Linear Homogeneous Differential Equations - Millersville University of ...

Linear Homogeneous Differential Equations

3-23-2008

The full description of these equations is: Linear constant coefficient homogeneous equations. The equations 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 the

derivatives of y, each multiplied by x stuff. (In this case, the x stuff is constant.) Homogeneous means that

the right side is 0 -- there's no term involving only x.

It's convenient to let D = d stand for the operation of differentiating with respect to x. (Note that

dx

D = d is the operation of differentiation, whereas Dy = dy is the derivative.) In this notation, D2

dx

dx

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 - )y = 0.

A solution to the equation is a function y = f (x) which satisfies the equation. Equivalently, if you think of anDn + ? ? ? 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 emx = remx, dx

d dx2

emx

=

m2emx,

...,

dn dxn

emx

=

mnemx.

Plug emx into

y(n) + bn-1y(n-1) + ? ? ? + b2y + b1y + b0y = 0.

The result:

mnemx + bn-1mn-1emx + ? ? ? + b2m2emx + b1memx + b0emx = 0.

Factor out emx and cancel it. This leaves

mn + bn-1mn-1 + ? ? ? + b2m2 + b1m + b0 = 0.

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

1

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 - c2e-x, y = 36c1e6x + c2e-x.

Therefore, y - 5y + 6y = 36c1e6x + c2e-x - 5 6c1e6x - c2e-x - 6 c1e6x + c2e-x = 0.

Example. Solve (D4 - 9D2 + 20)y = 0.

It's easy to write down the characteristic equation: just replace the D's with m's:

m4 - 9m2 + 20 = 0,

(x2 - 4)(x2 - 5) = 0,

(x - 2)(x + 2)(x - 5)(x + 5) = 0.

The roots are ?2 and ? 5. (Don't fall into the trap of assuming that roots must be integers, or even

rationals!) 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 characteristic equation is x2 - 2x + 1 = 0, which has x = 1 as a double root. It is true that ex is a solution, but it would be incorrect to write y = c1ex + c2ex.

The terms c1ex and c2ex are redundant -- you could combine them to get y = (c1 + c2)ex = c3ex. To put 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 + c2xex.

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

2

In general, if m is a repeated root of multiplicity k in the characteristic polynomial, you get terms emx, xemx, . . . , xk-1emx in the general solution.

Example.

Solve

d3y dx3

-

3

d2y 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 + c2xex + c3x2ex.

Example. Solve (D2 - 1)(D2 - D - 2)y = 0. The characteristic equation is (m2 - 1)(m2 - m - 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 + c2e2x + c3e-x + c4xe-x.

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

Example. equations

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

dx = x = x + 4y, dy = y = 2x + 3y.

dt

dt

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 and

m = -1. Therefore, y = c1e5t + c2e-t.

Now y = 5c1e5t - c2e-t, so

x

=

1 2

(y

- 3y)

=

1 2

5c1e5t - c2e-t - 3 c1e5t + c2e-t

= c1e5t - 2c2e-t.

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

3

Now suppose the characteristic equation has a complex root a + bi. From basic algebra, complex roots of real polynomials come in conjugate pairs: a + bi and a - bi. It's reasonable to expect solutions

c1e(a+bi)x and c2e(a-bi)x.

However, these are complex solutions, and you should have real solutions to the original real differential equation. I'll 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(a-bi)x = c1eaxeibx + c2eaxe-ibx =

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(a-bi)x = eax(c3 cos bx + c4 sin bx).

Each pair of conjugate complex roots a?bi in the characteristic equation generates a pair of independent solutions of this form.

Example. Solve y + y = 0. The characteristic equation m2 + 1 = 0 has roots ?i. The solution is y = c1 cos x + c2 sin x.

Example. Solve (D2 + 2D + 5)y = 0. The characteristic equation m2 + 2m + 5 = 0 has roots m = -1 ? 2i. The solution is y = e-x(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 and

m=

1 2

?i

3 2

.

The solution

is

y = c1e-x + ex/2

c2 cos

3 2

x

+

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). The solution is

y = c1 cos 2x + c2 sin 2x + c3x cos 2x + c4x sin 2x. (What's the solution to (D2 + 2D + 5)2y = 0?)

c 2008 by Bruce Ikenaga

4

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download