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Reduction of Order
We shall take a brief break from developing the general theory for linear differential equations to discuss one method (the “reduction of order method”) for finding the general solution to any linear differential equation. In some ways, this method may remind you of the material in chapter
- Indeed, part of the method involves solving a higher-order equation via first-order methods as discussed in chapter 11. The general theory developed in chapter 12 will not, however, be used to any great extent. Instead, the material developed here will help us finish that general theory (at least partially confirming the suspicions raised at the end of the chapter), and will help lead us to the complete result on constructing general solutions from particular solutions. But why worry about completing that general theory if any linear differential equation can be completely solved by this “reduction of order method”? Because this method requires that one solution to the differential equation already be known. This limits the method’s applicability. Also, serious practical difficulties arise when the differential equation to be solved is of order three or more. Still, there are situations where the method is of practical value, and it will help us confirm suspicions we already have about general solutions. Oh yes, there is another reason to develop this method: A rather powerful method for solving nonhomogeneous equations, the “variation of parameters” method described in chapter 23, is simply a clever refinement of the reduction of order method.
13.1 The General Idea
The “reduction of order method” is a method for converting any linear differential equation to another linear differential equation of lower order, and then constructing the general solution to the original differential equation using the general solution to the lower-order equation. In general, to use this method with an N th-order linear differential equation
a 0 y ( N^ )^ + a 1 y ( N^ −^1 )^ + · · · + aN − 2 y ′′^ + aN − 1 y ′^ + aN y = g ,
we need one known nontrivial solution y 1 = y 1 ( x ) to the corresponding homogeneous differ- ential equation. We then try a substitution of the form
y = y 1 u
where u = u ( x ) is a yet unknown function (and y 1 = y 1 ( x ) is the aforementioned known solution). Plugging this substitution into the differential equation then leads to a linear differential
283
284 Reduction of Order
equation for u. As we will see, because y 1 satisfies the homogeneous equation, the differential equation for u ends up being of the form
A 0 u ( N^ )^ + A 1 u ( N^ −^1 )^ + · · · + AN − 2 u ′′^ + AN − 1 u ′^ = g
— remarkably, there is no “ AN u ” term. This means we can use the substitution
v = u ′^ ,
as discussed in chapter 11, to rewrite the differential equation for u as a ( N − 1 )th-order differential equation for v ,
A 0 v( N^ −^1 )^ + A 1 v( N^ −^2 )^ + · · · + AN − 2 v′^ + AN − 1 v = g.
So we have reduced the order of the equation to be solved. If a general solution v = v( x ) for this equation can be found, then the most general formula for u can be obtained from v by integration (since u ′^ = v ). Finally then, by going back to the original substitution formula y = y 1 u , we can obtain a general solution to the original differential equation. This method is especially useful for solving second-order homogeneous linear differential equations since (as we will see) it reduces the problem to one of solving relatively simple first- order differential equations. Accordingly, we will first concentrate on its use in finding general solutions to second-order, homogeneous linear differential equations. Then we will briefly discuss using reduction of order with linear homogeneous equations of higher order, and with nonhomogeneous linear equations.
13.2 Reduction of Order for Homogeneous Linear
Second-Order Equations
The Method
Here we lay out the details of the “reduction of order method” for second-order homogeneous linear differential equations. To illustrate the method, we’ll use the differential equation
x^2 y ′′^ − 3 x y ′^ + 4 y = 0.
Note that the first coefficient, x^2 , vanishes when x = 0. From comments made in chapter 12 (see page 261), we should suspect that x = 0 ought not be in any interval of interest for this equation. So we will be solving over the intervals ( 0 , ∞) and (−∞, 0 ). Before starting the reduction of order method, we need one nontrivial solution y 1 to our differential equation. Ways for finding that first solution will be discussed in later chapters. For now let us just observe that if y 1 ( x ) = x^2 ,
then
x^2 y 1 ′′^ − 3 x y 1 ′^ + 4 y 1 = x^2 d^2 dx^2
[
x^2
]
− 3 x d dx
[
x^2
]
[
x^2
]
= x^2 [ 2 · 1 ] − 3 x [ 2 x ] + 4 x^2
= x^2 [ ︸ 2 − ( (^3) ︷︷ · 2 ) + (^4) ︸] 0!
286 Reduction of Order
Plugging the formulas just computed above for y , y ′ and y ′′ into equation (13.1), we get
0 = x^2 y ′′^ − 3 x y ′^ + 4 y = x^2
[
2 u + 4 xu ′^ + x^2 u ′′
]
− 3 x
[
2 xu + x^2 u ′
]
[
x^2 u
]
= 2 x^2 u + 4 x^3 u ′^ + x^4 u ′′^ − 6 x^2 u − 3 x^3 u ′^ + 4 x^2 u = x^4 u ′′^ +
[
4 x^3 − 3 x^3
]
u ′^ +
[
2 x^2 − 6 x^2 + 4 x^2
]
u = x^4 u ′′^ + x^3 u ′^ + 0 · u.
Notice that the u term drops out! So the resulting differential equation for u is simply x^4 u ′′^ + x^3 u ′^ = 0 , which we can further simplify by dividing out x^4 ,
u ′′^ + 1 x u ′^ = 0
In general, plugging in the formulas for y and its derivatives into the given differential equation yields
0 = ay ′′^ + by ′^ + cy = a
[
y 1 ′′ u + 2 y 1 ′ u ′^ + y 1 u ′′
]
[
y 1 ′ u + y 1 u ′
]
[
y 1 u
]
= a y 1 ′′ u + 2 a y 1 ′ u ′^ + ay 1 u ′′^ + by 1 ′ u + by 1 u ′^ + cy 1 u = ay 1 u ′′^ +
[
2 a y 1 ′^ + by 1
]
u ′^ +
[
a y 1 ′′^ + by 1 ′^ + cy 1
]
u.
That is, the differential equation becomes
Au ′′^ + Bu ′^ + Cu = 0
where
A = ay 1 , B = 2 a y 1 ′^ + by 1 and C = a y 1 ′′^ + by 1 ′^ + cy 1.
But remember, y 1 is a solution to the homogeneous equation
ay ′′^ + by ′^ + cy = 0.
Consequently, C = a y 1 ′′^ + by 1 ′^ + cy 1 = 0 , and the differential equation for u automatically reduces to
Au ′′^ + Bu ′^ = 0.
The u term always drops out!
3. Now find the general solution to the second-order differential equation just obtained for u , Au ′′^ + Bu ′^ = 0 , via the substitution method discussed in chapter 11:
Reduction of Order for Homogeneous Linear Second-Order Equations 287
(a) Let u ′^ = v (and, thus, u ′′^ = v′^ = d v/ dx ) to convert the second-order differential equation for u to the first-order differential equation for v ,
A d v dx
(It is worth noting that this first-order differential equation will be both linear and separable.) (b) Find the general solution v( x ) to this first-order equation. (Since it is both linear and separable, you can solve it using either the procedure developed for first-order linear equations or the approach developed for first-order separable equations.) (c) Using the formula just found for v , integrate the substitution formula u ′^ = v to obtain the formula for u ,
u ( x ) =
v( x ) d x.
Don’t forget all the arbitrary constants. In our example, we obtained
u ′′^ + 1 x u ′^ = 0.
Letting v = u ′ and, thus, v′^ = u ′′ this becomes d v dx
1 x v = 0.
Equivalently, d v dx
1 x v. This is a relatively simple separable first-order equation. It has one constant solution, v = 0. To find the others, we divide through by v and proceed as usual with such equations: 1 v
d v dx
1 x
H⇒
1 v
d v dx d x = −
1 x d x
H⇒ ln |v| = − ln | x | + c 0
H⇒ v = ± e −^ ln| x | +^ c^0
H⇒ v = ± ec^0 x −^1.
Letting A = ± ec^0 , this simplifies to
v = A x
which also accounts for the constant solution (when A = 0 ). Since u ′^ = v , it then follows that
u ( x ) =
v( x ) d x =
A x d x = A ln | x | + B.
Reduction of Order for Nonhomogeneous Linear Second-Order Equations 289
13.3 Reduction of Order for Nonhomogeneous Linear
Second-Order Equations
If you look back over our discussion in section 13.2, you will see that the reduction of order method applies almost as well in solving a nonhomogeneous equation
ay ′′^ + by ′^ + cy = g ,
provided that “one solution y 1 ” is a solution to the corresponding homogeneous equation
ay ′′^ + by ′^ + cy = 0.
Then, letting y = y 1 u in the nonhomogeneous equation and then replacing u ′^ with v leads to an equation of the form A v′^ + B v = g
instead of
A v′^ + B v = 0.
So we don’t end up with a first-order equation for v which is both separable and linear; it is just linear. Still, we know how to solve such equations. Solving that first-order linear differential equation for v and continuing with the method already described finally yields the general solution to the desired nonhomogeneous differential equation. We will do one example. Then I’ll tell you why the method is rarely used in practice.
! ◮ Example 13.1: Let us try to solve the second-order nonhomogeneous linear differential equation x^2 y ′′^ − 3 x y ′^ + 4 y =
x (13.2)
over the interval ( 0 , ∞). As we saw in our main example in the section 13.2, the corresponding homogeneous equation x^2 y ′′^ − 3 x y ′^ + 4 y = 0. has y 1 ( x ) = x^2 as one solution (in fact, from that example, we know the entire general solution to this homogeneous equation, but we only need this one particular solution for the method). Let y = y 1 u = x^2 u
where u = u ( x ) is the function yet to be determined. The derivatives of y are
y ′^ =
x^2 u
= 2 xu + x^2 u ′
and y ′′^ = ( y ′)′^ =
2 xu + x^2 u ′
= 2 u + 2 xu ′^ + 2 xu ′^ + x^2 u ′′ = 2 u + 4 xu ′^ + x^2 u ′′.
290 Reduction of Order
Plugging these into equation (13.2) yields √ x = x^2 y ′′^ − 3 x y ′^ + 4 y = x^2
[
2 u + 4 xu ′^ + x^2 u ′′
]
− 3 x
[
2 xu + x^2 u ′
]
[
x^2 u
]
= 2 x^2 u + 4 x^3 u ′^ + x^4 u ′′^ − 6 x^2 u − 3 x^3 u ′^ + 4 x^2 u = x^4 u ′′^ +
[
4 x^3 − 3 x^3
]
u ′^ +
[
2 x^2 − 6 x^2 + 4 x^2
]
u = x^4 u ′′^ + x^3 u ′^ + 0 · u.
As before, the u term drops out. In this case, we are left with
x^4 u ′′^ + x^3 u ′^ =
x.
That is, x^4 v′^ + x^3 v = x^1 /^2 with v = u ′. This is a relatively simple first-order linear equation. To help find the integrating factor, we now divide through by x^4 , obtaining d v dx
1 x
v = x −^7 /^2.
Thus, the integrating factor is
μ = e
∫ (^1) x dx^ = e ln| x |^ = | x |.
Since we are just attempting to solve over the interval ( 0 , ∞) , we really just have
μ = x.
Multiplying the last differential equation for v and proceeding as usual when solving first- order linear differential equations:
x
[
d v dx
1 x v
]
= x
[
x −
7 / 2 ]
H⇒ x d v dx
H⇒ d dx
[
x v
]
= x −^5 /^2
H⇒
d dx
[
x v
]
d x =
x − (^5) / 2 d x
H⇒ x v = − 2 3 x − (^3) / 2
H⇒ v = − 2 3 x − (^5) / 2
c 1 x Recalling that v = u ′ , we can rewrite the last line as du dx
2 3 x −^5 /^2 + c 1 x
292 Reduction of Order
13.4 Reduction of Order in General
In theory, reduction of order can be applied to any linear equation of any order, homogeneous or not. Whether it’s application is useful is another issue.
! ◮ Example 13.2: Consider the third-order homogeneous linear differential equation
y ′′′^ − 8 y = 0. (13.4)
If you rewrite this equation as y ′′′^ = 8 y , and think about what happens when you differentiate exponentials, you will realize that
y 1 ( x ) = e^2 x
is ‘obviously’ a solution to our differential equation (verify it yourself). Letting
y = y 1 u = e^2 x^ u
and repeatedly using the product rule, we get
y ′^ =
e^2 x^ u
= 2 e^2 x^ u + e^2 x^ u ′ ,
y ′′^ =
e^2 x^ u
2 e^2 x^ u + e^2 x^ u ′
= 4 e^2 x^ u + 2 e^2 x^ u ′^ + 2 e^2 x^ u ′^ + e^2 x^ u ′′ = 4 e^2 x^ u + 4 e^2 x^ u ′^ + e^2 x^ u ′′ ,
and y ′′′^ =
e^2 x^ u
4 e^2 x^ u + 4 e^2 x^ u ′^ + e^2 x^ u ′′
= 8 e^2 x^ u + 4 e^2 x^ u ′^ + 8 e^2 x^ u ′^ + 4 e^2 x^ u ′′^ + 2 e^2 x^ u ′′^ + e^2 x^ u ′′′ = 8 e^2 x^ u + 12 e^2 x^ u ′^ + 6 e^2 x^ u ′′^ + e^2 x^ u ′′′.
So, using y = e^2 x^ u ,
y ′′′^ − 8 y = 0
H⇒
[
8 e^2 x^ u + 12 e^2 x^ u ′^ + 6 e^2 x^ u ′′^ + e^2 x^ u ′′′
]
[
e^2 x^ u
]
H⇒ e^2 x^ u ′′′^ + 6 e^2 x^ u ′′^ + 12 e^2 x^ u ′^ +
[
8 e^2 x^ − 8 e^2 x^
]
u = 0.
Again, the u term cancel out, leaving us with
e^2 x^ u ′′′^ + 6 e^2 x^ u ′′^ + 12 e^2 x^ u ′^ = 0.
Letting v = u ′ and dividing out the exponential, this becomes the second-order differential equation v′′^ + 6 v′^ + 12 v = 0. (13.5)
Additional Exercises 293
Thus we have changed the problem of solving the third-order differential equation to one of solving a second-order differential equation. If we can now correctly guess a particular solution v 1 to that second-order differential equation, we could again use reduction of order to get the general solution v( x ) to that second-order equation, and then use that and the fact that y = e^2 x^ u with v = u ′ to obtain the general solution to our original differential equation. Unfortunately, even though the order is less, “guessing” a solution to equation (13.5) is a good deal more difficult than was guessing a particular solution to the original differential equation, equation (13.4).
As the example illustrates, even if we can, somehow, obtain one particular solution to a given N th-order linear homogeneous linear differential equation, and then use it to reduce the problem to solving an ( N − 1 )th-order differential equation, that lower order differential equation may be just as hard to solve as the original differential equation (unless N = 2 ). In fact, we will learn how to solve differential equations such as equation (13.5), but those methods can also be used to find the general solution to the original differential equation, equation (13.4), as well. Still it does no harm to know that the problem of solving an N th-order linear homogeneous linear differential equation can reduced to that of solving an ( N − 1 )th-order differential equation, especially since we may refer to this fact in the next chapter. For the record, here is a theorem to that effect:
Theorem 13.1 (reduction of order in homogeneous equations) Let y be any solution to some N^ th-order homogeneous differential equation
a 0 y ( N^ )^ + a 1 y ( N^ −^1 )^ + · · · + aN − 2 y ′′^ + aN − 1 y ′^ + aN y = g (13.6)
where g and the ak ’s are known functions on some interval of interest I , and let y 1 be a nontrivial particular solution to the corresponding homogeneous equation
a 0 y ( N^ )^ + a 1 y ( N^ −^1 )^ + · · · + aN − 2 y ′′^ + aN − 1 y ′^ + aN y = 0.
Set
u = y y 1 (so that y = y 1 u ).
Then v = u ′ satisfies an ( N − 1 )^ th-order differential equation
A 0 v( N^ −^1 )^ + A 1 v( N^ −^2 )^ + · · · + AN − 2 v′^ + AN − 1 v = g.
where the Ak ’s are functions on the interval I that can be determined from the ak ’s along with y 1 and its derivatives.
The proof is relatively straightforward: You see what happens when you repeatedly use the product rule with y = y 1 u , and plug the results into the equation (13.6). I’ll leave the details to you (see exercise 13.3).
Additional Exercises 295
given y 1 is a solution to the given differential equation or to the corresponding homo- geneous equation (as appropriate), and then find the general solution to the differential equation using the given y 1 with the method of reduction of order. a. y ′′′^ − 9 y ′′^ + 27 y ′^ − 27 y = 0 , y 1 = e^3 x b. y ′′′^ − 9 y ′′^ + 27 y ′^ − 27 y = e^3 x^ sin( x ) , y 1 = e^3 x c. y (^4 )^ − 8 y ′′′^ + 24 y ′′^ − 32 y ′^ + 16 y = 0 , y 1 = e^2 x d. x^3 y ′′′^ − 4 y ′′^ + 10 y ′^ − 12 y = 0 , y 1 = x^2
