(y')^4 - y^4"( y+ xy') = 0
find the genral solution of the
equations
Answers
Answer: c1, c2
Explanation:
The differential equation
y
'
'
+
4
y
=
0
is what we call second order differential equation.
So we need to find its characteristic equation which is
r
2
+
4
=
0
This equation will will have complex conjugate roots, so the final answer would be in the form of
y
=
e
α
x
⋅
(
c
1
⋅
sin
(
β
x
)
+
c
2
⋅
cos
(
β
x
)
)
where
α
equals the real part of the complex roots and
β
equals the imaginary part of (one of) the complex roots.
We need to use the quadratic formula
r
=
−
b
±
√
b
2
−
4
a
c
2
⋅
a
when
a
⋅
r
2
+
b
⋅
r
+
c
=
0
In this equation
a
=
1
,
b
=
0
, and
c
=
4
Hence the roots are
r
1
=
2
i
and
r
2
=
−
2
i
Now the form of
y
=
e
α
x
⋅
(
c
1
⋅
sin
(
β
x
)
+
c
2
⋅
cos
(
β
x
)
)
where
a
=
0
and
β
=
2
becomes
y
=
e
0
⋅
x
⋅
(
c
1
⋅
sin
(
2
⋅
x
)
+
c
2
⋅
cos
(
2
⋅
x
)
)
Finally
y
=
(
c
1
⋅
sin
(
2
⋅
x
)
+
c
2
⋅
cos
(
2
⋅
x
)
)
The coefficients
c
1
,
c
2
can be determined if we have initial conditions for the differential equation.
The differential equation
y
'
'
+
4
y
=
0
is what we call second order differential equation.
So we need to find its characteristic equation which is
r
2
+
4
=
0
This equation will will have complex conjugate roots, so the final answer would be in the form of
y
=
e
α
x
⋅
(
c
1
⋅
sin
(
β
x
)
+
c
2
⋅
cos
(
β
x
)
)
where
α
equals the real part of the complex roots and
β
equals the imaginary part of (one of) the complex roots.
We need to use the quadratic formula
r
=
−
b
±
√
b
2
−
4
a
c
2
⋅
a
when
a
⋅
r
2
+
b
⋅
r
+
c
=
0
In this equation
a
=
1
,
b
=
0
, and
c
=
4
Hence the roots are
r
1
=
2
i
and
r
2
=
−
2
i
Now the form of
y
=
e
α
x
⋅
(
c
1
⋅
sin
(
β
x
)
+
c
2
⋅
cos
(
β
x
)
)
where
a
=
0
and
β
=
2
becomes
y
=
e
0
⋅
x
⋅
(
c
1
⋅
sin
(
2
⋅
x
)
+
c
2
⋅
cos
(
2
⋅
x
)
)
Finally
y
=
(
c
1
⋅
sin
(
2
⋅
x
)
+
c
2
⋅
cos
(
2
⋅
x
)
)
The coefficients
c
1
,
c
2
can be determined if we have initial conditions for the differential equation.