GATE XE (Engineering Sciences) 2012 - Question 19

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Section: Engineering Mathematics | Marks: 2

The general solution of 4 3 2 4 3 2 2 2 2 0 d y d y d y dy y dx dx dx dx − + − + = is

Figure for question 19, GATE XE (Engineering Sciences) 2012

  • A. 1 2 3 4 cosh( ) sinh( ) x x c e c xe c x c x + + +
  • B. 1 2 3 4 x x ix ix c e c e c e c e − − + + +
  • C. 1 2 3 4 cos( ) sin( ) x x c e c xe c x c x + + +
  • D. 1 2 3 4 x x ix ix c e c xe c e c e − − + + +

Answer: C. 1 2 3 4 cos( ) sin( ) x x c e c xe c x c x + + +

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