So I go this problem: $\ y1[t]\text{ solves } y^\prime[t]+9.7 y[t]= e^{-0.4t} \cos[t],\text{ with } y[0]=0,$
and
$y2[t]\text{ solves } y^\prime[t]+9.7 y[t]=0,\text{ with } y[0]=1. $
What numbers $p$ and $q$ do you pick to make $y[t]=p y1[t] + q y2[t]$
solve $ y^\prime[t]+9.7 y[t]=3 e^{-0.4 t} \cos[t],\text{ with } y[0]=3\text{?} $ I found that: $ \ y1[t] = 0.0980487e^{-9.7t} (1 + \sin(t) + \cos(t)) $ and $ \ y2[t] = e^{-9.7} $ I tried solving $y[t]$ and getting two equations but keep getting stuck on how to find $p$ and $q$ ...any hints :) ?