Math 308, Sections 301, 302,
Summer 2008
Lecture 9.
06/18/2008
Cauchy-Euler Equations
A linear second order differential equation that can be expressed in
the form
d 2y
dy
ax 2 2 + bx
+ cy = 0,
dx
dx
where a, b, and c are constants is called a homogeneous
Cauchy-Euler equation.
x(t) = et
a
dy
d 2y
+ (b − a)
+ cy = 0
2
dt
dt
Example Find the general solution to the equation
x 2 y ′′ (x) + 7xy ′ (x) − 7y (x) = 0
for x > 0.
ay ′′ + by ′ + cy = 0,
(1)
where a, b, c are constants. The associated auxiliary equation is
ar 2 + br + c = 0.
(2)
The equation
√ (4) is a quadratic, and√its roots are
−b + b 2 − 4ac
−b − b 2 − 4ac
r1 =
, r2 =
2a
2a
.
When b2 − 4ac > 0, then r1 , r2 ∈ R and r1 6= r2 . So, {er1 x , er2 x } is
the FSS to (1) and the general solution to (1) is
y (x) = c1 er1 x + c2 er2 x
If b2 − 4ac = 0, then the equation (2) has a repeated root r ∈ R,
b
. Then, {y1 (x) = erx , y2 (x) = xerx } is the FSS to (1) and
r = − 2a
the general solution to (1) is
y (x) = c1 erx + c2 xerx = (c1 + c2 x)erx
Section 4.6 Auxiliary Equation with Complex Roots
If b2 − 4ac < 0, then the equation (2) has two complex conjugate
roots
√
ac − b 2
b
r1 = − + i
= α + i β,
2a
2a
√
ac − b 2
b
r2 = − − i
= α − i β = r¯1 ,
2a
2a
b
,β=
here i 2 = −1, α = − 2a
√
ac−b2
,
2a
α, β ∈ R.
We’d like to assert that the functions er1 x and er2 x are solutions to
the equation (1). If we assume that the law of exponents applies
to complex numbers, then
e(α+i β)x = eαx ei βx
ei β = cos β + i sin β
ei βx −?
e(α+i β)x = eαx (cos βx + i sin βx)
Lemma 1.
Let z(x) = u(x) + iv (x) be a complex valued function of the real
variable x, here u(x) and v (x) are real valued functions. And let
z(x) be a solution to the equation (1). Then, the functions u(x)
and v (x) are real-valued solutions to the equation (1).
When we apply Lemma 1 to the solution
e(α+i β)x = eαx (cos βx + i sin βx),
we obtain the following.
Complex conjugate roots.
If the auxiliary equation has complex conjugate roots α ± i β, then
two linearly independent solutions to (1) are eαx cos βx and
eαx sin βx and a general solution is
y (x) = c1 eαx cos βx + c2 eαx sin βx,
where c1 and c2 are arbitrary constants.
Example 1. Find the general solution to the following equations.
1. y ′′ + 4y = 0
2. y ′′ + 4y ′ + 13y = 0
3. y ′′ − 6y ′ + 25y = 0
Example 2. Solve the given IVPs.
1. y ′′ + 9y = 0,
y (0) = −2, y ′ (0) = 3
2. 3y ′′ − 6y ′ + 4y = 0,
1
y (0) = 0, y ′ (0) = − √
3
Section 4.7 Superposition and nonhomogeneous equations
Let L be a linear differential operator
L[y ](x) = y ′′ + p(x)y ′ + q(x)y .
L maps function y into the function y ′′ + p(x)y ′ + q(x)y .
Suppose that y1 (x) an y2 (x) are two functions, and
L[y1 ](x) = y1′′ (x) + p(x)y1′ (x) + q(x)y1 (x) = g1 (x),
L[y2 ](x) = y2′′ (x) + p(x)y2′ (x) + q(x)y2 (x) = g2 (x).
Since L is a linear operator,
L[c1 y1 + c2 y2 ](x) = c1 L[y1 ](x) + c2 L[y2 ](x) = c1 g1 (x) + c2 g2 (x).
Theorem 1 (superposition principle) Let y1 be a solution to a
differential equation
L[y1 ](x) = y1′′ (x) + p(x)y1′ (x) + q(x)y1 (x) = g1 (x)
and let y2 be a solution to a differential equation
L[y2 ](x) = y2′′ (x) + p(x)y2′ (x) + q(x)y2 (x) = g2 (x),
where L is a linear differential operator. Then for any constant c1
and c2 , the function c1 y1 + c2 y2 is the solution to the differential
equation
L[y ](x) = c1 g1 (x) + c2 g2 (x).
Example 3. Given that y1 (x) = cos x is a solution to
y ′′ − y ′ + y = sin x,
2x
and y2 (x) = e3 is a solution to
y ′′ − y ′ + y = e2x ,
find solutions to the following differential equations:
1. y ′′ − y ′ + y = 5 sin x;
2. y ′′ − y ′ + y = 4 sin x + 18e2x
Theorem 2 (representation of solutions for nonhomogeneous
equations). Let yp (x) be particular solution to the
nonhomogeneous equation
y ′′ + p(x)y ′ + q(x)y = g (x)
(3)
on the interval (a, b) and let y1 (x) and y2 (x) be linearly
independent solutions on (a, b) of the corresponding homogeneous
equation
y ′′ + p(x)y ′ + q(x)y = 0.
Then a general solution of (5) on the interval (a, b) can be
expressed in form
y (x) = yp (x) + c1 y1 (x) + c2 y2 (x).
(4)
Procedure for solving nonhomogeneous equations.
To solve y ′′ + p(x)y ′ + q(x)y = g (x):
(a) Determine the general solution c1 y1 (x) + c2 y2 (x) of the
corresponding homogeneous equation.
(b) Find the particular solution yp (x) of the given
nonhomogeneous equation.
(c) Form the sum of the particular solution and a general solution
to the homogeneous equation
y (x) = yp (x) + c1 y1 (x) + c2 y2 (x),
to obtain the general solution to the given equation.
Example 4. A nonhomogeneous equation and a particular
solution are given. Find a general solution to the equation.
1. y ′′ − 2y ′ + y = 8ex , yp (x) = 4x 2 ex .
2. y ′′ + 4y = sin x − cos x, yp (x) = 31 (sin x − cos x).
2
3. y ′′ + y ′ − 12y = x 2 − 1, yp (x) = − x2 −
x
2
+ 54 .
Section 4.8 Method of Undetermined Coefficients
In this section, we give a simple procedure for finding a particular
solution to the equation
ay ′′ + by ′ + cy = g (x),
(5)
when the nonhomogeneous term g (x) is of a special form
g (x) = eαx (Pm1 (x) cos βx + Qm2 (x) sin βx),
where
Pm1 (x) = p0 x m1 + p1 x m1 −1 + p2 x m1 −2 + . . . + pm1 −1 x + pm1
is a polynomial of degree m1 and
Qm2 (x) = q0 x m2 + q1 x m2 −1 + q2 x m2 −2 + . . . + qm2 −1 x + qm2
is a polynomial of degree m2 , α, β ∈ R.
To apply the method of undetermined coefficients, we first have to
solve the auxiliary equation for the corresponding homogeneous
equation
ar 2 + br + c = 0
Let α = β = 0, then
g (x) = p0 x m + p1 x m−1 + p2 x m−2 + . . . + pm−1 x + pm1 .
We seek a particular solution of the form
yp (x) = Ax m + Bx m−1 + Cx m−2 + . . . + Dx + F ,
if r = 0 is not a root to the auxiliary equation. Here A, B, C , D,
and F are unknown numbers.
If r = 0 is one of two roots of the auxiliary equation, then the
particular solution is
yp (x) = x(Ax m + Bx m−1 + Cx m−2 + . . . + Dx + F )
To find unknowns A, B, C ,..., D, and F , we have to substitute
yp (x), yp′ (x), and yp′′ (x) into equation (5). Set the corresponding
coefficients from both sides of this equation to each other to form
a system of linear equations with unknowns A, B, C ,..., D, and F .
Solve the system of linear equation for A, B, C ,..., D, and F .
Example 1. Find the general solution to the equation
y ′′ + y ′ = 1 − 2x 2 .