Ответы
Відповідь: я
Покрокове пояснення:
Explanation:
y
'
'
+
9
y
=
0
is a linear homogeneous differential equation with constant parameters. The general solution for an equation of this kind is
y
=
e
λ
x
substituting we get
(
λ
2
+
9
)
e
λ
x
=
0
but
e
λ
x
≠
0
for all
x
∈
R
so
λ
2
+
9
=
0
⇒
λ
=
±
3
i
so
y
=
C
1
e
3
i
x
+
C
2
e
−
3
i
x
but using the de Moivre's identity
e
i
x
=
cos
x
+
i
sin
x
and assuming that the solution is real, we easily could establish analogously
y
=
C
3
sin
(
3
x
)
+
C
4
cos
(
3
x
)
Answer link
Cem Sentin
Dec 16, 2017
y
=
c
1
⋅
cos
3
x
+
c
2
⋅
sin
3
x
Explanation:
If
y
1
=
sin
3
x
i solution of this differential equation, I used
y
=
u
sin
3
x
and
y
'
'
=
u
'
'
sin
3
x
+
6
u
'
cos
3
x
−
9
u
sin
3
x
transformation.
Hence,
u
'
'
sin
3
x
+
6
u
'
cos
3
x
−
9
u
sin
3
x
+
9
u
sin
3
x
=
0
u
'
'
sin
3
x
+
6
u
'
cos
3
x
=
0
It reduced to linear differential equation in terms of
u
'
Consequently,
u
'
'
u
'
+
6
cos
3
x
sin
3
x
=
0
After integrating both sides,
ln
u
'
+
2
ln
(
sin
3
x
)
=
ln
(
−
3
c
1
)
ln
[
u
'
⋅
(
sin
3
x
)
2
]
=
ln
(
−
3
c
1
)
u
'
⋅
(
sin
3
x
)
2
]
=
−
3
c
1
u
'
=
−
3
c
1
(
sin
3
x
)
2
After integrating both sides,
u
=
c
1
⋅
cot
3
x
+
c
2
Thus,
y
=
u
sin
3
x
=
sin
3
x
⋅
(
c
1
⋅
cot
3
x
+
c
2
)
=
c
1
⋅
cos
3
x
+
c
2
⋅
sin
3
x
Answer link