2
− n(n + 1)R = 0
6, Section 0.1: ρ2 ddρR2 + 2ρ dR
dρ
We look for the solutions in the form R = ρm . We get a characteristic
equation m(m − 1) + 2m − n(n − 1) = 0.
Using the formula for the roots of quadratic equation, solutions are
m = n and m = −(n − 1). Since we assume, that n is an integer the
solutions are different. Therefore, the general solution, is
R = c1 ρn + c2 ρ−n+1 .
8, Section 0.1: (ex φ0 )0 + λ2 ex φ = 0
By the product rule, ex φ00 + ex φ0 + ex λ2 φ = 0
Factoring out ex , we get a second order linear differential equation with
constant coefficients, φ00 + φ0 + λ2 φ = 0.
We look for the solutions in the form φ = xµ .
The characteristic equation: µ2 + µ + λ2 = 0.
Solutions are µ1 =
√
−1+ 1−4λ2
,
2
µ2 =
√
−1− 1−4λ2
.
2
For −1/2 < λ < 1/2, the general solution is φ = c1 eµ1 x + c2 eµ2 x .
For λ = ±1/2, the general solution is φ = c1 xe±1/2x + c2 e±1/2x .
√
For λ > 1/2 √
or λ < −1/2, the general solution is φ = c1 e−1/2x cos( 1 − 4λ2 x/2)+
c2 e−1/2x sin( 1 − 4λ2 x/2)
16, Section 0.1:
d2 u
dt2
+ 2a du
+ a2 u = 0,
dt
u1 = e−at
Set u = u1 v. After we plug u in this form into the original equation,
we get an equation v 00 = 0, so v = At + B
So the general solution is u = Ate−at + Be−at
4, Section 0.2:
d2 u
+ u = cos(wt),
dt2
(w 6= 1)
Then the general solution of the homogeneous equation
u = c1 cos t + c2 sin t.
1)We assume w 6= −1.
1
d2 u
dt2
+ u = 0 is
We find a particular solution of non-homogeneous equation by method
of undetermined coefficients
u1 = A cos(wt) + B sin(wt),
u001 +u1 = −Aw2 cos(wt)−Bw2 sin(wt)+A cos(wt)+B sin(wt) = cos(wt)
A=
1
,
−w2 +1
B=0
So the general solution is u = c1 cos(t) + c2 sin(t) +
1
−w2 +1
cos(wt)
2)When w = −1, we need to look for particular solutions in the form
u1 = At sin(t) + Bt cos(t). We find A = 1/2, B = 0
So the general solution is u = c1 sin(t) + c2 cos(t) + 1/2t sin(t)
d
r du
= −1; u1 (r) = 1, u2 = ln(r)
18, Section 0.2: 1r dr
dr
First, we write the equation in the standard form
d2 u 1 du
+
= −1
dr2
r dr
The Wronskian is W (r) = 1/r
Therefore, a particular solution is
R r R −1
Rr
dt + ln(r) 1 1/t
dt = −r2 /4 + 1/2 ln(r)
up = − 1 ln(t)(−1)
1/t
12, Section 0.3: First, we look for a particular solution in the form up = Ax2 + Bx + C,
and we find
A=
w
,
2P
B = − wL
, C = − wEI
P
2P 2
So the general solution is
q
q
P
P
u = c1 sin( EI
x) + c2 cos( EI
x) +
u(0) = 0, so c2 =
u(L) = 0, so c1 =
wEI
,
2P 2
√P
− wEI
2 (1+cos(
2P
sin(
√P
EI
L)
EI
2
L))
w 2
x
2P
−
wL
x
P
−
wEI
2P 2