0.2 Nonhomogeneous Linear Equations 25
- d
(^2) u
dt^2
=−1, u 1 (t)=1, u 2 (t)=t.
18.^1 rdrd
(
rdudr
)
=−1, u 1 (r)=1, u 2 (r)=ln(r).
19.t^2
d^2 u
dt^2 +t
du
dt−u=1, u^1 (t)=t, u^2 (t)=
1
t.
In Exercises 20–22, use Theorem 3 to develop the formula shown for a partic-
ular solution of the differential equation.
- d
(^2) u
dt^2
+γ^2 u=f(t), up(t)=^1
γ
∫t
0
sin
(
γ(t−z)
)
f(z)dz.
21.
du
dt+au=f(t), up(t)=
∫t
0
e−a(t−z)f(z)dz.
22.
d^2 u
dt^2 −γ
(^2) u=f(t), up(t)=^1
γ
∫t
0
sinh
(
γ(t−z)
)
f(z)dz.
23.In “Model for temperature estimation of electric couplings suffering heavy
lightning currents” [A.D. Polykriti et al.,IEE Proceedings — Generation,
Transmission and Distribution, 151 (2004): 90–94], the authors model the
temperature rise above ambient in a coupling with this initial value prob-
lem:
ρcdTdt =i^2 (t)R( 1 +αT), T( 0 )= 0.
Parameters:ρis density,cis specific heat,i(t)is the current due to a light-
ening strike,Ris the resistance of the coupling at ambient temperature,
and the factor( 1 +αT)shows how resistance increases with temperature.
Simplify the differential equation algebraically to get
dT
dt =Ki
(^2) (t)(β+T), T( 0 )= 0 ,
and identifyβandKin terms of the other parameters.
24.(Continuation) The authors model the lightning current with the func-
tioni(t)=Imax(e−λt−e−μt)/n,wherenis a factor to makeImaxthe ac-
tual maximum. Obtain graphs of this function and the simpler function
i(t)=Imaxe−λt, using these values:Imax=100 kA,n= 0 .93,λ= 2 .1,
μ=150. The unit for time is milliseconds. Graph fortfrom 0 to 2 ms,
which is the range of interest.
25.(Continuation) Solve the initial value problem using the simpler function
for current. (Don’t forget to square.) Graph the result fortfrom 0 to 2 ms,
usingβ= 0 .26 andK=13.