Question

All real solution of the de y'' +2ay'+by=cosx are periodic if

Answer 1

yp=cos(x)−bcos(x)−2asin(x)(1+2a2+2aa2−b−−−−−√−b)(−1−2a2+2aa2−b−−−−−√+b)
y
p
=
cos
⁡
(
x
)
−
b
cos
⁡
(
x
)
−
2
a
sin
⁡
(
x
)
(
1
+
2
a
2
+
2
a
a
2
−
b
−
b
)
(
−
1
−
2
a
2
+
2
a
a
2
−
b
+
b
)
.

And the complimentary solution is

yc=c1e(−a−a2−b√)x+c2e(−a+a2−b√)x
y
c
=
c
1
e
(
−
a
−
a
2
−
b
)
x
+
c
2
e
(
−
a
+
a
2
−
b
)
x
.

So the solution of the DE is

y=c1e(−a−a2−b√)x+c2e(−a+a2−b√)x+cos(x)−bcos(x)−2asin(x)(1+2a2+2aa2−b−−−−−√−b)(−1−2a2+2aa2−b−−−−−√+b)
Now as you can see if a2≥b
a
2
≥
b
, then all solutions are real and has a periodic part and a non-periodic part. To cancel out the non-periodic part there are two ways. If the arbitrary constants c1
c
1
and c2
c
2
becomes zero then the solution would be periodic. But that would not be a general solution. Also if the exponents becomes zero then again the solution would be periodic as the exponential part would be reduced to a constant. This would be a general solution. So for the exponents to be zero the following is required,

−a−a2−b−−−−−√=0
−
a
−
a
2
−
b
=
0
and −a+a2−b−−−−−√=0
−
a
+
a
2
−
b
=
0
.

It is easy to notice that both of them can not be zero at the same time unless both a
a
and b
b
are zero. When both a
a
and b
b
are zero the DE reduces to

y′′=cos(x)
y
″
=
cos
⁡
(
x
)

which is very easy to solve and the general solution is

y=−cos(x)
y
=
−
cos
⁡
(
x
)
.