Exercises 6.1
Integration yields
v
1
(
x
) =
w
1
(
x
)
dx
=
(
−
3
e
−
3
x
)
dx
=
e
−
3
x
,
v
2
(
x
) =
w
2
(
x
)
dx
=
(
2
e
2
x
)
dx
=
e
2
x
,
where we have chosen zero integration constants.
(d)
With functions
v
1
(
x
) and
v
2
(
x
) obtained in (c), we have
y
1
(
x
) =
v
1
(
x
)
e
x
=
e
−
3
x
e
x
=
e
−
2
x
,
y
2
(
x
) =
v
2
(
x
)
e
x
=
e
2
x
e
x
=
e
3
x
.
To show that the functions
e
x
,
e
−
2
x
, and
e
3
x
are linearly independent on (
−∞
,
∞
), we
can use the approach similar to that in Problem 7. Alternatively, since these functions
are solutions to the differential equation (32), one can apply Theorem 3, as we did in
Problem 15. To this end,
W e
x
, e
−
2
x
, e
3
x
(
x
) =
e
x
e
−
2
x
e
3
x
e
x
−
2
e
−
2
x
3
e
3
x
e
x
4
e
−
2
x
9
e
3
x
=
e
x
e
−
2
x
e
3
x
1
1
1
1
−
2
3
1
4
9
=
−
30
e
2
x
= 0
on (
−∞
,
∞
) and so the functions
e
x
,
e
−
2
x
, and
e
3
x
are linearly independent on (
−∞
,
∞
).
33.
Let
y
(
x
) =
v
(
x
)
e
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Derivative, e2x

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