Solve (D2 – 2D + 1) y = x log x, using the method of variation parameters; where D =
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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 =
dx = dt
y'' - 2y' + y = t .
=> (y'' - 2y' + y) = t
=> (y'' - y' - y' + y) = t
=> (y'' - y') - (y' - y) = t
=> (y')' - (y )' = t
integrating both sides
(y )' = t²/2 + C
integrating again
y = t³/6 + Ct + D
=> y = ( t³/6 + Ct + D )
y = x ( (log x)³/6 + Clog(x) + D )
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