Solving a Recurrence Equation: What Did I Do Wrong?

[yoran]

Hi, I'm trying to solve this recurrence equation.

Homework Statement

Solve y[k+2]+y[k]=sin(k).
The answer is already given, it's
y[k]=c_1sin(\frac{\pi}{2}k) + c_2cos(\frac{\pi}{2}k) + \frac{sin(k)+sin(k-2)}{2(1+cos(2))}

Homework Equations

The Attempt at a Solution

First I solve the homogeneous equation.
y[k+2]+y[k]=0.
The characteristic function is
r^2+1=0
This has two complex roots r=i and r=-i.
Thus the general solution is
y[k]=c_1(i)^k+c_2(-i)^k
If you change the notation for the complex numbers, then it gives
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))
After grouping together it gives
(c_1+c_2)cos(\frac{\pi}{2}k) + (c_1-c_2)isin(\frac{\pi}{2}k)
Apparently, the answer is without complex numbers so I guess they only keep the real part. Then it must be so that
c_1-c_2=0\Leftrightarrow c_1=c_2
Then I get as solution
y[k]=2c_1cos(\frac{\pi}{2}k)
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
y[k]=\alpha sin(k)
with \alpha the parameter to determine. I don't this is the right solution. Can anyone help me with this problem?

Thanks,

Yoran

 

[jhicks]

They don't "only keep the real part". Plug the solution back into the homogeneous equation; the sine part falls out without imposing any constraint on c1 and c2.

 

[yoran]

Hi,

Thanks for your quick reply. I tried to do what you told me but when I plug it into the homogeneous equation, everything cancels out! I get 0=0. This is what I did:
We have that
y[k]=(c_1+c_2)cos(\frac{\pi}{2}k)+(c_1-c_2)isin(\frac{\pi}{2}k)
I plug it into
y[k+2]+y[k]=0
I get
(c_1+c_2)cos(\frac{\pi}{2}(k+2))+(c_1-c_2)isin(\frac{\pi}{2}(k+2))+(c_1+c_2)cos(\frac{\pi}{2}k)+(c_1-c_2)isin(\frac{\pi}{2}k)=0
Because
cos(\theta + \pi)=-cos(\theta) and
sin(\theta + \pi)=-sin(\theta)
I get that
-(c_1+c_2)cos(\frac{\pi}{2}k)-(c_1-c_2)isin(\frac{\pi}{2}k)+(c_1+c_2)cos(\frac{\pi}{2}k)+(c_1-c_2)isin(\frac{\pi}{2}k)=0
which makes
0=0
How come?

 

[jhicks]

Actually, you are correct. That was a silly suggestion of me, because 0=0 simply confirms your solution satisfies the equation from which you derived the solution. I will edit this post if I come up with something.

Edit: Wow I had been reading it wrong for a good 45 minutes, sorry! Consider complex coefficients and make the substitution d1=c1+c2 and d2=i(c1-c2).

Can you find the particular solution from here?

 

[yoran]

Hi,

Yeah of course... just give different names as they are just parameters... Thanks a lot.
I tried the substitution \alpha sin(k) because there is sin(k) on the right side of the equation. However, I can't find the right answer. This is what I have done:
Try
y[k]=\alpha sin(k)
Plug it into the equation:
\alpha sin(k+2) + \alpha sin(k) = sin(k)
Then we get
\alpha = \frac{sin (k)}{2sin(k+1)cos(1)}
by using the identity
sin(\theta) + sin(\alpha) = 2sin(\frac{\theta + \alpha}{2})cos(\frac{\theta - \alpha}{2})
The answer for \alpha is obviously not the right answer.
Did I make the wrong substitution?

 

[yoran]

Anyone?

 

[tiny-tim]

Try
y[k]=\alpha sin(k)

Hi yoran!

Try y[k]=\alpha\,sin(k\,-\,1) .

 

[yoran]

Ok, I tried
y[k]=\alpha \sin(k-1).
Then I get that
\alpha = \frac{1}{2\cos 1}
So given the solution for the homogeneous equation and this solution, I have that
y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{1}{2\cos{1}}\sin{k-1}
But the answer should be
y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}
Where does it go wrong in my solution?

 

[tiny-tim]

Hi yoran!

…I have that
y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{1}{2\cos{1}}\sin{k-1}
But the answer should be
y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}
Where does it go wrong in my solution?

But they're the same! ​

Standard trig equation: 1 + cos(2) = 2cos²(1).

So \frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})} = … ? ​

 

[yoran]

Hehe, of course... Thanks a lot!
\frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}=\frac{2\sin(k-1)\cos(1)}{4\cos^2(1)}=\frac{\sin(k-1)}{2\cos 1}