Question

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

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)$e^{-t}$

=>  (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 )

Learn more:

Integration of x²×sec²(x³) dx - Brainly.in

https://brainly.in/question/18063473

pls don't spam...if u spam , the answer will be reported​ - Brainly.in

https://brainly.in/question/17519995