques

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Exactly, 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
 one year ago
Best ResponseYou've already chosen the best response.1I see I see, thanks!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I've never really used the product rule in reverse, this is nifty

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4** realized up to a constant term

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1This is so amazing, omg

Empty
 one year ago
Best ResponseYou've already chosen the best response.1yup it's definitely one of my favorite tricks I wish all ODEs could be solved by integrating factors haha

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Basically 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
 one year ago
Best ResponseYou've already chosen the best response.4It is my fav too, and it really becomes powerful in cases where you want to make a DE exact...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Sweet 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
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.1I feel I should've known this

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[(4+t^2)y=\int\limits 4t dt \implies (4+t^2)y=2t^2+C\] to be more clear

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1This is so Newtonian!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Haha I think it was euler...
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