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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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I fixed the equation. This is all I was given on the superposition principle: The superposition principle says: If y1[t] solves y'[t]+r y[t]=f[t] with y[0]=p and if y2[t] solves y'[t]+r y[t]=g[t] with y[0]=q then y3[t]=y1[t]+y2[t] solves y'[t]+r y[t]=f[t]+g[t] with y[0]=p+q. I dont understand how you get py1'(t) + qy2'(t) in the equation. –  riotburn Sep 23 '11 at 12:36
    
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
    
Oh okay but isn't p = 0, q = 1? The 3 in the y3[t] equation is really throwing me off. I can find the solutions to y1[t] and y2[t]. So if I plug in those solutions into the y3[t] and y3'[t] can't I solve for p and t? –  riotburn Sep 23 '11 at 22:11
    
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
    
oh, so is it simply just p = 3 and q = 0 if y1(t) solves y′[t]+9.7y[t]=e−0.4tcos[t]? –  riotburn Sep 25 '11 at 15:17
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