Question

Solve (D2
– 2D + 1) y = x log x, using the method of variation parameters; where D = ௗ
ௗ௫

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 )

Substitute x =  $e^{t}$   and log x = t

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

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