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If y = x cos x is a solution of an nth order linear differential equation

With real constant coefficients, then the least possible value of n is:-
  • a)
    1
  • b)
    2
  • c)
    3
  • d)
    4
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
If y = x cos x is a solution of an nth order linear differential equat...

y = x cos x
y 1 = - x sinx + cosx
yn = - sin x - x cos x - sin x
yn + y = - 2 sin x...(i)
yn1 + y1 = - 2cosx
y +yn = 2sinx...(ii)
(1)+(2) gives
y+ 2 yn + y = 0
so, ⇒ n = 4
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Most Upvoted Answer
If y = x cos x is a solution of an nth order linear differential equat...

y = x cos x
y 1 = - x sinx + cosx
yn = - sin x - x cos x - sin x
yn + y = - 2 sin x...(i)
yn1 + y1 = - 2cosx
y +yn = 2sinx...(ii)
(1)+(2) gives
y+ 2 yn + y = 0
so, ⇒ n = 4
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Community Answer
If y = x cos x is a solution of an nth order linear differential equat...

y = x cos x
y 1 = - x sinx + cosx
yn = - sin x - x cos x - sin x
yn + y = - 2 sin x...(i)
yn1 + y1 = - 2cosx
y +yn = 2sinx...(ii)
(1)+(2) gives
y+ 2 yn + y = 0
so, ⇒ n = 4
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If y = x cos x is a solution of an nth order linear differential equationWith real constant coefficients, then the least possible value of n is:-a)1b)2c)3d)4Correct answer is option 'D'. Can you explain this answer?
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