Superposition question given two diffeq's

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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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 $p y_1(0) + q y_2(0)$?
How do I get $p y_1'(t) + q y_2'(t)$? If $y_3(t) = p y_1(t) + q y_2(t)$, then $y_3'(t) = p y_1'(t) + q y_2'(t)$. Anyway, just use the superposition principle exactly as you stated it. In this case $r = 9.7$, $f(t) = e^{-0.4t} \cos(t)$, $p = 0$, ... – Robert Israel Sep 23 '11 at 21:14
Actually you have two different $p$'s and $q$'s, which makes things confusing. Let me restate the superposition principle with better notation. If $y_1(t)$ solves $y'(t) + r y(t) = f(t)$ with $y(0) = a$ and $y_2(t)$ solves $y'(t) + r y(t) = g(t)$ with $y(0) = b$, then $y_3(t) = p y_1(t) + q y_2(t)$ solves $y'(t) + r y(t) = p f(t) + q g(t)$ with $y(0) = p a + q b$. – Robert Israel Sep 25 '11 at 5:22