1.

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

Answer»

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

 (ye^{-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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