eliminate the arbitrary constant:

y=c_{1}e^{x} + c_{2}xe^{x}

y'=c_{1}e^{x} + c_{2}(xe^{x }+ e^{x})

y''=c_{1}e^{x} + c_{2}(xe^{x }+ 2e^{x})

by comparing eq.1 and eq,2 = eq,4

by comparing eq.2 and eq.3 = eq.5

by comparing eq.4 and eq.5 = y'' + 2y - 3y'

am i doing it right??

the roots are 1 and 1....

Your answer has the root of -1 and -2.......

Hope you not solving for the DE

$y = c_1e^x + c_2xe^x$ ← eq (1)

$y' = c_1e^x + c_2(xe^x + e^x)$ ← eq (2)

$y'' = c_1e^x + c_2(xe^x + 2e^x)$ ← eq (3)

eq (2) - eq (1)

$y' - y = c_2e^x$ ← eq (4)

eq (3) - eq (2)

$y'' - y' = c_2e^x$ ← eq (5)

eq (5) - eq (4)

$(y'' - y') - (y' - y) = 0$

$y'' - 2y' + y = 0$

answer