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LECTURE 15
Reduction of Order
Recall that the general solution of a second order homogeneous linear differential equation
(15.1) L[y] = y′′ + p(x)y′ + q(x)y = 0
is given by
(15.2) y(x) = c 1 y 1 (x) + c 2 y 2 (x)
where y 1 and y 2 are any two solutions such that
(15.3) W y 1 , y 2 = y 1 (x)y′ 2 (x) − y′ 1 (x)y 2 (x) = 0.
In this section we shall assume that we have already found one solution y 1 of (15.1)and that we are seeking to find another solution y 2 so that we can write down the general solution as in (15.2).
So suppose we have one non-trivial solution y 1 (x)of (15.1)and suppose there is another solution of the form
(15.4) y 2 (x) = v(x)y 1 (x).
Then
W [y 1 , y 2 ] = y 1 y 2 ′ − y′ 1 y 2
= y 1 (v′ y 1 + vy′ 1 ) − y 1 ′ (vy 1 )
= (y 1 )^2 v′
unless v′ = 0. Thus, any solution we construct by multiplying our given solution y 1 (x)by a non-constant function v(x)will give us another linearly independent solution.
The question we now wish to address is: how does one find an appropriate function v(x)?
Certainly, we want to choose v(x)so that y 2 (x) = v(x)y 1 (x)satisfies (15.1). So let us insert y(x) = v(x)y 1 (x) into (15.1):
0 = d
2
dx^2 (vy^1 ) +^ p(x)^
d
dx (vy^1 ) +^ q^ (vy^1 )
= v′′ y 1 + 2v′ y′ 1 + vy′′ 1 + p(x)v′ y 1 + p(x)vy′ 1 + qvy 1
= v (y′′ 1 + p(x)y′ 1 + q(x)y 1 ) + v′′ y 1 + (2y′ 1 + p(x)y 1 ) v′
The first term vanishes since y 1 is a solution of (15.1), so v(x)must satisfy
(15.7) 0 = y 1 v′′ + (2y′ 1 + p(x)y 1 ) v′
or
v′′ +
p(x) +
2 y′ 1
y 1
(15.8) v′ = 0.
Now set
(15.9) u(x) = v′ (x).
1
- REDUCTION OF ORDER 2
Then we have
u′ +
p(x) +
2 y′ 1 (x)
y 1 (x)
(15.10) u = 0.
This is a first order linear differential equation which we know how to solve. Its general solution is
u(x) = C exp
[
−
∫ x^ (
p(t) + 2 y
′ 1 (t) y 1 (t)
dt
]
= C exp
[
−
∫ x
(p(t)) dt − 2
∫ x y′ 1 (t)
y 1 (t) dt
(15.11)^ ]
Now note that
d
dt
ln [y 1 (t) ] =
y′ 1 (t)
y 1 (t)
so
exp
[
∫ x y′ 1 (t)
y 1 (t) dy
]
= exp
[
∫ x d
dt (ln [y^1 (t)])^ dt
]
= exp [−2 ln [y 1 (x)]] = exp
[
ln
[
(y 1 (x))−^2
]]
= (^) (y^1 1 (x))^2
Thus, (15.11)can be written as
u(x) =
C
(y 1 (x))^2
exp
[
−
∫ x
p(t)dt
]
Now recall from (15.9)that u(x)is the derivative of the factor v(x)which we originally sought out to find.
So
v(x) =
∫ x
u(t) dt + D
∫ x^ [^ C
(y 1 (t))^2 exp
[
−
∫ t
p(t′ )dt′
]]
+ D
It is not too difficult to convince oneself that it is not really necessary to carry along the constants of
integration C and D. For the constant D can be absorbed into the constant c 1 of the general solution
y(x) = c 1 y 1 (x) + c 2 y 2 (x), while the factor C can be absorbed into the constant c 2. Thus, without loss of
generality, we can take C = 1 and D = 0. So given one solution y 1 (x)of (15.1), a second solution y 2 (x)of
(15.1)can be formed by computing
v(x) =
∫ x
(15.16) u(t) dt
where
u(x) =
(y 1 (x))^2
exp
[
−
∫ x
p(t)dt
]
and then setting
(15.18) y 2 (x) = v(x)y 1 (x).
The general solution of (15.1)is then
(15.19) y(x) = c 1 y 1 (x) + c 2 v(x)y 1 (x).
This technique for constructing the general solution from single solution of a second order linear homogene- neous differential equation is called reduction of order.
For those of you who like nice tidy formulae we can write
y 2 (x) = y 1 (x)
∫ x
(y 1 (s))^2
exp
[
−
∫ s
p(t)dt
]
(15.20) ds
for the second solution.
