Solve: y'''+ y' = sec x
SOLUTION USING VARIATION OF PARAMETERS
The solutions for the homogeneous y'''+ y' = 0 are y1 =1, y2 = cos x and y3 = sin x and therefore
y c = Α + Β cos x + Γ sin x.
Then we are suggesting a particular solution of the form y p = y1 u1 + y2 u2 + y3u3
and for then for the u1, u2, and u3 to be determined we must solve the system
y1 u'1 + y2 u'2 + y3u'3 = 0
y'1 u'1 + y'2 u'2 + y'3u'3 = 0
y''1 u'1 + y''2 u'2 + y''3u'3 = 0 or equivalently
u'1 + cos x u'2 + sin xu'3 = 0
-sin x u'2 + cos xu'3 = 0
-cos xu'2 -sin x u'3 = sec x and solving for u'1 , u'2 , and u'3 we get
u'1 = sec x ⇒ u1 = ln | sec x + tan x |
u'2 = -1 ⇒ u2 = -x
u'3 = - tan x ⇒ u3 = - ln | cos x |
Then the general solution is
y = y c + y p = Α + Β cos x + Γ sin x. + sec x - x cos x - sin x ln | cos x |.
