dr/d?=3*cos(pi/4-?)^2, r(0)=(pi/8), solve the intial value problem
%#
(
dr
= 3cos2 ' $ " *
&4
)
d"
%#
(
dr = 3cos2 ' $ " *d"
&4
)
r=
%#
(
+ 3cos '& 4 $ " *)d"
2
Using the identity cos2 ( x ) =
!
1+ cos(2x )
:
2
%"
(
1+ cos' # 2$ *
%"
(
(
&2
) 1 1 %"
! 2' # $ * =
cos
= + cos' # 2$ *
&4
)
)
2
2 2 &2
The integral is then:
!
%"
+1
(
(.
r=
1 -, 2 + 2 cos'& 2 # 2$ *)0/d$
2
+3
3
%"
(.
Substituting r(0) = π/8:
'
3
3 $#
#
(0) " sin& " 2(0)) + C =
(
2
4 %2
8
3 $#'
#
" sin& ) + C =
4 %2(
8
!
%"
1 3cos '& 4 # $ *)d$ = 1 3-, 2 + 2 cos'& 2 # 2$ *)0/d$
(
3
3 %"
r = $ # sin' # 2$ * + C
)
2
4 &2
!
1
r=
3
#
" +C =
4
8
C=
!
# 3
+
8 4
The complete solution is then:
( $ 3
3
3 %$
r(" ) = " # sin' # 2" * + +
) 8 4
2
4 &2
!