1.

D*2-2d+1)y=xlogx using the method of variation parameters ;where d=d/dx

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}  )'   -  (ye^{-t} )' = t

integrating both sides

 (ye^{-t} )' =  t²/2  + C

integrating again

 ye^{-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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