ISRO Electrical Engineering (EE) 2015 - Question 65

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The solution of the differential equation (dy/dx) = ky, Y(0) = C is

  • A. x = c e^-xy
  • B. y = c e^kx
  • C. x = k e^cy
  • D. y = c e^-kx

Answer: B. y = c e^kx

Explanation: Separating variables: dy/y = k dx => ln y = kx + C1 => y = C·e^(kx). Given y(0) = C, the exact solution is y = C·e^(kx).

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