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 May 20th, 2019, 09:39 AM #1 Member   Joined: Jan 2016 From: / Posts: 43 Thanks: 1 non-homogeneous recurrence problem Hello everyone, I have this problem $a_n + a_{n-1} + 6a_{n-2} = 5n(-1)^n + 2^n$ 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: $\displaystyle x_{1,2} = -1 \pm \frac{\sqrt{1-24}}{2}$ 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 $x_{1,2}$. Thanks from topsquark 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 $\displaystyle x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ 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
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Quote:
 Originally Posted by SuperNova1250 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?
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 $\displaystyle \frac{-1 \pm \sqrt{1-24}}{2}$ 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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post reza767 Number Theory 0 April 10th, 2015 02:42 PM BonaviaFx Differential Equations 1 March 29th, 2015 08:06 AM LouArnold Algebra 0 September 24th, 2012 10:37 PM arron1990 Calculus 5 February 21st, 2012 07:22 AM westaf Differential Equations 3 April 7th, 2011 09:29 PM

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