## Calculus Of One Real Variable By Pheng Kim Ving Chapter 16: Differential Equations Group 16.2: Second-Order Linear Homogeneous Equations Section 16.2.2: Equations With Variable Coefficients – Reduction Of Order

16.2.2
Equations With Variable Coefficients – Reduction Of Order

 1. Second-Order Linear Homogeneous Differential Equations With     Variable Coefficients

In this section, we'll use the abbreviations:

 “DE” for “differential equation”, “GS” for “general solution”, “HDE” for “homogeneous differential equation”, and “HE” for “homogeneous equation”.

Recall that a second-order linear homogeneous differential equation with constant coefficients is one of the form:

y'' + by' + cy = 0,

where the coefficients b and c are constants. When b is replaced by a non-constant function b(x) or c is replaced by a
non-constant function c(x) or both, one or both coefficients of course vary, they become variable. So naturally an equation of
the form:

y'' + b(x) y' + c(x) y = 0,

where b(x) or c(x) or both are non-constant functions of x, is said to be an equation with variable coefficients. Note that x is
the independent variable of the function y, and thus of the functions y' and y'' also. All the functions in the equation are of the
same variable.

Definition 1.1 – Second-Order Linear Homogeneous Differential Equations With Variable
Coefficients

 A differential equation of the form:     where b(x) or c(x) or both are non-constant functions, is called a second-order linear homogeneous differential equation with variable coefficients.

Recall that while the equation is linear, each function y, y', and y'' doesn't have to be linear. For example, y = 2x + 3 is linear
while y = x2 + 3 and y = e2x + 3 aren't.

 2. General Solution

Just like the GS of the constant-coefficients HE:

y'' + by' + cy = 0

as presented in Section 16.2.1 Theorem 2.2, the GS of the variable-coefficients HE:

y'' + b(x) y' + c(x) y = 0

is the linear combination of any two of its linearly independent PSs. The proof of the following theorem is given in a course on
differential equations and is omitted here.

Theorem 2.1 – General Solution

 If y1 and y2 are linearly independent particular solutions of the HE:   y'' + b(x) y' + c(x) y = 0,   then the GS of this equation is:   y = c1 y1 + c2 y2,   where c1 and c2 are arbitrary constants.

 3. The Method Of Reduction Of Order

Recall that to solve a DE is to find its GS. Let's discuss the solving of our equation:

By Theorem 2.1, we obtain the GS of Eq. [3.1] provided we get two linearly independent solutions y1 and y2 of this equation.
The GS is then given by the linear combination:

y = c1 y1 + c2 y2,

where c1 and c2 are arbitrary constants. Clearly y1 = 0 is a solution. But it's not linearly independent of any other solution (0 =
0y2 whatever y2 is). So of course we're interested in non-0 solutions.

Unlike the case of constant-coefficients HEs as developed in Section 16.2.1 Part 3, there's no standard procedure to find two
linearly independent solutions y1 and y2 of the variable-coefficients HE [3.1]. However, there's a general procedure for finding
another solution if one solution is known. It's often possible to find one solution by inspecting the equation or by trial and error.

Let y1 be a non-0 solution of Eq. [3.1]. If y2 is another solution that's linearly independent of y1, then y2/y1 isn't a constant.
Thus:

Indeed y2 is a solution. Now let's verify that it's linearly independent of y1. We have:

We've reduced a second-order equation to a first-order equation. The order of the derivative is reduced. So this approach of
using one solution to find another is called the method of reduction of order.

We've of course proved the following theorem.

Theorem 3.1 –A Solution By The Method Of Reduction Of Order

 Suppose y1 is a solution of the DE:     where c1 and c2 are arbitrary constants.

##### Example 3.1

Consider the DE:

x2y''xy' + y = 0,     x > 0,

a. Verify that y1 = x is a solution.

b. Find a second solution y2 that's linearly independent of y1 by using the formula obtained by the method of reduction of order.
c. Find y2 by imitating the method of reduction of order itself, not using the formula obtained by it.
d. Are the answers in parts b and c the same?
e. Verify that y2 is indeed a solution that's linearly independent of y1.

Solution

xu'' + u' = 0,
divide both sides by xu':

y2 = uy1 = (ln x) x = x ln x.

d. Yes.

Since ln x isn't a constant, y2 is linearly independent of y1. Thus indeed y2 is a solution that's linearly independent of y1.

## EOS

##### Remarks 3.2 – About Example 3.1

a. In part b, we utilize the formula for y2, which contains the function b(x), so we should write the given equation in the form
y'' + b(x) y' + c(x) y = 0 (coefficient of y'' is 1), so that it's clear what b(x) is. Caution: b(x) isn't –x, it's –1/x.

b. Part c illustrates the following general procedure of the method of reduction of order:

– let y2 = uy1 = g(x)u, where y1 = g(x), use g(x) only, not y1,
– substitute y2 = g(x)u into the original equation,
– transform the term containing u'' in an equation into the term u''/u' by division,

– let v = u',
– find v, then u, then y2, see Fig. 3.1.

### Fig. 3.1

Process From y2 To u To v Then Back To u To y2.

d. If we're not asked to verify a solution, then we don't have to verify.

##### Remark 3.3

It's recommended that you remember the procedure of the method of reduction of order, as outlined in part b of Remarks 3.2
above. In case you forget its formula or don't remember the formula clearly, you can utilize the procedure of the method.
Sometimes you may even be required to use the method itself, not the formula obtained by it.

Choosing y2

Suppose it's determined that y2 is a constant multiple of a function, ie that it's of the form y2 = k f (x), where k is a non-0
constant. For example, y2 = 5x3 or y2 = –3(sin 2x + 7e4x). We now verify that f (x) is also a solution that's linearly independent
of y1. We have:

which isn't constant, since y2/y1 isn't. Thus f (x) is linearly independent of y1. That completes the verification.

We've shown that if k f (x), where k is a non-0 constant, is a solution that's linearly independent of y1, then so is f (x). We
conclude that if we determine that y2 = k f (x), then it's possible to simply choose y2 = f (x).

Let's look at this possibility from another angle. If y2 = k f (x), then the GS is y = c1y1 + c2k f (x). Re-using the letter c2 for the
product c2k, which is valid because this product, just like the factor c2 in it, is an arbitrary constant, the GS is y = c1y1 +
c2 f (x). This is the same as simply choosing y2 = f (x).

##### Example 3.2

One solution of the DE:

xy''y' + 4x3y = 0,     x > 0,

is y1 = sin (x2). Solve the equation.

Solution
xy''y' + 4x3y = 0,

Let u = x2. So du = 2x dx. Then:

So, choosing y2 = cos (x2), the GS is y = c1 sin (x2) + c2 cos (x2).
EOS

 4. For Constant-Coefficients Equations

The procedure of the method of reduction of order doesn't require that the coefficients be variable functions. So the method
also applies to constant-coefficients equations.

##### Example 4.1

Consider the constant-coefficients DE:

y'' + 5y' + 6y = 0.

a. Solve it by using the method of characteristic equation.
b. Given that one solution is y1 = e–2x, solve it by using the method of reduction of order.
c. Are the answers in parts a and b the same?

Solution

So, choosing y2 = e–3x, the GS is y = c1e–2x + c2e–3x.

c. Yes.
EOS

# Problems & Solutions

1. Consider the Legendre equation of order 1:

(1 – x2) y'' – 2xy' + 2y = 0,     |x| < 1.

a. Verify that y1 = x is a solution.

b. Find a second solution y2 that's linearly independent of y1 by using the formula obtained by the method of reduction of order.
c. Find y2 by imitating the method of reduction of order itself, not using the formula obtained by it.
d. Are the answers in parts b and c the same?
e. Verify that y2 is indeed a solution that's linearly independent of y1.

##### Solution

Let u = 1 – x2. So du = –2x dx. Then:

Using the method of partial fractions we have:

C = 0,
D = –1,
0 = A + B + C = A + B + 0 = A + B,     B = –A,
0 = – A + B + D = – AA – 1 = – 2A – 1,     A = –1/2,
B = –(–1/2) = 1/2,

Using the method of partial fractions we have:

Using the method of partial fractions we have:

d. Yes.

e) Let's show that y2 is a solution:

So indeed y2 is a solution. Now let's show that it's linearly independent of y1:

which clearly isn't constant. Thus indeed y2 is linearly independent of y1.

2. Consider the DE:

x2y'' – 2xy' + (x2 + 2) y = 0,     x > 0.

a. Verify that y1 = x sin x is a solution.
b. Solve the equation.

##### Solution

Choose y2 = x cos x. So the GS of the given equation is y = c1 x sin x + c2 x cos x, or:

y = x(A cos x + B sin x),

where A = c2 and B = c1 are arbitrary constants.

3. Consider the constant-coefficients DE:

y'' + y' + 7y = 0.

a. Verify that:

is a solution.

b. Use y1 to solve the equation.

## Note

We're required to use y1 to solve the equation. So we can't apply the method of characteristic equation, which doesn't use y1.
We must apply the method of reduction of order, which uses y1 to find y2.

##### Solution

Indeed y1 is a solution.

b. Let y2 be another solution that's linearly independent of y1. Then by the method of reduction of order we have:

where A = c2 and B = c1 are arbitrary constants.

4. Consider the Bessel DE:

x2y'' + xy' + (x2p2) y = 0,     p a constant.

For p = 1/2 and x > 0:

b. Solve the equation.

##### Solution

a. When p = 1/2 the given equation becomes:

where A and B are arbitrary constants.

5. Given the DE:

xy'' – (x + n) y' + ny = 0,     x > 0,     integer constant n > 0,

verify that y1 = ex is a solution and determine another solution linearly independent of y1.

Solution

Let's verify that y1 is a solution. We have:

Thus:

In = – xnex + nIn–1

= – ex(xn) + n(– xn–1ex + (n – 1)In–2)

= – ex(xn + nxn–1) + n(n – 1)(– xn–2ex + (n – 2)In–3)

= – ex(xn + nxn–1 + n(n – 1)xn–2) + n(n – 1)(n – 2)(– xn–3ex + (n – 3)In–4)

= – ex(xn + nxn–1 + n(n – 1)xn–2 + n(n – 1)(n – 2)xn–3) + n(n – 1)(n –2)(n – 3)(– xn–4ex + (n – 4)In–5)