24 Chapter 0 Ordinary Differential Equations
1.du
dt+a(u−T)=0.- dudt+au=e−at.
- d
(^2) u
dt^2
+u=cos(t).
- d
(^2) u
dt^2
- 3 du
dt - 2 u=cosh(t).
9.^1
ρ^2
d
dρ(
ρ^2 du
dρ)
=−1.
2.
du
dt+au=eat.4.
d^2 u
dt^2 +u=cos(ωt)(ω=^1 ).- d
(^2) u
dx^2 −γ
(^2) (u−U)= 0
(U,γ^2 are constants).
8.^1
r
d
dr(
rdu
dr)
=−1.
- d
(^2) u
dt^2 =−1.
11.Leth(t)be the height of a parachutist above the surface of the earth. Con-
sideration of forces on his body leads to the initial value problem forh:
Md
(^2) h
dt^2
+Kdh
dt
=−Mg,
h( 0 )=h 0 , dhdt( 0 )= 0
(M=mass,g=acceleration of gravity,K=parachute constant). Solve
the problem, takingg=32 ft/s^2 andK/M= 0 .1/s.
12.Solve this initial value problem for forced vibrations,
d^2 u
dt^2 +ω
(^2) u=f 0 sin(μt),
u( 0 )= 0 , du
dt
( 0 )= 0 ,
in two cases: (a)μ=ω,(b)μ=ω.
In Exercises 13–19, use variation of parameters to find a particular solution of
the differential equation. Be sure that the differential equation is in the correct
form.
- dudt+au=e−at, uc(t)=e−at.
14.tdu
dt=−1, uc(t)=1.15.d^2 y
dx^2 +y=tan(x), y^1 (x)=cos(x), y^2 (x)=sin(x).- d
(^2) y
dx^2 +y=sin(x), y^1 (x)=cos(x), y^2 (x)=sin(x).