G(t) = | G(v)d(t—v)av. (6.58) 
0 
Then thinking the characters of derivatives in Eq. 6.57, at r= 0, G’’(r) that is 
D*G(t) contains the same discontinuity as does 0(t). Therefore, G’(t) must contain the 
same discontinuity as does the unit step function, and similarly G(r), which is the integral 
of G'(t), behaves like the integral of the unit step function, which is a ramp function. 
Therefore, G(r) is continuous at r = 0, and the initial conditions are 
G'(th=G@a@=0, t<0 
G'(0) = 1, G(0) = 0. (6.59) 
From these initial conditions (Eq. 6.59), the coefficients C,, C2 of the solution of the 
homogeneous equation 
(D? + aD +a)G(t) = 0, (6.60) 
that is, of 
G(t) = Cy" + Coe", (6.61) 
are determined as 
1 =] 
Pe oe 6.62 
1 Tote 2 Tot (6.62) 
Therefore 
wv _ Art 
Be. for t>0 
Ay Az 
G(t) = (6.63) 
0 otherwise. 
Here A ,A2 are the eigenvalues or the roots of the characteristic equation equated to 
ZeTO, as 
fz) = Z* +a;Z+ ao = 0, (6.64) 
or 
Z* + 2kw,Z+ 0% = (Z—-A1)\(Z—A>) = 0. (6.64’) 
Then 
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