Question

(d²+4)y=ex+sin2x+cos2x

​

Answer 1

sorry

Answer 2

The characteristic polynomial of the equation is 2+4
t
2
+
4
, with roots ±2
±
2
i
. Hence the homogeneous equation (2+4)=0
(
d
2
+
4
)
y
=
0
has solutions cos(2)+sin(2)
A
cos
⁡
(
2
x
)
+
B
sin
⁡
(
2
x
)
.

To find a particular solution to the heterogeneous equation, we first note that sin2=1−cos(2)2
sin
2
⁡
x
=
1
−
cos
⁡
(
2
x
)
2
, since

1−cos(2)2=1−(cos2−sin2)2=(1−cos2)+sin22=sin2+sin22=2sin22=sin2
1
−
cos
⁡
(
2
x
)
2
=
1
−
(
cos
2
⁡
x
−
sin
2
⁡
x
)
2
=
(
1
−
cos
2
⁡
x
)
+
sin
2
⁡
x
2
=
sin
2
⁡
x
+
sin
2
⁡
x
2
=
2
sin
2
⁡
x
2
=
sin
2
⁡
x
.

Therefore, we’d expect to find a particular solution of the form =+cos(2)+sin(2)
y
p
=
C
+
D
cos
⁡
(
2
x
)
+
E
sin
⁡
(
2
x
)
. However, cos(2)
cos
⁡
(
2
x
)
and sin(2)
sin
⁡
(
2
x
)
are already annihilated by the homogeneous equation, so they won’t make a particular solution any more possible. We must multiply such terms by
x
when using undetermined coefficients.

Thus we take:

=+cos(2)+sin(2)
y
p
=
C
+
D
x
cos
⁡
(
2
x
)
+
E
x
sin
⁡
(
2
x
)

and we compute:

′=(+2)cos(2)+(−2)sin(2)
y
p
′
=
(
D
+
2
E
x
)
cos
⁡
(
2
x
)
+
(
E
−
2
D
x
)
sin
⁡
(
2
x
)

″=(4−4)cos(2)−(4+4)sin(2)
y
p
″
=
(
4
E
−
4
D
x
)
cos
⁡
(
2
x
)
−
(
4
D
+
4
E
x
)
sin
⁡
(
2
x
)

″+4=4+4cos(2)−4sin(2)
y
p
″
+
4
y
p
=
4
C
+
4
E
cos
⁡
(
2
x
)
−
4
D
sin
⁡
(
2
x
)

and we want it to equal 1−cos(2)2
1
−
cos
⁡
(
2
x
)
2
. For this we take =18
C
=
1
8
, =0
D
=
0
and =−18
E
=
−
1
8
. Therefore, =18−18sin(2)
y
p
=
1
8
−
1
8
x
sin
⁡
(
2
x
)
.

Hence, our general solution is:
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xx