3-23-2008

Linear Homogeneous Differential Equations

The full description of these equations is: Linear constant coefficient homogeneous equations. Theequations described in the title have the form

any(n) + + a2y + a1y + a0y = 0.

Here y is a function of x, and an, . . . , a0 are constants. Linear means the equation is a sum of thederivatives of y, each multiplied by x stuff. (In this case, the x stuff is constant.) Homogeneous means thatthe right side is 0 theres no term involving only x.

Its convenient to let D =d

dxstand for the operation of differentiating with respect to x. (Note that

D =d

dxis the operation of differentiation, whereas Dy =

dy

dxis the derivative.) In this notation, D2

computes the second derivative, D3 computes the third derivative, and so on. The equation above becomes

(anDn + a2D2 + a1D + a0)y = 0.

Example. The following equations are linear homogeneous equations with constant coefficients:

y + 3y + 3y + y = 0,

y 5y 6y = 0,((D 1)(D 2)(D pi)y = 0.

A solution to the equation is a function y = f(x) which satisfies the equation. Equivalently, if you thinkof anD

n+ a2D2+a1D+a0 as a linear transformation, it is an element of the kernel of the transformation.The general solution is a linear combination of the elements of a basis for the kernel, with the

coefficients being arbitrary constants.The form of the equation makes it reasonable that a solution should be a function whose derivatives are

constant multiples of itself. emx is such a function:

d

dxemx = remx,

d

dx2emx = m2emx, . . . ,

dn

dxnemx = mnemx.

Plug emx intoy(n) + bn1y

(n1) + + b2y + b1y + b0y = 0.The result:

mnemx + bn1mn1emx + + b2m2emx + b1memx + b0emx = 0.

Factor out emx and cancel it. This leaves

mn + bn1mn1 + + b2m2 + b1m+ b0 = 0.

Thus, emx is a solution to the original equation exactly when m is a root of this polynomial. Thepolynomial is called the characteristic polynomial; as the derivation showed, its obtained by building apolynomial using the coefficients of the original differential equation.

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Example. Solve y 5y 6y = 0.The characteristic polynomial is m2 5m 6; solving m2 5m 6 = 0 yields (m 6)(m+ 1) = 0, so

m = 6 or m = 1. The general solution is

y = c1e6x + c2e

x.

You can check this by plugging back in. Here are the derivatives:

y = c1e6x + c2e

x, y = 6c1e6x c2ex, y = 36c1e6x + c2ex.

Therefore,

y 5y + 6y = (36c1e6x + c2ex) 5 (6c1e6x c2ex) 6 (c1e6x + c2ex) = 0.

Example. Solve (D4 9D2 + 20)y = 0.Its easy to write down the characteristic equation: just replace the Ds with ms:

m4 9m2 + 20 = 0, (x2 4)(x2 5) = 0, (x 2)(x+ 2)(x5)(x+

5) = 0.

The roots are 2 and 5. (Dont fall into the trap of assuming that roots must be integers, or evenrationals!) The solution is

y = c1e2x + c2e

2x + c3e

5x + c4e

5x.

What happens if there are repeated roots? Look at the equation y 2y + y = 0. The characteristicequation is x2 2x+ 1 = 0, which has x = 1 as a double root. It is true that ex is a solution, but it wouldbe incorrect to write y = c1e

x + c2ex.

The terms c1ex and c2e

x are redundant you could combine them to get y = (c1 + c2)ex = c3e

x. Toput it another way, as function ex and ex are linearly dependent.

It is reasonable to suppose that for a second order equation you should have two different solutions. ex

is one; how can you find another?The idea is to guess the form that such a solution might take. Guess:

y = f(x)ex,

i.e. something times the known solution ex. What should f be?To find f , plug f(x)ex into the equation. Here are the derivatives:

y = f(x)ex, y = f (x)ex + f(x)ex, y = f (x)ex + 2f (x)ex + f(x)ex.

Plug them in:

y 2y + y = (f (x)ex + 2f (x)ex + f(x)ex) 2(f (x)ex + f(x)ex) + f(x)ex = f (x)ex = 0.

Hence, f (x) = 0. Integrate twice and obtain f(x) = c1 + c2x. Thus,

y = c1ex + c2xe

x.

In fact, this is the general solution notice the two arbitrary constants. The functions ex and xex areindpendent solutions to the original equation.

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In general, if m is a repeated root of multiplicity k in the characteristic polynomial, you get terms emx,xemx, . . . , xk1emx in the general solution.

Example. Solved3y

dx3 3d

2y

dx2+ 3

dy

dx y = 0.

The characteristic equation is m3 3m2 + 3m 1 = 0, which has m = 1 as a root with multiplicity 3.The general solution is

y = c1ex + c2xe

x + c3x2ex.

Example. Solve (D2 1)(D2 D 2)y = 0.The characteristic equation is (m2 1)(m2m 2) = 0, or (m 1)(m+1)2(m 2) = 0. The roots are

1, 2, and 1 (double). The general solution is

y = c1ex + c2e

2x + c3ex + c4xe

x.

Note: You can write the terms in the solution in any order you please. Nor does it matter which cgoes with which term, since they are arbitrary constants.

Example. (Linear systems) Suppose x and y are functions of t. Consider the system of differentialequations

dx

dt= x = x+ 4y,

dy

dt= y = 2x+ 3y.

I want to solve for x and y in terms of t.Solve the second equation for x:

x =1

2(y 3y) .

Differentiate:

x =1

2(y 3y) .

Plug the expressions for x and x into the first equation:

1

2(y 3y) = 1

2(y 3y) + 4y.

Simplify:y 4y 5y = 0.

The characteristic equation is m2 4m 5 = 0, or (m 5)(m + 1) = 0. The roots are m = 5 andm = 1. Therefore,

y = c1e5t + c2e

t.

Now y = 5c1e5t c2et, so

x =1

2(y 3y) = 1

2

[(5c1e

5t c2et) 3 (c1e5t + c2et)] = c1e5t 2c2et.

There are other ways of solving linear systems, but for small systems brute force works reasonably well!

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Now suppose the characteristic equation has a complex root a+ bi. From basic algebra, complex rootsof real polynomials come in conjugate pairs: a+ bi and a bi. Its reasonable to expect solutions

c1e(a+bi)x and c2e

(abi)x.

However, these are complex solutions, and you should have real solutions to the original real differentialequation. Ill use the complex exponential formula

ei = cos + i sin , R.You can derive this formula by considering the Taylor series for ei, cos , and sin .Now

c1e(a+bi)x + c2e

(abi)x = c1eaxeibx + c2e

axeibx =

eax (c1(cos bx+ i sin bx) + c2(cos bx i sin bx)) = eax ((c1 + c2) cos bx+ i(c1 c2) sin bx) .Let c3 = c1 + c2 and c4 = i(c1 c2). Observe that c1 and c2 can be solved for in terms of c3 and c4, so

no generality is lost with this substitution. Then

c1e(a+bi)x + c2e

(abi)x = eax(c3 cos bx+ c4 sin bx).

Each pair of conjugate complex roots abi in the characteristic equation generates a pair of independentsolutions of this form.

Example. Solve y + y = 0.

The characteristic equation m2 + 1 = 0 has roots i. The solution isy = c1 cosx+ c2 sinx.

Example. Solve (D2 + 2D + 5)y = 0.

The characteristic equation m2 + 2m+ 5 = 0 has roots m = 1 2i. The solution isy = ex(c1 cos 2x+ c2 sin 2x).

Example. Solve (D3 + 1)y = 0.

The characteristic polynomial m3 + 1 factors into (m + 1)(m2 m + 1). The roots are m = 1 andm =

1

2 i

3

2. The solution is

y = c1ex + ex/2

(c2 cos

3

2x+ c3 sin

3

2x

).

Example. Solve (D2 + 4)2y = 0.

The characteristic equation (m2 + 4)2 = 0 has repeated complex roots: m = 2i (each double). Thesolution is

y = c1 cos 2x+ c2 sin 2x+ c3x cos 2x+ c4x sin 2x.

(Whats the solution to (D2 + 2D + 5)2y = 0?)

c2008 by Bruce Ikenaga 4