Question

Solve (D2 – 2D + 1) y = x log x, using the method of variation parameters; where D =

Answer 1

Given :   (D² - 2D + 1)y = x log x    

To find : solve for y

Solution:

(D² - 2D + 1)y = x log x

=> y'' - 2y' + y = x log x

let say log x = t   => x =  $e^{t}$

dx =    $e^{t}$  dt

y'' - 2y' + y =   t . $e^{t}$

=> (y''  - 2y'  + y)  $e^{-t}$  = t

=> (y''  - y'    - y'  + y) $e^{-t}$  = t

=> (y''  - y')$e^{-t}$ - (y'  - y)$e^{-t}$ = t

=>  (y'$e^{-t}$)'   -  (y $e^{-t}$ )'  = t

integrating both sides

 (y$e^{-t}$ )' =  t²/2  + C

integrating again

 y $e^{-t}$  =  t³/6  + Ct + D

=> y =   $e^{t}$ ( t³/6  + Ct + D )

y =    x  ( (log x)³/6  + Clog(x) + D )

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