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d

2

d

–2x

----

y +

----

y – 6y = e

dt

dt

First solve for the homogeneous part,

d

2

d

try

y = e

Bx

----

y + ----

y – 6y = 0

dt

dt

d

Bx

Be

----

y =

dt

d

2

y =

B

2

e

Bx

----

B2eBx + BeBx – 6eBx = 0

dt

B2 + B – 6 = 0

B = –3,2

yh

= e–3x + e2x

Next, solve for the particular part. We will guess the function below.

y = Ce–2x
d					–2x
---- y = –2Ce
dt
d	2	y =	4Ce	–2x
----
dt

4Ce–2x + –2Ce–2x – 6Ce–2x = e–2x

4C – 2C – 6C = 1

C = 0.25

yp = 0.25e–2x

Finally,

y = e–3x + e2x + 0.25e–2x

2.6.3.3 - Higher Order Differential Equations