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

574 Chapter 19 Ordinary differential equations I
Solution
(a) With the given parameter values the differential equation becomes
d 2 i
dt + di
2 dt +i=e−2t
Itisfirstnecessarytofind thecomplementary function. Lettingi = e kt , theauxiliaryequation
isk 2 +k+1 = 0 which has complex solutions
k = − 1 √
3
2 ± 2 j
The complementary function istherefore
√ √ ) 3 3
i =e
(Acos
−t/2 2 t+Bsin 2 t
For a particular integral we try a solution of the form i = αe −2t . Substitution
into the differential equation gives
so that
4αe −2t −2αe −2t + αe −2t = e −2t
3α=1, thatis α= 1 3
Hence a particular integral isi = 1 3 e−2t . The general solution is the sum of the
complementary function and the particular integral:
√ √ ) 3 3
i =e
(Acos
−t/2 2 t+Bsin 2 t + 1 3 e−2t
We now apply the initial conditions to find the constantsAandB. Giveni = 1
whent = 0 means
1=A+ 1 so that A = 2 3 3
To apply the second condition weneed tofind di
dt :
( √ √ √ √ )
di 3 3 3 3
dt = e−t/2 −
2 Asin 2 t + 2 Bcos 2 t
− 1 2 e−t/2 (Acos
Given di = 2 whent = 0 means
dt
√
3
2 =
2 B − 1 2 (A) − 2 3
√
3
2 =
2 B − 1 ( ) 2
− 2 2 3 3
√
3
2 t+Bsin √
3
2 t )
− 2 3 e−2t
2 =
3 =
√
3
2 B −1
√
3
2 B