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 June 2nd, 2017, 04:56 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4 Linear Differential equation with constant coefficient A solution of $(D^2+1)^3y=\sin x$ is A) $\displaystyle \sum_{n=1}^{3} (c_nx^{n-1}\sin x+d_nx^{n-1}\cos x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$. B) $\displaystyle \sum_{n=1}^{3} (c_n \sin^n x+d_n \cos^n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$. C) $\displaystyle \sum_{n=1}^{3} (c_n \sin n x+d_n \cos n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$. D) None of the above. My answer is Option B. Please check and let me know if wrong. Thanks. Last edited by skipjack; June 2nd, 2017 at 11:39 PM. June 3rd, 2017, 03:32 AM   #2
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 Originally Posted by Lalitha183 A solution of $(D^2+1)^3y=\sin x$ is A) $\displaystyle \sum_{n=1}^{3} (c_nx^{n-1}\sin x+d_nx^{n-1}\cos x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$. B) $\displaystyle \sum_{n=1}^{3} (c_n \sin^n x+d_n \cos^n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$. C) $\displaystyle \sum_{n=1}^{3} (c_n \sin n x+d_n \cos n x)$ for some $\displaystyle c_n, d_n \in R, 1 \leq n \leq 3$. D) None of the above. My answer is Option B. Please check and let me know if wrong. Thanks.
Tomorrow I have an exam  June 4th, 2017, 12:28 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Are you really required to use a "series solution" method? I would use that to try to solve a linear differential equation with variable coefficients, but this, $\displaystyle (D^2+ 1)^3y= \sin(x)$, has constant coefficients. The "characteristic equation" for the associated homogeneous equation ($\displaystyle (D^2+1)^3= 0$) is $\displaystyle (r^2+ 1)^3= 0$. $r= i$ is a triple root and $r= -i$ is another triple root. That means that the general solution to the associated homogeneous equation is $\displaystyle y(x)= A \cos(x)+ B \sin(x)+ C x \cos(x)+ D x \sin(x)+ E x^2 \cos(x)+ F x^2 \sin(x)$. The "non-homogeneous" part of the equation is $\displaystyle \sin(x)$ which is already a solution to the homogeneous equation, as well as $x$ and $\displaystyle x^2$ times $\displaystyle \sin(x)$. So try $\displaystyle y(x)= Gx^3 \cos(x)+ Hx^3 \sin(x)$. The calculations from this point are tedious but straightforward. Last edited by skipjack; June 5th, 2017 at 12:36 AM. Tags coefficient, constant, differential, equation, linear 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 zylo Calculus 5 November 30th, 2016 06:56 PM MaRoVy Differential Equations 0 May 16th, 2015 10:33 AM rody Differential Equations 2 November 29th, 2013 04:59 PM ashubharatbansal Differential Equations 2 April 27th, 2012 08:45 AM jokes_finder Differential Equations 3 November 20th, 2010 09:52 AM

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