Question

solve( D²- 2d +1)y = x log x , using the method of variation parameters; where D = d/dx

Answer 1

Answer:

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Answer 2

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)

=>  (y'$e^{-t}$ )'   -  (y

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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