Chapter 7

Nonconstant Coefficients

We return to second-order linear ODEs, but with nonconstant coefficients. Thatis, we consider

(7.1) y′′ + p(t)y′ + q(t)y = 0,

with not both p(t) and q(t) constant. The theory developed in Chapter 3 still holds,and in particular Theorem 3.4 is still valid. Specifically, the general solution is still

y(t) = c1y1 + c2y2

with y1, y2 independent solutions to (7.1) and c1, c2 arbitrary constants. Unfor-tunately, there is no method to find explicit formulas for y1 and y2. There arehowever some special cases.

7.1. Reduction of Order

If somehow we are lucky enough to fine one solution to (7.1), then a method existsto find the other. The method is similar to the technique we employed in therepeated root section of Chapter 3.1.3, and in deriving the variation of parametersin Chapter 3.3. Suppose somehow we find y1 solving (7.1). We try to find a secondindependent solution by modifying it; y2 = u(t)y1(t). Substituting this back into(7.1), we find

u′′y1 + (2y′

1 + py1)u′ = 0.

Since u does not appear in the ODE, we may immediately integrate. Indeed, setz = u′. Then z satisfies the first-order ODE

(7.2) z′ +

(

p(t) + 2y′

1

y1

)

z = 0.

In principle this can be solved explicitly using methods in Chapter 1.2.

81

82 7. Nonconstant Coefficients

Example 7.1. Consider the second-order, linear, nonconstant coefficient ODE

y′′ −1 + x

xy′ +

1

xy = 0

Note that y1 = ex is a solution. To find another independent solution, we reducethe order. Here p(x) = −(1 + x)/x. Equation (7.2) is

z′ +x − 1

xz = 0.

The integrating factor is (Chapter 1.2)

µ = eR

x−1

xdx =

ex

x.

Thus (exz/x)′ = 0. That is, u′ = z = cxe−x or u = c(xe−x + e−x), and y2 =u(x)y1(x) = u(x)ex = 1 + x. The general solution is therefore

y = c1ex + c2(1 + x).

Homework 7.1 (Reduction of Order)

Find the general solution.

1. t2y′′ − 4ty′ + 6y = 0, y1(t) = t2

2. t2y′′ + 3ty′ + y = 0, y1(t) = t−1

3. xy′′ − y′ + 4x3y = 0, y1(t) = sin x2

4. xy′′ − (x + 1)y′ + y = 0, y1(t) = ex

5. x2y′′ − 2xy′ − 4y = 0, y1(t) = 1/x

Answers

1. y2 = t3

2. y2 = t−1 ln t

3. y2 = cos t2

4. y2 = x + 1

5. y2 = x4

7.2. Cauchy-Euler Equations

The Cauchy-Euler equation is a special form of Equation (7.1). It is given by

(7.3) x2y′′ + axy′ + by = 0,

where aand b are real numbers. Note that the powers of x match the number ofderivatives. The may seem fortuitous, but the ODE arises in many physical settingsincluding heat conduction and electrostatics.

The form of the equation suggests y = xα may be a solution. Indeed, y′ =αxα−1 and y′′ = α(α − 1)xα−2. Putting these in (7.3), we find

α(α − 1) + aα + b = 0.

7.3. Series Solutions 83

Using the quadratic formula we can solve for α. If the roots, r1, r2 are real, thegeneral solution is

y = c1xr1 + c2x

r2 .

If the roots are complex, λ ± iµ, then

y = c1xλxiµ + c2x

λx−iµ

= xλ(

c1eiµ lnx + c2e

−iµ lnx)

= C1xλ cos(µ lnx) + C2x

λ sin(µ lnx),

where C1 = c1 + c2 and C2 = i(c1 − c2).

Suppose the roots are repeated, r1 = r2 = r. Then we know one solution,y1 = xr. To find the other solution we employ the method of the previous section.We find in this case

y = c1xr + c2 ln(x)xr .

Homework 7.2 (Cauchy-Euler Equations)

Find the (general) solution.

1. 2x2y′′ + xy′ − 3y = 0, y(1) = 1, y′(1) = 4

2. 4x2y′′ + 8xy′ + 17y = 0

3. x2y′′ − 3xy′ + 4y = 0, y(−1) = 2, y′(−1) = 3

4. By making the substitution x = et show that (7.3) can be written y′′ + (a −1)y′ + by = 0, where differentiation is now with respect to t. Note the coeffi-cients are now constant.

Answers

1. y = 2x3/2 − x−1

2. y = c1x−1/2 cos(2 ln x) + c2x

−1/2 sin(2 lnx)

3. y = 2x2 − 7x2 ln |x|

7.3. Series Solutions

There is a general method of finding solutions to (7.1). Let us suppose that thesolution to (7.1) is analytic. That is, we suppose the solution y has a Taylor serieswhich converges on an interval. Specifically, we suppose

y(x) = a0 + a1x + a2x2 + a3x

3 + · · · + anxn + · · ·(7.4)

=

∞∑

n=0

anxn,

on some interval containing the origin, or more generally

y(x) = a0 + a1(x − x0) + a2(x − x0)2 + a3(x − x0)

3 + · · · + an(x − x0)n + · · ·

=

∞∑

n=0

an(x − x0)n,

84 7. Nonconstant Coefficients

on some interval containing x0. We suppose p(x) and q(x) are analytic as well.Familiar examples might include

ex = 1 + x +x2

2!+

x3

3!+ · · · =

∞∑

n=0

xn

n!

sin x = x −x3

3!+

x5

5!+ · · · =

∞∑

n=0

(−1)n

(2n + 1)!

n

x2n+1

cosx = 1 −x2

2!+

x4

4!+ · · · =

∞∑

n=0

(−1)n−1

(2n)!

n

x2n.

The idea is to put (7.4) into the ODE and find the coefficients. To do so we willneed some shifting formulas. One can easily verify the following.

y(x) =

∞∑

n=0

anxn,

y′(x) =∞∑

n=1

nanxn−1 =∞∑

n=0

(n + 1)an+1xn,

y′′(x) =

∞∑

n=2

n(n − 1)anxn−2 =

∞∑

n=1

(n + 1)nan+1xn−1,

=

∞∑

n=0

(n + 2)(n + 1)an+2xn.(7.5)

Example 7.2. Find the general solution to

y′′ + y = 0.

Solution. We just have to compute the an in (7.4). Using the formula (7.5), weget

∞∑

n=2

(n)(n − 1)anxn−2 +

∞∑

n=0

anxn =

∞∑

n=0

(

(n + 2)(n + 1)an+2 + an

)

xn = 0.

The polynomials on either side of the equation must balance. So, (n + 2)(n +1)an+2 + an = 0 for n = 0, 1, 2, . . .. This last equation is called an RecurrenceRelation. We can use it to find the coefficients in (7.4). It would also be easy touse in computations. If we know a0, then the recurrence relation provides us witha0, a2, a4 . . . and knowledge of a1 gives us a1, a3, a5, . . .. In particular, if we seta0 = c1, then

a0 = c1

a2 =a0

(0 + 2)(0 + 1)=

c1

2

a4 =a2

(2 + 2)(2 + 1)=

1

4 · 3a2 =

c1

4!

a2n =a2n

(2n + 2)(2n + 1)=

c1

(2n)!

7.3. Series Solutions 85

Similarly, if we set a1 = c2, we find

a1 = c2

a3 =a1

(1 + 2)(1 + 1)=

c1

3 · 2

a5 =a3

(3 + 2)(3 + 1)=

1

5 · 4a3 =

c1

5!

a2n+1 =a2n+1

(2n + 3)(2n + 2)=

c1

(2n + 1)!.

Thus the general solution is

y(x) = c1

(

1 −x2

2!+

x4

4!+ · · ·

)

+ c2

(

x −x3

3!+

x5

5!+ · · ·

)

= c1 cosx + c2 sin x

as expected.

Example 7.3. Find the series solution to Airy’s equation

y′′ − xy = 0, y(0) = 1, y′(0) = 1.

Solution. Again we put (7.4) into the ode. We find

∞∑

n=2

n(n − 1)anxn−2 − x

∞∑

n=0

anxn = 0

or∞∑

n=0

(n + 2)(n + 1)an+2xn −

∞∑

n=1

an−1xn = 0.

That is,

2a2 +

∞∑

n=1

(

(n + 2)(n + 1)an+2 − an−1

)

xn = 0.

We see a2 = 0, and the recurrence relation is

(n + 2)(n + 1)an+2 − an−1 = 0 n ≥ 1.

Knowledge of a0, provides a3, a6, a9 . . ., and knowledge of a1 gives us a4, a7, a10 . . ..Thus,

a0 = a0

a3 =a0

(1 + 2)(1 + 1)=

a0

3 · 2

a6 =a3

(4 + 2)(4 + 1)=

a0

2 · 3 · 5 · 6

a9 =a6

(7 + 2)(7 + 1)=

a0

2 · 3 · 5 · 6 · 8 · 9

86 7. Nonconstant Coefficients

and

a1 = a1

a4 =a1

(2 + 2)(2 + 1)=

a0

4 · 3

a7 =a4

(5 + 2)(5 + 1)=

a0

3 · 4 · 6 · 7

a10 =a7

(8 + 2)(8 + 1)=

a0

3 · 4 · 6 · 7 · 9 · 10

and the general solution is y = a0y1 + a1y2. That is,

y(x) = a0

(

1 +x3

2 · 3+

x6

2 · 3 · 5 · 6+

x9

2 · 3 · 5 · 6 · 8 · 9+ · · ·

)

+a1

(

x +x4

3 · 4+

x7

3 · 4 · 6 · 7+

x10

3 · 4 · 6 · 7 · 9 · 10+ · · ·

)

-10 -8 -6 -4 -2 0 2X

-2

-1

0

1

2

Y

n=15n=30n=45n=60n=75n=90

Figure 1. Polynomial approximations of the solution y1 in Airy’s Equation.

7.3. Series Solutions 87

Example 7.4. Find the solution to

y′′ + xy′ + y = 0, y(0) = 1, y′(0) = 0.

We put (7.4) back into the ode and find∞∑

n=2

n(n − 1)anxn−2 + x

∞∑

n=1

nanxn−1 +

∞∑

n=0

anxn = 0

or∞∑

n=0

(n + 2)(n + 1)an+2xn +

∞∑

n=0

nanxn +∞∑

n=0

anxn = 0.

That is,∞∑

n=0

(

(n + 2)(n + 1)an+2 + nan + an

)

xn = 0,

and the recurrence relation is

(n + 2)(n + 1)an+2 + (n + 1)an = 0

or

an+2 =−an

n + 2.

Knowledge of a0 reveals a2, a4, . . ., and knowledge of a1 gives us a3, a5, . . ..

a0 = a0

a2 =−a0

(0 + 2)=

−a0

2

a4 =−a2

2 + 2=

a0

2 · 4

a6 =−a4

4 + 2=

−a0

2 · 4 · 6

and

a1 = a1

a3 =−a1

1 + 2=

−a1

3

a5 =−a3

3 + 2=

a1

3 · 5

a7 =−a5

5 + 2=

−a1

3 · 5 · 7.

The general solution is

y(x) = a0

(

1 −x2

2+

x4

2 · 4−

x6

2 · 4 · 6+ · · ·

)

+a1

(

x −x3

3+

x5

3 · 5−

x7

3 · 5 · 7+ · · ·

)

= a0

∞∑

n=0

(−1)n

2nn!x2n + a1

∞∑

n=0

(−1)n2nn!

(2n + 1)!x2n+1.

88 7. Nonconstant Coefficients

Homework 7.3 (Power Series)

Find the recurrence relation on the first four terms in each of two solutions y1 andy2 (unless the series truncates sooner).

1. y′′ − y′ = 0

2. y′′ − xy′ − y = 0

3. (1 − x)y′′ + y = 0

4. (1 + x2)y′′ − 4xy′ + 6y = 0

Partial Answers

1. an+2 = an/(n + 2)(n + 1)

2. an+2 = an/(n + 2)

3. (n + 2)(n + 1)an+2 − n(n + 1)an+1 + an = 0 for n ≥ 1; a2 = − 12a0

4. y1 = 1 − 3x2, y2 = x − x3/3.

Index

asymptotically stable, 76

backward Euler, 12

Center, 59Characteristic Polynomial, 20Complex roots, 21Convolution, 50critical point, 75Critically Damped, 32

Delta function, 47

Eigenvalues, 55Eigenvectors, 55Euler Formula, 21exponential order, 37exponentially bounded, 37

Focus, 59forward Euler, 10

Heaviside function, 42

improved Euler, 14integrating factor, 4

Jacobian, 76

Laplace transform, 37linear, 17

ODE, 1Over Damped, 31

Partial Differential Equation, 1Phase Portrait, 56piecewise continuous, 37piecewise smooth, 37

Real distinct roots, 20Recurrence Relation, 84Repeated roots, 23

Saddle, 56

second order linear ODE, 17stable critical point, 76

trajectory, 75

unstable critical point), 76

Wronskian, 30

89