Second-Order Linear ODE Solutions

Chapter 2. Second-Order Linear ODEs 1
§2.1 Homogeneous Linear ODEs of Second Order
y + p(x)y + q(x)y = r(x), (1)
y + p(x)y + q(x)y = 0, (2)
y(x0) = K0, y (x0) = K1.
Note:
• (2) is called homogeneous.
• If r(x) = 0, then (1) is called nonhomogeneous.
• c1y1 + c2y2 : a linear combination of y1 and y2
• linear, nonlinear
Set L = d2
dx2 + p(x) d
dx
+ q(x), then we write the ODE (1) as
L(y) = r.
L(c1y1 + c2y2) = c1L(y1) + c2L(y2).
Ex. 1. xy + y + xy = 0 : homogeneous linear ODE
2. y y + y 2
= 0 : nonlinear ODE
Theorem 1. Superposition Principle
For a homogeneous linear ODE (2) on open interval I, any linear combina-
tion of two solutions on I is again a solution of (2) on I.
Proof. Let y1 and y2 be solutions of (2) and y = c1y1 + c2y2. Then
y + py + qy
= (c1y1 + c2y2) + p(c1y1 + c2y2) + q(c1y1 + c2y2)
= c1(y1 + py1 + qy1) + c2(y2 + py2 + qy2)
= c10 + c20 = 0.
Hence y = c1y1 + c2y2 is a solution on I.
Computational Science & Engineering (CSE) C. K. Ko

Example 1. Superposition of Solutions
y1 = cos x and y2 = sin x are solutions of a homogeneous linear
ODE
y + y = 0
for all x. Moreover,
y = c1 cos x + c2 sin x
is also a solution of the ODE.
Example 2. A nonhomogeneous Linear ODE
y1 = 1 + cos x and y2 = 1 + sin x are solutions of a linear nonhomo-
geneous ODE
y + y = 1
for all x. But
y1 + y2 = 2 + cos x + sin x, 2y1 = 2 + 2 cos x
are not solutions of the ODE.
Example 3. A nonlinear ODE
y1 = x2
and y2 = 1 are solutions of a nonlinear ODE
y y − xy = 0
for all x. But
y1 + y2 = x2
+ 1, − y1 = −x2
are not solutions of the ODE.

Deﬁnition (General Solution, Basis, Particular Solution)
• y1 and y2 are called linearly independent on an interval I if
k1y1(x) + k2y2(x) = 0, ∀x ∈ I ⇒ k1 = k2 = 0.
Otherwise, that is, for some numbers k, l
y1 = ky2 or y2 = ly1, ∀x ∈ I,
y1 and y2 are called linearly dependent on an interval I.
• In ODE (2), y = c1y1 + c2y2 is a general solution where y1
and y2 are solutions of (2) on an open interval I that are not propor-
tional(linearly independent).
• These y1, y2 (a pair of linearly independent) are called a basis(or
fundamental system) of solutions of (2) on I.
• For speciﬁc values of c1 and c2 we obtain a particular solution of (2)
on I.
Example 6. Basis, General Solution, Particular Solution
Solve
y − y = 0, y(0) = 6, y (0) = −2.
Sol. y1(x) = ex
, y2(x) = e−x
: solutions of the ODE(Why!)
y1/y2 = e2x
= const : y1 and y2 are linearly independent.
Hence they form a basis of solutions for all x.
y = c1ex
+ c2e−x
: general solution
y(0) = c1 + c2 = 6, y (0) = c1 − c2 = −2 ⇒ c1 = 2, c2 = 4
∴ y = 2ex
+ 4e−x
: particular solution

[Find a Basis if One Solution is Known.
Reduction of Order]
y + p(x)y + q(x)y = 0 (∗)
Assume that y1 is a solution of (∗) on I.
A second linearly independent solution y2 of (∗) on I ?
(Method of Reduction of Order to a homogeneous linear ODE)
• Set y2 = uy1.
y2 = u y1 + uy1, y2 = u y1 + 2u y1 + uy1 .
Then
(u y1 + 2u y1 + uy1 ) + p(u y1 + uy1) + quy1 = 0
⇒ u y1 + u (2y1 + py1) + u(y1 + py1 + qy1) = 0.
Since y1 + py1 + qy1 = 0, we get
u y1 + u (2y1 + py1) = 0.
• Set U = u . Then
U +
2y1
y1
+ p U = 0.
⇒
dU
U
= −
2y1
y1
+ p dx,
⇒ ln |U| = −2 ln |y1| − p dx, U =
1
y2
1
e− p dx
⇒ u = U dx =
1
y2
1
e− p dx
dx
∴ y2 = y1u = y1
1
y2
1
e− p dx
dx.

Example 7. Reduction of Order if a Solution is known. Basis
Find a basis of solutions of
(x2
− x)y − xy + y = 0.
Note that y1 = x is a solution.
Sol. Since y1 = x is a solution, we set y2 = uy1 = ux.
Notice that
y −
1
x − 1
y +
1
x2 − x
y = 0.
Using the formula, we have
u =
1
y2
1
e− p dx
dx
=
1
x2
e
1
x−1 dx
dx
=
|x − 1|
x2
dx
=
x − 1
x2
dx (why!)
= ln |x| +
1
x
.
∴ y2 = x ln |x| + 1
Since y1 = x and y2 = x ln |x| + 1 are linearly independent, they form
a basis of solutions.

§2.2 Homogeneous Linear ODEs with Constant Coef-
ﬁcients
Consider second-order homogeneous linear ODEs with constant coeﬃcients
y + ay + by = 0. (∗)
Note: y + ky = 0 has a solution y = ce−kx
.
Try y = eλx
as a solution of (∗).
=⇒ (λ2
+ aλ + b)eλx
= 0, ∀x ∈ R.
=⇒ λ2
+ aλ + b = 0 : characteristic equation.
λ = λ1, λ2.
Case I. Two distinct real roots λ1 and λ2.
Since y1 = eλ1x
and y2 = eλ2x
are linearly independent solutions of the
ODE, the general solution of the ODE is
y = c1eλ1x
+ c2eλ2x
.
Example 2. Case of Distinct Real Roots
Solve
y + y − 2y = 0, y(0) = 4, y (0) = −5.
Sol. characteristic equation : λ2
+ λ − 2 = 0, λ = 1, −2.
∴ y = c1ex
+ c2e−2x
.
y(0) = c1 + c2 = 4, y (0) = c1 − 2c2 = −5
=⇒ c1 = 1, c2 = 3
∴ y = ex
+ 3e−2x
.

Case II. Real double root λ = −a
2
.
y1 = e−a
2x
, y2 =?
Let y2 = uy1. Using the formula, we have
u =
1
y2
1
e− p dx
dx
= eax
e−ax
dx
= x.
Thus y2 = xe−a
2x
(another independent solution), and the general solution
is
y = (c1 + c2x)e−ax/2
.
Example 3. Case of a Double Root
Solve
y + 6y + 9y = 0.
+ 6λ + 9 = 0, λ = −3
∴ y = (c1 + c2x)e−3x
.

Case III. Complex roots λ1 = −1
2
a + iω and λ2 = −1
2
a − iω.
Note: Euler formula eit
= cos t + i sin t
eit
=
∞
n=0
(it)n
n!
=
∞
n=0
(it)2n
(2n)!
+
∞
n=0
(it)2n+1
(2n + 1)!
=
∞
n=0
(−1)n
t2n
(2n)!
+ i
∞
n=0
(−1)n
t2n+1
(2n + 1)!
= cos t + i sin t.
We have two linearly independent solutions
y1 = eλ1x
= e−1
2ax
(cos wx + i sin wx),
y2 = eλ2x
= e−1
2ax
(cos wx − i sin wx).
Since the ODE is linear and homogeneous,
1
2
(y1 + y2) = e−1
2ax
cos wx,
1
2i
(y1 − y2) = e−1
2ax
sin wx
are linearly independent solutions.
The general solution is
∴ y = e−1
2ax
(A cos ωx + B sin ωx) .
Example 5. Case of Complex Roots
Solve
y + 0.4y + 9.04y = 0.
+ 0.4λ + 9.04 = 0, λ = −0.2 ± 3i.
∴ y = e−0.2x
(A cos 3x + B sin 3x) .

§2.4 Modeling : Free Oscillations
(Mass-Spring System)
[Setting Up the Model]
• Newton’s second law : Force = Mass × Acceleration = my .
• Hook’s law: F0 = −κs0 where κ > 0 is the spring constant
(or spring modulus).
• In static equilibrium, −κs0 + mg = 0.
• From the position y = 0, the displacement y(t) we pull it down,
my = −ky ⇒ my + κy = 0.
[Undamped System : my + κy = 0]
characteristic equation : mλ2
+ κ = 0
=⇒ λ2
+ ω2
0 = 0, ω0 = κ
m
, λ = ±iω0
∴ y(t) = A cos ω0t + B sin ω0t.
The corresponding motion is called a harmonic oscillation.
Since the trigonometric functions have the period
2π
ω0
, the frequency
(or natural frequency) is
ω0
2π
.

Note:
• cycles/sec = hertz(Hz)
•
y(t) = A cos(ω0t) + B sin(ω0t)
= C[cos(ω0t)
A
C
+ sin(ω0t)
B
C
]
= C[cos(ω0t) cos δ + sin(ω0t) sin δ]
= C cos(ω0t − δ)
where C =
√
A2 + B2, δ = arctan (B/A) .
• A = y(0), B = y (0)/ω0.
• B = 0 ⇒ δ = 0.

Example 1. Undamped Motion. Harmonic Oscillation
• If an iron ball of weight W = 98 nt stretches a spring 1.09 m, how
many cycles per minute will this mass-spring system execute?
• What will its motion be if we pull down the weight an additional 16 cm
and let it start with zero initial velocity?
Sol. By Hook’s law,
98 = 1.09κ ⇒ κ =
98
1.09
= 90(kg/sec).
The mass is m = W/g = 98/9.8 = 10(kg).
∴ frequency:
ω0
2π
=
κ
m
2π
=
3
2π
(Hz)
harmonic oscillation
y(t) = A cos ω0t + B sin ω0t
y(0) = A = 0.16m,
y (0) = ω0B = 0 ⇒ B = 0
∴ y(t) = 0.16 cos 3t (m)

[Damped System]
A damping force is F2 = −cy , c = damping constant.
my = −ky − cy =⇒ my + cy + κy = 0
=⇒ λ2
+
c
m
λ +
κ
m
= 0
=⇒ λ1 = −α + β, λ2 = −α − β,
where α = c
2m
, β = 1
2m
√
c2 − 4mκ.
• Case I. c2
> 4mκ : distinct real roots λ1, λ2 (overdamping)
∴ y(t) = c1e−(α−β)t
+ c2e−(α+β)t
• Case II. c2
= 4mκ : a real double root (critical damping)
∴ y(t) = c1e−αt
+ c2te−αt
• Case III. c2
< 4mκ : complex conjugate roots (underdamping)
β = iω∗
, where ω∗
=
1
2m
4mκ − c2 (> 0)
∴ y(t) = e−αt
(A cos ω∗
t + B sin ω∗
t)
= Ce−αt
cos(ω∗
t − δ),
where C =
√
A2 + B2, tan δ = B/A.
Note that the frequency is
ω∗
2π
and ω∗
−→ ω0 as c −→ 0.

§2.5 Euler-Cauchy Equations
x2
y + axy + by = 0 : Euler-Cauchy equation.
Try y = xm
. Then
[m(m − 1) + am + b]xm
= 0.
m2
+ (a − 1)m + b = 0, m = m1, m2.
• Case 1. Two distinct real roots : m1, m2.
Since y1 = xm1 and y2 = xm2 are linearly independent solutions of
the ODE, the general solution of the ODE is
y = c1xm1 + c2xm2.
• Case 2. Real double root : m1 = m2 = 1
2
(1 − a)
y1 = x
1
2(1−a)
, y2 =?
Let y2 = uy1.
x2
y + axy + by = 0 =⇒ y +
a
x
y +
b
x2
= 0.
Using the formula, we have
u =
1
y2
1
e− p dx
dx = xa−1
x−a
dx = ln x.
Thus y2 = x
1
2(1−a)
ln x (another independent solution) and the gen-
eral solution is
y = (c1 + c2 ln x)x
1
2(1−a)
.
Example 1-2.
Solve (a) x2
y + 1.5xy − 0.5y = 0, (b) x2
y − 5xy + 9y = 0.
Sol. (a) m(m − 1) + 1.5m − 0.5 = m2
+ 0.5m − 0.5 = 0,
m = 0.5, −1 ∴ y = c1
√
x + c2x−1
.
(b) m(m − 1) − 5m + 9 = m2
− 6m + 9 = 0, m = 3.
∴ y = (c1 + c2 ln x)x3
.

• Case 3. Complex roots m1 = 1
2
(1−a)+ωi, m2 = 1
2
(1−a)−ωi.
We have two linearly independent solutions
y1 = xm1 = x
1
2(1−a)
eiω ln x
= x
1
2(1−a)
(cos(ω ln x) + i sin(ω ln x)),
y2 = xm2 = x
1
2(1−a)
e−iω ln x
= x
1
2(1−a)
(cos(ω ln x) − i sin(ω ln x)).
Since the ODE is linear and homogeneous,
1
2
(y1 + y2) = x
1
2(1−a)
cos(ω ln x),
1
2i
(y1 − y2) = x
1
2(1−a)
sin(ω ln x)
are linearly independent solutions.
The general solution is
y = x
1
2(1−a)
[A cos(ω ln x) + B sin(ω ln x)].
Example 3. Case of Complex Roots
Solve
x2
y + 0.6xy + 16.04y = 0.
Sol. m(m − 1) + 0.6m + 16.04 = m2
− 0.4m + 16.04 = 0
⇒ m1 = 0.2 + 4i, m2 = 0.2 − 4i.
∴ y = x0.2
[A cos(4 ln x) + B sin(4 ln x)] .

§2.6 Existence and Uniqueness of Solutions.
Wronskian
y + p(x)y + q(x)y = 0, (1)
y(x0) = K0, y (x0) = K1. (2)
Theorem 1. (Existence and Uniqueness Theorem for IVP)
If p(x) and q(x) are continuous on I and x0 ∈ I,
then the IVP of (1) and (2) has a unique solution y(x) on I.
[Recall]
y1 and y2 are called linearly independent on an interval I if
k1y1(x) + k2y2(x) = 0, ∀x ∈ I ⇒ k1 = k2 = 0.
Otherwise, that is, for some numbers k, l
y1 = ky2 or y2 = ly1, ∀x ∈ I,
y1 and y2 are called linearly dependent on an interval I.
Theorem 2.
Let p(x) and q(x) in (1) be continuous on I. Then
1. Two solutions y1, y2 of (1) on I are linearly dependent on I.
⇐⇒ W (y1, y2) ≡
y1 y2
y1 y2
= y1y2 − y2y1 = 0
at some x0 ∈ I, (W (y1, y2) :Wronskian).
2. If W = 0 at an x0 ∈ I, then W ≡ 0 on I.
Hence, if ∃ x1 ∈ I at which W is not 0, then y1 and y2 are linearly
independent on I.

Proof. 1.(⇒) Let y1, y2 be linearly dependent on I and y1 = ky2.
Then W (y1, y2) = y1y2 − y2y1 = ky2y2 − ky2y2 = 0.
(⇐) Let W (y1, y2) = 0 for some x = x0 ∈ I.
We consider the linear system in the unknowns K1, K2.
K1y1(x0) + K2y2(x0) = 0
K1y1(x0) + K2y2(x0) = 0
(3)
=⇒
y1(x0) y2(x0)
y1(x0) y2(x0)
K1
K2
=
0
0
.
Since
y1(x0) y2(x0)
y1(x0) y2(x0)
= y1y2 − y2y1 = W (y1(x0), y2(x0)) = 0,
∃ not all zero K1 and K2, say, K2
1 + K2
2 = 0.
Then for these constants K1, K2,
˜y(x) = K1y1(x) + K2y2(x)
is a solution of (1) and ˜y(x0) = ˜y (x0) = 0 from (3).
Notice that ˆy ≡ 0 is also a a solution of (1) and ˆy(x0) = ˆy (x0) = 0.
By the uniqueness theorem of solutions,
˜y = ˆy, i.e, K1y1 + K2y2 = 0
=⇒ y1 = −
K2
K1
y2 or y2 = −
K1
K2
y1.
∴ y1 and y2 are linearly dependent.
2. Let W (x0) = 0 at x0 ∈ I. By part (⇐), y1 and y2 are linearly
dependent. Hence W ≡ 0.
Example 1.
W (cos ωx, sin ωx) =
cos ωx sin ωx
−ω sin ωx ω cos ωx
= ω.
∴ cos ωx and sin ωx are linearly independent ⇔ ω = 0

Theorem 3. (Existence of a General Solution)
If p(x) and q(x) are continuous on I, then (1) has a general solution on
I.
Theorem 4. ( A General Solution Includes All Solutions)
If p(x) and q(x) are continuous on I, then every solution y = Y (x) of
(1) is of the form
Y (x) = C1y1(x) + C2y2(x)
where y1, y2 is any basis of solutions of (1) on I.
Hence (1) does not have singular solutions.

§2.7 Nonhomogeneous ODEs
y + p(x)y + q(x)y = r(x). (1)
y + p(x)y + q(x)y = 0. (2)
Definition
A general solution of (1) is a solution of the form
y(x) = yh(x) + yp(x).
Here, yh = c1y1 + c2y2 is a general solution of (2) and yp is any solution
of (1).
Theorem 1.
• solution of (1)+ solution of (2)= solution of (1).
• solution of (1)− solution of (1)= solution of (2).
Theorem 2.
Let p, q, r in (1) be continuous on some open interval I.
Every solution of (1) is obtained by assigning suitable values to c1, c2 in a
solution of the form y(x) = yh(x) + yp(x).
Here, yh is a general solution of (2) and yp is any solution of (1).
[Method of Undetermined Coefficients]
• Choice Rules for yp
(a) Basic rule
r(x) yp(x)
keγx
Ceγx
kxn
(n = 0, 1, , · · · ) Knxn
+ Kn−1xn−1
+ · · · + K1x + K0
k cos ωx ,k sin ωx K cos ωx + M sin ωx
keαx
cos ωx, keαx
sin ωx keαx
(K cos ωx + M sin ωx)
(b) Modification rule(Multiply by x or x2
, · · · )
(c) Sum rule (r(x) = r1(x) + r2(x))

Example 1.
Solve
y + y = 0.001x2
, y(0) = 0, y (0) = 1.5.
Sol. step1. y + y = 0 ⇒ yh = A cos x + B sin x.
step2. yp = K2x2
+ K1x + K0.
yp + yp = 2K2 + K2x2
+ K1x + K0 = 0.001x2
.
∴ yp = 0.001x2
− 0.002.
step3. y = yh + yp = A cos x + B sin x + 0.001x2
− 0.002.
y(0) = 0, y (0) = 1.5 =⇒ A = 0.002, B = 1.5.
∴ y = 0.002 cos x + 1.5 sin x + 0.001x2
− 0.002.
Example 2.
Solve
y + 3y + 2.25y = −10e−1.5x
.
Sol. step1. y +3y +2.25y = 0, λ2
+3λ+2.25 = 0, λ = −1.5.
⇒ yh = (c1 + c2x)e−1.5x
.
step2. yp = Cx2
e−1.5x
(why!).
∴ yp = −5x2
e−1.5x
.
step3. y = yh + yp = (c1 + c2x − 5x2
)e−1.5x
.

Example 3.
Solve the ODE
y + 2y + 5y = e0.5x
+ 40 cos 10x − 190 sin 10x,
y(0) = 0.16, y (0) = 40.08.
Sol. step1. y + 2y + 5y = 0
⇒ λ2
+ 2λ + 5 = 0, λ = −1 ± 2i.
∴ yh = e−x
(A cos 2x + B sin 2x).
step2. yp = Ce0.5x
+ D cos 10x + E sin 10x.
∴ yp = 0.16e0.5x
+ 2 sin 10x.
step3. y = yh + yp = e−x
(A cos 2x + B sin 2x)
+0.16e0.5x
+ 2 sin 10x.
From the initial conditions, A = 0, B = 10, and the solution is
y = 10e−x
sin 2x + 0.16e0.5x
+ 2 sin 10x.

§2.8 Modelling: Forced Oscillations. Resonance
my + cy + ky = r(t) (1)
internal force external force
• r(t) = 0 ⇒ (1): free motion.
• r(t) = 0 ⇒ (1): forced motion with forcing function r(t) (input
or driving force).
• y(t): output or response of the system to the driving force.
[Solving the Nonhomogeneous ODE]
Consider the ODE with a periodic external force r(t) = F0 cos ωt.
my + cy + ky = F0 cos ωt (2)
• Put yp(t) = a cos ωt + b sin ωt.
• Substituting yp, yp, yp into (2), we get
[(k − mω2
)a + ωcb] cos ωt + [−ωca + (k − mω2
)b] sin ωt
=F0 cos ωt
=⇒(k − mω2
)a + ωcb = F0, − ωca + (k − mω2
)b = 0.

• If (k − mω2
)2
+ ω2
c2
= 0,
a =
(k − mω2
)F0
(k − mω2)2 + ω2c2
, b =
ωcF0
(k − mω2)2 + ω2c2
.
• If k/m = ω0(> 0), then k = mω2
0 and we obtain
a =
m(ω2
0 − ω2
)F0
m2(ω2
0 − ω2)2 + ω2c2
, b =
ωcF0
m2(ω2
0 − ω2)2 + ω2c2
.
[Case 1. Undamped Forced Oscillations. Resonance]
• c = 0 ⇒ a = F0/[m(ω2
0 − ω2
)], b = 0
∴ yp(t) =
F0
m(ω2
0 − ω2)
cos ωt =
F0
k[1 − (ω/ω0)2]
cos ωt.
• In equations above, we must assume that ω2
= ω2
0; physically,
the frequency ω/(2π)[cycles/sec] of the driving force
= the natural frequency ω0/(2π)[cycles/sec] of the system.
• From the results in Sec. 2.4 we have the general solution of the ”un-
damped system”
y(t) = C cos(ω0t − δ) +
F0
m(ω2
0 − ω2)
cos ωt.
• Resonance.
• The maximum amplitude of yp is (put cos ωt = 1)
a0 =
F0
k
ρ where ρ =
1
1 − (ω/ω0)2
.
• If ω → ω0, then ρ → ∞ and a0 → ∞.
• This excitation of large oscillations by matching input and natural
frequencies (ω = ω0) is called resonance. ρ: resonance factor.

• ρ/k = a0/F0: the ratio of the amplitudes of yp and of the input
F0 cos ωt.
• If ω = ω0, then the nonhomogeneous ODE becomes
y + ω2
0y =
F0
m
cos ω0t.
From the Modiﬁcation Rule in Sec. 2.7
yp(t) = t(a cos ω0t + b sin ω0t) =
F0
2mω0
t sin ω0t.

[Case 2. Damped Forced Oscillations]
• If the damping is not negligibly small, we have c > 0 and a damping
term cy . Then the general solution yh of the homogeneous ODE
approaches zero as t → ∞, that is, the trasient solution y =
yh+yp approaches the steady-state solution yp as t → ∞.
• Amplitude of the Steady-State Solution. Practical Resonance
The amplitude will always be ﬁnite. But it may have a maximum for
some ω depending on the damping constant c. This may be called
practical resonance.
• Let yp(t) = a cos ωt + b sin ωt = C∗
cos(ωt − η).
C∗
is called the amplitude of yp and η the phase angle or phase
lag. Here,
C∗
= a2 + b2 =
F0
m2(ω2
0 − ω2)2 + ω2c2
,
tan η(ω) =
b
a
=
ωc
m(ω2
0 − ω2)
.
• Let R = m2(ω2
0 − ω2)2 + ω2c2. Then
dC∗
dω
= F0 −
1
2
R−3/2
[2m2
(ω2
0 − ω2
)(−2ω) + 2ωc2
] = 0
⇒ c2
= 2m2
(ω2
0 − ω2
) (ω2
0 = k/m) (15)
⇒ 2m2
ω2
= 2m2
ω2
0 − c2
= 2mk − c2
.
• c2
> 2mk ⇒ (15) has no real solution.
c2
< 2mk ⇒ (15) has a real solution ω = ωmax, where
ω2
max = ω2
0 −
c2
2m2
,
C∗
(ωmax) =
2mF0
c 4m2ω2
0 − c2
=
2mF0
c
√
4mk − c2
.

Example
• Take m = 1, k = 1, hence ω0 = 1, and various values of the
damping constant c.
• Fig. 56 shows the ampliﬁcation C∗
/F0 as a function of ω.
• Fig. 57 shows the phase angle η:
ω < ω0 ⇒ η < (π/2); ω > ω0 ⇒ η > (π/2).

§2.9 Modelling: Electric Circuits
• Voltage Drops.
• RI Voltage drop for a resistor of resistance R (Ω) (Ohm’s law).
• LI Voltage drop for an inductor of capacitance L (H).
•
Q
C
Voltage drop for a capacitor of capacitance C (F).
Here Q coulombs is the charge on the capacitor; Q(t) = I(t)dt.
• Kirchhoff’s Voltage Law (KVL). The voltage impressed on a
closed loop is equal to the sum of the voltage drops across the
other elements of the loop.
• According to KLV we have in Fig. 60 the ”integro-differential equa-
tion”
LI + RI +
1
C
Idt = E(t) = E0 sin ωt.
Differentiating the above equation, we obtain
LI + RI +
1
C
I = E (t) = E0ω cos ωt. (1)

[Solving the ODE (1) for the Current.
Discussion of Solution]
• A general solution of (1) is the sum I = Ih + Ip.
• Let Ip = a cos ωt + b sin ωt. Then
Lω2
(−a) + Rωb + a/C = E0ω (Cosine terms)
Lω2
(−b) + Rω(−a) + b/C = 0 (Sine terms)
• Introducing the reactance S = ωL − 1
ωC
, we get
− Sa + Rb = E0
− Ra − Sb = 0
=⇒ a =
−E0S
R2 + S2
, b =
E0R
R2 + S2
.
• Ip(t) = I0 sin(ωt − θ) where
I0 = a2 + b2 =
E0
√
R2 + S2
, tan θ = −
a
b
=
S
R
.
• The quantity
√
R2 + S2 is called impedance.
Note that
√
R2 + S2 = E0/I0 is analogous to R = E/I.
• Let Ih = c1eλ1t
+ c2eλ2t
. Then
λ2
+
R
L
λ +
1
LC
= 0, λ1 = −α + β, λ2 = −α − β,
where α =
R
2L
, β =
R2
4L2
−
1
LC
=
1
2L
R2 −
4L
C
.
• Since R > 0, the transient current I = Ih + Ip tends to the
steady-state current Ip as t → ∞.

Example 1. RLC-Circuit
Find the current I(t) in an RLC-circuit with R = 11Ω, L = 0.1H,
C = 10−2
F and E(t) = 100 sin 400t. Assume that current and charge
are zero when t = 0.
Sol. 0.1I + 11I + 100I = 100 · 400 cos 400t.
Ih = c1e−10t
+ c2e−100t
Ip = a cos 400t + b sin 400t
=⇒ Ip = −2.3368 cos 400t + 0.6467 sin 400t
I = Ih + Ip = c1e−10t
+ c2e−100t
−2.3368 cos 400t + 0.6467 sin 400t
initial condition : I(0) = 0, Q(0) = 0
LI + RI +
1
C
Idt = E(t) = E0 sin ωt.
Putting t = 0, LI (0) + RI(0) + 1
C
· 0 = 0 =⇒ I (0) = 0.
I = −0.2776e−10t
+2.6144e−100t
−2.3368 cos 400t+0.6467 sin 400t
[Analogy of Electrical and Mechanical Quantities]
LI + RI +
1
C
I = E0ω cos ωt my + cy + ky = F0 cos ωt

§2.10 Solution by Variation of Parameters
y + p(x)y + q(x)y = r(x) (1)
When yh(x) = c1y1 + c2y2 is the general solution of the ODE
y + p(x)y + q(x)y = 0, (2)
ﬁnd a solution yp of (1).
• Setting yp = uy1 + vy2, determine u and v. Then
yp = u y1 + uy1 + v y2 + vy2.
• We may impose a second condition:
u y1 + v y2 = 0.
Thus yp = uy1 + vy2 and yp = u y1 + uy1 + v y2 + vy2 .
Substituting yp, yp and yp into (1), we obtain
u(y1 + py1 + qy1
=0
) + v(y2 + py2 + qy2
=0
) + u y1 + v y2 = r
and
u y1 + v y2 = r
u y1 + v y2 = 0.
We solve the linear system:
u = −
y2r
W
, v =
y1r
W
, where W =
y1 y2
y1 y2
.
By integration,
u = −
y2r
W
dx, v =
y1r
W
dx.
∴ yp(x) = −y1
y2r
W
dx + y2
y1r
W
dx
(method of variation of parameters).

Example 1. Method of Variation of Parameters
Solve y + y = sec x.
Sol. The general solution of the homogeneous ODE y + y = 0 is
yh = c1 cos x + c2 sin x.
Let y1 = cos x and y2 = sin x.
Then
W =
cos x sin x
− sin x cos x
= 1.
⇒ yp = −y1
y2r
W
dx + y2
y1r
W
dx
= − cos x sin x
1
cos x
dx + sin x cos x
1
cos x
dx
= cos x ln | cos x| + x sin x.
∴ y = c1 cos x + c2 sin x + cos x ln | cos x| + x sin x.

Second-Order Linear ODE Solutions

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