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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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)
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.
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