Use the method of undetermined coefficients to solve
y" – 5y' + 6y = e^3x + sin x.
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Step-by-step explanation:
y"-5y'+6y=0
First find the General Solution:
λ
2
−
5
λ
+
6
=
0
⇒
(
λ
−
2
)
(
λ
−
3
)
=
0
So,
λ
1
=
2
and
λ
2
=
3
Since these zeros of the characteristic equation are different, we use the general form:
y
(
x
)
=
C
1
e
λ
1
x
+
C
2
e
λ
2
x
y
(
x
)
=
C
1
e
2
x
+
C
2
e
3
x
Now we want a particular solution, so we will use your given IVs that:
1. y(0) = 1
2. y'(0) = -2
So:
y
′
(
x
)
=
2
C
1
e
2
x
+
3
C
2
e
3
x
y
′
(
0
)
=
2
C
1
+
3
C
2
=
−
2
C
1
=
1
2
(
−
3
C
2
−
2
)
Also,
y
(
0
)
=
C
1
+
C
2
=
1
C
1
=
1
−
C
2
Therefore, \\
1
−
C
2
=
1
2
(
−
3
C
2
−
2
)
2
−
2
C
2
=
−
3
C
2
−
2
2
+
C
2
=
−
2
C
2
=
−
4
S
o
,
C
1
=
5
Finally, we have:
y
(
x
)
=
5
e
2
x
−
4
e
3
x
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