Differential Equation: Eliminate $c_1$ and $c_2$ from $y = c_1 e^x + c_2 xe^x$

Submitted by omrcsr on

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??

Tags
Elimination of Arbitrary Constant

real75 Fri, 06/24/2016 - 05:31

the roots are 1 and 1....
Your answer has the root of -1 and -2.......
Hope you not solving for the DE

Jhun Vert Sat, 06/25/2016 - 13:30

$y = c_1e^x + c_2xe^x$   ←   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