Solve the second order Differential equation xy” + y’ = 0; where y1 = ln x is a solution.?

2 Answers

• xy'' + y' = 0

  y1 = lnx

  Y = v(x) * y1

  Y = v ln(x) ( i wanna make it simpler but keeping in mind that v is a function of x)

  dY/dx = v (1/x) + ln(x) *dV/dx

  Y' = v/x + v'ln(x)

  Y'' = 2v'/x - v/x² + v''ln(x)

  replace in the original diff equa; xy'' + y' =- 0

  xy'' + y' = 0

  x{2v'/x - v/x² + v''ln(x) } + v/x + v'ln(x) = 0

  2v' - v/x + xv''ln(x) + v/x + v'ln(x) = 0

  2v' + xv''ln(x) + v'ln(x) = 0

  let u = v'

  ```````````````` keeping in mind that u = u(x)

  u' = v''

  2v' + xv''ln(x) + v'ln(x) = 0

  2u + xu' ln(x) + uln(x) = 0

  xu'lnx = -ulnx - 2u

  `````````````````````````` separable

  u' (xlnx) = -u(2 + lnx)

  u' / u = -(2 + lnx) / (xlnx)

  (1/u) du/dx = -(2 + lnx) / (xlnx)

  (1/u) du = -(2 + lnx) / (xlnx) dx

  integrate both sides to get

  ∫ (1/u) du = -∫ (2 + lnx) / (xlnx) dx

  u = C / (xln²x)

  but u = v'

  so

  v(x) = A + B/lnx

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