Math 308, Sections 301, 302,
Summer 2008
Lecture 8.
06/16/2008
Chapter 4. Linear Second Order Equations
Section 4.4 Reduction of Order
A general solution to a linear second order homogeneous equation
is given by a linear combination of two linearly independent
solutions.
Let f be nontrivial solution to equation
y ′′ + p(x)y ′ + q(x)y = 0.
Let’s try to find solution of the form
y (x) = v (x)f (x),
where v (x) is an unknown function.
y ′ = v ′ f + vf ′ ,
y ′′ = v ′′ f + 2v ′ f ′ + vf ′′ .
Substituting these expression into equation gives
v ′′ f + 2v ′ f ′ + vf ′′ + p(v ′ f + vf ′ ) + qvf = 0
or
(f ′′ + pf ′ + qf )v + fv ′′ + (2f ′ + pf )v ′ = 0.
Since f is the solution,
fv ′′ + (2f ′ + pf )v ′ = 0.
Let’s w (x) = v ′ (x), then we have
fw ′ + (2f ′ + pf )w = 0,
separating the variables and integrating gives
f′
dw
= (−2 − p)dx,
w
f
Z
Z ′
Z
dw
f
= −2
dx − pdx,
w
f
Z
ln |w | = ln |f −2 | − pdx,
w =±
e−
R
e−
R
pdx
,
f2
which holds on any interval where f (x) 6= 0.
′
v =±
v =±
Z
pdx
f2
R
,
e− p(x)dx
.
[f (x)]2
Examples 1.
(a) Given that f (x) = e3x is a solution to
y ′′ + 2y ′ − 15y = 0,
determine the second linear independent solution.
(b) Given that f (x) =
1
x
is a solution to
x 2 y ′′ − 2xy ′ − 4y = 0,
x >0
determine the second linear independent solution.
(c) The equation
xy ′′′ − xy ′′ + y ′ − y = 0
has f (x) = ex as a solution. Use the substitution y (x) = v (x)f (x)
to reduce this third order equation to a second order equation.
Section 4.5 Homogeneous Linear Equations with Constant
Coefficients
For the equation
ay ′′ + by ′ + cy = 0,
(1)
where a, b, c are constants, we try to find a solution of the form
y = erx . If we substitute y = erx into (1), we obtain
(ar 2 + br + c)erx = 0.
Since erx is never zero,
ar 2 + br + c = 0.
(2)
Consequently, y = erx is a solution to (1) if an only if r satisfies
(2). Equation (2) is called the auxiliary equation or
characteristic equation associated with equation (1).
So, the equation (2) is a quadratic, and its roots are:
√
√
−b + b 2 − 4ac
−b − b 2 − 4ac
r1 =
, r2 =
2a
2a
√
2
When b − 4ac > 0, then r1 , r2 ∈ R and r1 6= r2 . So,
y1 (x) = er1 x and y2 (x) = er2 x are two linearly independent
solutions to (1) and
y (x) = c1 er1 x + c2 er2 x
is the general solution to (1).
√
If b2 − 4ac = 0, then the equation (2) has a repeated root
b
. In this case, y1 (x) = erx and y2 (x) = xerx are two
r ∈ R, r = − 2a
linearly independent solutions to (1) and
y (x) = c1 erx + c2 xerx = (c1 + c2 x)erx
is the general solution to (1).
Example 2.
Find a general solution to the given equation
(a) y ′′ − y ′ − 2y = 0.
(b) y ′′ + 6y ′ + 9y = 0.
(c) y ′′ − 5y ′ + 6y = 0.
(d) 3y ′′ + 11y ′ − 7y = 0.
Example 3.
Solve the given initial value problems.
(a) y ′′ + y ′ = 0, y (0) = 2, y ′ (0) = 1.
(b) y ′′ − 4y ′ + 4y = 0, y (1) = 1, y ′ (1) = 1.
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,
(3)
dx
dx
where a, b, and c are constants is called a homogeneous
Cauchy-Euler equation.
To solve a homogeneous Cauchy-Euler equation, we make the
substitution x(t) = et . Because x(t) = et , it follows by the Chain
Rule that
dy
dy dx
dy t
dy
=
=
e =x ,
dt
dx dt
dx
dx
2
d dy
d
dy
d y
=
=
x
=
dt 2
dt dt
dt
dx
d dy
d 2 y dx
dx dy
dy
+x
+x 2
=
=
= et
dt dx
dt dx
dx
dx dt
et
d 2y
d 2y
d 2y
dy
dy
dy
+ xet 2 = x
+ x2 2 =
+ x2 2 .
dx
dx
dx
dx
dt
dx
x2
d 2y
d 2y
dy
=
−
.
2
2
dx
dt
dt
2
2d y
Substituting into (3) the expressions for x dy
dx and x dx 2 gives
a
d 2y
dy
−
dt 2
dt
+b
dy
+ cy = 0,
dt
dy
d 2y
+ (b − a)
+ cy = 0,
2
dt
dt
which is linear second order differential equation with constant
coefficients.
a
Example 4. Find the general solution to the equation
x 2 y ′′ (x) + 7xy ′ (x) − 7y (x) = 0
for x > 0.