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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 :) ?

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(This should be a comment, but for some strange reason I'm suddenly unable to make a comment)

First of all, the question is stated wrong: either there should be $e^{-.4 t} \cos(t)$ and $3 e^{-.4 t} \cos(t)$, or $e^{-.4 t} + \cos(t)$ and $3 (e^{-.4 t} + \cos(t))$, on the right sides of two of the equations.

Then you don't need to solve the differential equations (which you did wrong in any case), just use the superposition principle. If y_1'(t) + 9.7 y_1(t) = f(t) and y_2'(t) + 9.7 y_2(t) = 0, what is (p y_1'(t) + q y_2'(t)) + 9.7 (p y_1(t) + q y_2(t))? If $y_1(0) = 0$ and $y_2(0) = 1$, what is $p y_1(0) + q y_2(0)$?

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    @riotburn: no, because you need to also fit the initial condition.2011-11-22