Answer:
The general solution of the ordinary differential equation can be given by
y
(
x
)
=
C
.
F
.
+
P
.
I
.
y(x)=C.F.+P.I., where
C
.
F
.
C.F. is a complementary function and
P
.
I
.
P.I. is a particular solution. For the general solution of the differential equation, find out the
C
.
F
.
C.F. by taking the auxiliary equation. We can use the normal form method or the variation of parameters method to find out the particular solution of the differential equation.
Answer and Explanation: 1
Take
d
d
x
=
D
ddx=D. So, the given equation is
D
2
y
−
6
D
y
+
9
y
=
x
−
3
e
3
x
D2y−6Dy+9y=x−3e3x.
Simplify the left side of the obtained equation.
(
D
2
−
6
D
+
9
)
y
=
x
−
3
e
3
x
(D2−6D+9)y=x−3e3x
Find out the
C
.
F
.
C.F. by taking the auxiliary equation, that is,
m
2
−
6
m
+
9
=
0
m2−6m+9=0.
m
2
−
6
m
+
9
=
0
m
2
−
3
m
−
3
m
+
9
=
0
m
(
m
−
3
)
−
3
(
m
−
3
)
=
0
(
m
−
3
)
(
m
−
3
)
=
0
m2−6m+9=0m2−3m−3m+9=0m(m−3)−3(m−3)=0(m−3)(m−3)=0
m
−
3
=
0
m
=
3
m−3=0m=3
m
−
3
=
0
m
=
3
m−3=0m=3
As the roots are real and equal,
C
.
F
.
=
(
c
1
+
x
c
2
)
e
3
x
C.F.=(c1+xc2)e3x.
Now, find out the particular solution of the given equation,
P
.
I
.
=
1
D
2
−
6
D
+
9
x
−
3
e
3
x
P.I.=1D2−6D+9x−3e3x.
P
.
I
.
=
1
D
2
−
6
D
+
9
x
−
3
e
3
x
=
1
(
D
−
3
)
2
x
−
3
e
3
x
=
e
3
x
1
(
(
D
+
3
)
−
3
)
2
x
−
3
=
e
3
x
1
(
D
+
3
−
3
)
2
x
−
3
=
e
3
x
1
D
2
x
−
3
=
e
3
x
1
D
(
1
D
(
x
−
3
)
)
=
e
3
x
1
D
(
x
−
2
−
2
)
=
−
e
3
x
2
1
D
(
x
−
2
)
=
−
e
3
x
2
(
x
−
1
−
1
)
=
e
3
x
2
x
P.I.=1D2−6D+9x−3e3x=1(D−3)2x−3e3x=e3x1((D+3)−3)2x−3=e3x1(D+3−3)2x−3=e3x1D2x−3=e3x1D(1D(x−3))=e3x1D(x−2−2)=−e3x21D(x−2)=−e3x2(x−1−1)=e3x2x
Substitute the obtained values of
C
.
F
.
C.F. and
P
.
I
.
P.I. into the equation
y
(
x
)
=
C
.
F
.
+
P
.
I
.
y(x)=C.F.+P.I..
y
(
x
)
=
(
c
1
+
x
c
2
)
e
3
x
+
e
3
x
2
x
y(x)=(c1+xc2)e3x+e3x2x
Thus, the general solution of the given differential equation is
y
(
x
)
=
(
c
1
+
x
c
2
)
e
3
x
+
e
3
x
2
x
y(x)=(c1+xc2)e3x+e3x2x.
Explanation:
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