InterviewSolution
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InterviewSolution
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Solve ` (x log x ) ( dy)/(dx) + y = (2)/(x) log x `. |
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Answer» The given differential equation may be written as ` (dy)/(dx) + (1)/( (x logx )) * y = ( 2)/(x^(2)) `. This is of the form ` (dy)/(dx) + Py =Q, ` where `P = (1)/((x log x )) and Q = (2)/(x ^(2)) ` Thus, the given differential equation is linear. ` IF = e ^(int Pdx) = e ^( int (1)/(x log x) dx ) = e ^( int (1)/(t) dt)`, where ` log x = t ` `" " = e ^( log t) = t = log x ` So, the solution of the given differential equation is ` y xx IF = int {Q xx (IF)}dx + C`, i.e., `y (log x ) = int(( 2)/( x^(2)) log x )dx +C ` `" " = 2 int ( log x ) * (1)/(x ^(2)) dx+ C ` `" " = 2 [ (logx ) (- (1)/(x)) - int (1)/(x) * (- (1)/(x)) dx ] + C`. `" " ` [ integrating by parts] `" " = ( - 2log x ) /(x) - (2)/(x) + C` Hence, `y (log x ) = (-2)/(x) (log x + 1 ) +C ` is the required solution. |
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