Which method do I use to solve

(1 + t^2) dy/dt + 2ty = -picos(pi*t) y(0) = 0

I couldn't find a method to match this DE

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- Sep 1st 2010, 04:06 AMCookieCfirst order DE
Which method do I use to solve

(1 + t^2) dy/dt + 2ty = -picos(pi*t) y(0) = 0

I couldn't find a method to match this DE - Sep 1st 2010, 04:13 AMKhonics89
Well I don't if you can see it or not, but the LHS says this

ay'+a'y = ay (product rule)

since d/dt(1+t^2) = 2t and y->y'

we can say LHS = (1+t^2)*y = -pic0s(pi*t)

then y=-picos(pi*t)/1+t^2

or you can simply use the p(x)=e^int(something) - Sep 1st 2010, 04:31 AMProve It
I think you actually mean

$\displaystyle ay' + a'y = \frac{d}{dt}(ay)$.

So that means

$\displaystyle \frac{d}{dt}[(1 + t^2)y] = -\pi \cos{(\pi t)}$.

Go from here.