text: MAT244-2013F > Quiz 3
  
  Problem 1 (Day section)
  
  (1/1)
  
  Razak Pirani:
  4.2 #18 Find the general solution of the given differential equation.
  
  y(6) - y'' = 0
  
  r6 - r2 = 0
  r2(r4 - 1) = 0
  r2(r2 - 1)(r2 + 1) = 0
  r2(r - 1)(r + 1)(r - i)(r + i) = 0
  
  r1,2 = 0
  r3 = 1
  r4 = -1
  r5 = i
  r6 = -i
  
  y(t) = c1 + c2t + c3et + c4e-t + c5cost + c6sint
  
  Mark Kazakevich:
  For the differential equation:
  $$y^{(6)} - y''$$
  
  We assume that $y = e^{rt}$.
  Therefore, we must solve the characteristic equation:
  
  $$r^6 - r^2 = 0$$
  
  We find:
  $r^6 - r^2 = 0 \implies r^2(r^4-1) \implies r^2(r^2+1)(r^2-1) = 0 \implies r^2(r^2+1)(r-1)(r+1) = 0$
  
  This means the roots of this equation are:
  
  $r_1 = 0, r_2=0, r_3=i, r_4=-i, r_5=1,r_6=-1$
  (We have a repeated root at r = 0)
  
  So the general solution to (1) is:
  $$y(t) = c_1 + c_2t + c_3\cos{t} + c_4\sin{t} + c_5e^{t} + c_6e^{-t}$$
explanation: The document contains a math question and its solution. The question asks for the general solution of a given differential equation, and two students, Razak Pirani and Mark Kazakevich, provide detailed steps to solve it, arriving at the same general solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.content: MAT244-2013F > Quiz 3
  
  Problem 1 (Day section)
  
  (1/1)
  
  Razak Pirani:
  4.2 #18 Find the general solution of the given differential equation.
  
  y(6) - y'' = 0
  
  r6 - r2 = 0
  r2(r4 - 1) = 0
  r2(r2 - 1)(r2 + 1) = 0
  r2(r - 1)(r + 1)(r - i)(r + i) = 0
  
  r1,2 = 0
  r3 = 1
  r4 = -1
  r5 = i
  r6 = -i
  
  y(t) = c1 + c2t + c3et + c4e-t + c5cost + c6sint
  
  Mark Kazakevich:
  For the differential equation:
  $$y^{(6)} - y''$$
  
  We assume that $y = e^{rt}$.
  Therefore, we must solve the characteristic equation:
  
  $$r^6 - r^2 = 0$$
  
  We find:
  $r^6 - r^2 = 0 \implies r^2(r^4-1) \implies r^2(r^2+1)(r^2-1) = 0 \implies r^2(r^2+1)(r-1)(r+1) = 0$
  
  This means the roots of this equation are:
  
  $r_1 = 0, r_2=0, r_3=i, r_4=-i, r_5=1,r_6=-1$
  (We have a repeated root at r = 0)
  
  So the general solution to (1) is:
  $$y(t) = c_1 + c_2t + c_3\cos{t} + c_4\sin{t} + c_5e^{t} + c_6e^{-t}$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its solution. The question asks for the general solution of a given differential equation, and two students, Razak Pirani and Mark Kazakevich, provide detailed steps to solve it, arriving at the same general solution.
  
  Label: question-and-answer