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**1. Homework Statement**

Solve [tex]y[k+2]+y[k]=sin(k)[/tex].

The answer is already given, it's

[tex]y[k]=c_1sin(\frac{\pi}{2}k) + c_2cos(\frac{\pi}{2}k) + \frac{sin(k)+sin(k-2)}{2(1+cos(2))}[/tex]

**2. Homework Equations**

**3. The Attempt at a Solution**

First I solve the homogeneous equation.

[tex]y[k+2]+y[k]=0[/tex].

The characteristic function is

[tex]r^2+1=0[/tex]

This has two complex roots [tex]r=i[/tex] and [tex]r=-i[/tex].

Thus the general solution is

[tex]y[k]=c_1(i)^k+c_2(-i)^k[/tex]

If you change the notation for the complex numbers, then it gives

[tex]y[k]=c_1(cos(\frac{\pi}{2}k)+isin(\frac{\pi}{2}k)) + c_2(cos(\frac{-\pi}{2}k)+isin(\frac{-\pi}{2}k))[/tex]

After grouping together it gives

[tex](c_1+c_2)cos(\frac{\pi}{2}k) + (c_1-c_2)isin(\frac{\pi}{2}k)[/tex]

Apparently, the answer is without complex numbers so I guess they only keep the real part. Then it must be so that

[tex]c_1-c_2=0\Leftrightarrow c_1=c_2[/tex]

Then I get as solution

[tex]y[k]=2c_1cos(\frac{\pi}{2}k)[/tex]

which is not the same solution as the general solution in the answer. What did I do wrong?

I also tried to find a particular solution for the non-homogeneous equation. I tried as a solution

[tex]y[k]=\alpha sin(k)[/tex]

with [tex]\alpha[/tex] the parameter to determine. I don't this is the right solution. Can anyone help me with this problem?

Thanks,

Yoran