eliminate the arbitrary constant:
y=c1ex + c2xex
y'=c1ex + c2(xex + ex)
y''=c1ex + c2(xex + 2ex)
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??
eliminate the arbitrary constant:
y=c1ex + c2xex
y'=c1ex + c2(xex + ex)
y''=c1ex + c2(xex + 2ex)
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
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