082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

19.6 Constant coefficient equations 575
B = 6 √
3
= 2 √ 3
Finally, the required particular solution which gives the current through the
inductor atany time is
( √ √ )
2 3 3
i =e −t/2 3 cos 2 t+2√ 3sin
2 t + 1 3 e−2t
(b) The zero-input response is obtained by ignoring the source term i S
(t). From
(a)we see thatthis isjustthe complementary function:
√ √ ) 3 3
i =e
(Acos
−t/2 2 t+Bsin 2 t
Applying the given initial conditions to this solution givesA = 1 andB = √ 5 3
and so the zero-input response is
√
3
i =e
(cos
−t/2 2 t + √ 5 √ ) 3
sin
3 2 t
(c) Thezero-stateresponseoftheinhomogeneousequationisfoundbyapplyingthe
conditionsi = 0 and di = 0 att = 0 tothe general solution already obtained in
dt
(a). It is straightforward to show thatA = − 1 3 andB= √ 1 . So the zero-state
3
response is
(
i =e −t/2 − 1 √
3
3 cos 2 t + √ 1
√ ) 3
sin
3 2 t + 1 3 e−2t
(d) Inspectionofthepreviousworkingshowsthattheparticularsolutionobtainedin
(a)isthe sum of the zero-state response and zero-input response.
19.6.3 Inhomogeneoustermappearsinthecomplementaryfunction
In some examples, terms which form part of the complementary function also appear
in the inhomogeneous term. This gives rise to an additional complication. Consider
Example 19.29.
Example19.29 Considertheequationy ′′ −y ′ −6y = e 3x .Itisstraightforwardtoshowthatthecomplementaryfunctionis
y=Ae 3x +Be −2x
Findaparticular integraland deducethe generalsolution.
Solution Suppose we try to find a particular integral by using a trial solution of the form
y P
= αe 3x . Substitutioninto the l.h.s. ofthe inhomogeneous equation yields
9αe 3x −3αe 3x −6αe 3x whichsimplifies to0
so that αe 3x is clearlynotasolution of the inhomogeneousequation. Thereason is that
e 3x is part of the complementary function and so causes the l.h.s. to vanish. To obtain a