Solve a Linear Homogeneous Equation Initiate Value Problem (complex)

Question

I have this exercise on Linear Homogeneous Equation Initiate Value, and i'm kind of stuck. I don't know if what i'm about to ask exists already here as an answer, but i'm kind of newbie and i don't use the same symbols yet so i can't understand neither the question nor the answer. If you think it already exists, just point me that way. So here is the problem:

$y_{x+3} - y_{x+2} + 4y_{x+1} -4y_x=0, y_0=2, y_1=3, y_2=-3$

The corresponding characteristic equation is:

$l^3 - l^2 +4l -4 =0$, with roots $l=1$ or $l=2i$ or $l=-2i$

Now, i'm supposed to apply these:

$a=r\cos u$, $b=r\sin u$ where $l=a+bi$ $=>r=\sqrt(a^2+b^2)$ and $tgu=b/a$ , $r>0$, $u \in [0,2π).$

Pretty simple so far.

I said that $n=π/2 $, $r=2$ and this is my general solution:

$y_x=c_1*1^x + c_2*2^x*\cos(π*x/2) + c_3*2^x\sin(π*x/2)$, where $c_2*2^x*\cos(π*x/2) = 0$

Then, i apply the first conditions and i have:

$c_1 + c_3 = 2 = y_0$

$c_1 + 2c_3 = 3 = y_1$

$c_1 + 4c_3 =-3 = y_2$

So i cannot solve the system and i can't find out what i've done wrong. It is supposed to be possible to solve of course.

Thanks in advance

Answer 1

Thanks to everyone who took time to help me. I finally figured it out by myself,pretty dumm mistake. So here is the general solution:

$y_x=c_1*1^x + c_2*2^x*\cos(π*x/2) + c_3*2^x\sin(π*x/2)$, where $c_2*2^x*\cos(π*x/2) = 0$

I thought that:

$c_2*2^x*\cos(π*x/2) = 0$.

Well, that's wrong because:

e.g. $x=2\\c_2*2^2*\cos(π*2/2) = -4c_2$