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The general solution of the differential equation 
  • a)
    y = (c1 – c2x) ex + c3 cos x + c4 sin x
  • b)
    y = (c1 + c2x) ex – c3 cos x + c4 sin x
  • c)
    y = (c1 + c2x) ex + c3 cos x + c4 sin x
  • d)
    y = (c1 + c2x) ex + c3 cos x – c4 sin x
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The general solution of the differential equationa)y = (c1– c2x)...
Concept:
General equation for DE:
Then its corresponding Auxiliary equation will be
AE: Dn + k1 Dn-1 + … kn = 0
Then the solution of above DE will be as follows:
Calculation:
Given:

Its Auxillary equations:
(D4 – 2D3 + 2D2 – 2D + D) = 0
(D - 1)(D3 - D2 + D - 1) = 0
(D - 1)(D - 1)(D2 + 1) = 0
It roots = 1, 1, i, -i
Then it has two equal roots and one pair of imaginary roots.
Therefore, the solution is:
(c1 + c2x) ex + c3 cos x + c4 sin x = 0
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The general solution of the differential equationa)y = (c1– c2x) ex+ c3cos x + c4sin xb)y = (c1+ c2x) ex– c3cos x + c4sin xc)y = (c1+ c2x) ex+ c3cos x + c4sin xd)y = (c1+ c2x) ex+ c3cos x – c4sin xCorrect answer is option 'C'. Can you explain this answer?
Question Description
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