May 20th, 2019, 09:39 AM  #1 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1  nonhomogeneous recurrence problem
Hello everyone, I have this problem $a_n + a_{n1} + 6a_{n2} = 5n(1)^n + 2^n$ This is the given solution: $\displaystyle a_n = (\sqrt{6})^n(C_1\cdot \sin\phi n + C_2\cdot \cos\phi n) + \frac{1}{3} \cdot 2^n + \left(\frac{5}{6}\cdot n + \frac{55}{36}\right)(1)^n$ I used the discriminant to find the zeros of characteristic polynomial but then I got this: $\displaystyle x_{1,2} = 1 \pm \frac{\sqrt{124}}{2}$ and have no idea how to continue. Can someone please help me solve this step by step? Thanks in advance. Last edited by skipjack; May 21st, 2019 at 06:38 AM. 
May 20th, 2019, 11:42 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
That expression for $x_{1,2}$ is going to form the basis of your complementary solution (to the homogeneous equation), the $(\sqrt6)^n\big(c_1\sin (n\phi) + c_2 \cos (n\phi)\big)$, I guess. (The $\sqrt6$ is the norm of the complex roots). The rest of it you can get by the Method of Undetermined Coefficients using $A2^n + (Bn+C)(1)^n$ as your trial solution. I'm not convinced by your values for $x_{1,2}$. Last edited by v8archie; May 20th, 2019 at 11:51 AM. 
May 20th, 2019, 12:00 PM  #3 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
Well the determinant's formula is $\displaystyle x_{1,2} = \frac{b\pm\sqrt{b^24ac}}{2a}$ if I'm correct (btw, I did a mistake in the first post) regarding the determinant, where 1 should be in the numerator too. However still, plugging in the discriminant I get the expression with negative square root, is this ok? and if not how it is then? Last edited by skipjack; May 21st, 2019 at 06:47 AM. 
May 20th, 2019, 12:14 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra  That's the error I spotted. Your expression with the negative square root is OK. Since $x_1 = x_2 = \sqrt{6}$, can you make a guess as to what $\phi$ might be? If you are struggling with that, consider De Moivre's Theorem $$(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta$$
Last edited by skipjack; May 21st, 2019 at 07:20 AM. 
May 20th, 2019, 12:21 PM  #5 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
Wait, wait, you used the absolute value on $x_1$ and $x_2$, and got square root of 6. Can you please explain this part in detail, as the homogeneous part is the one I am struggling with? I haven't gotten yet to the particular solution. Last edited by skipjack; May 21st, 2019 at 07:21 AM. 
May 20th, 2019, 12:28 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
$x_1$ and $x_2$ are complex numbers $a_1 + ib_1$ and $a_2 + ib_2$. What are the absolute values (or modulus, or norm) of these? If you haven't yet met complex numbers, I don't know why you have this problem to solve. 
May 20th, 2019, 12:33 PM  #7 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
I have, the answer I got with the discriminant was $\displaystyle \frac{1 \pm \sqrt{124}}{2}$ which is $\displaystyle \frac{1 \pm 23i}{2}$ How to continue from here? Last edited by skipjack; May 21st, 2019 at 06:56 AM. 
May 20th, 2019, 04:18 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
That should be $$\frac{1 \pm i\sqrt{23}}{2}=\frac12 \pm i \frac{\sqrt{23}}2$$

May 20th, 2019, 10:23 PM  #9 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
Well.. and how does that become $\displaystyle \sqrt{6}$?
Last edited by skipjack; May 21st, 2019 at 06:58 AM. 
May 21st, 2019, 03:34 AM  #10 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
Come on, can anybody solve this step by step? I am really struggling with this problem. Thanks in advance. Last edited by skipjack; May 21st, 2019 at 07:04 AM. 

Tags 
homogenous, nonhomogeneous, problem, recurrence 
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