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Differential Equations Notes

[Last Updated 22 January 2011]


Linear Differential Equations with Constant Coefficients


Linear Differential Equations with Constant Coefficients

PROBLEM

Solve the following differential equations:
(a) d2y/dx2 - dy/dx - 2y = 0
(b) d3y/dx3 - d2y/dx2 - dy/dx + y = 0
(c) d2y/dx2 - 2 dy/dx + 2y = 0
(d) d4y/dx4 - 4 d3y/dx3 + 7 d2y/dx2 - 6 dy/dx + 2y = 0
(e) d3y/dx3 - y = 0
(f) d2y/dx2 - 2iy = 0 (i2 = - 1)

[from Hildebrand, Francis B. 1976, Advanced Calculus for Applications, Second Edition (Upper Saddle River, New Jersey: Prentice Hall), problem 1.18]

SOLUTION

(a) d2y/dx2 - dy/dx - 2y = 0

This is a homogeneous linear differential equation with constant coefficients. The solutions are of the form

y = erx

The first and second derivatives of y with respect to x are

dy/dx = rerx
d2y/dx2 = r2erx

Substituting these into the differential equation, we get

r2erx - rerx - 2erx = 0 => r2 - r - 2 = 0 => (r - 2)(r + 1) = 0 => r = - 1, 2

The general solution of the differential equation is a linear combination of the individual solutions,

y = c1e-x + c2e2x

(b) d3y/dx3 - d2y/dx2 - dy/dx + y = 0

Proceeding as in (a), we get the characteristic equation

r3 - r2 - r + 1 = 0

Now we make use of the theorem that, if the rational number a / b is a root of a polynomial with integral coefficients, then a is a factor of the constant term and b is a factor of the leading coefficient. The constant term of the polynomial in the characteristic equation is 1, while the leading coefficient is also 1. Thus, the possible roots of the characteristic equation are r = ± 1. If r = 1 is a root, then r - 1 must be a factor of r3 - r2 - r + 1. Use synthetic division to divide r3 - r2 - r + 1 by r - 1.

1  -1  -1  1 | 1
    1   0 -1 +---
____________
1   0  -1  0


The remainder term is zero. It follows that r - 1 is a factor of r3 - r2 - r + 1, and

r3 - r2 - r + 1 = (r - 1)(r2 - 1) = 0 => (r - 1)(r - 1)(r + 1) = 0 => (r - 1)2(r + 1) = 0 => r = 1, 1, - 1

The general solution of the differential equation is

y = (c1 + c2x)erx + c3e-rx

where the first term, (c1 + c2x)erx, comes from the double root r = 1.

(c) d2y/dx2 - 2 dy/dx + 2y = 0

Proceeding as above, the characteristic equation is

r2 - 2r + 2 = 0 => r = {2 ± sqrt[22 - (4)(1)(2)]} / [(2)(1)] = [2 ± sqrt(-4)] / 2 = (2 ± 2i) / 2 = 1 ± i

The general solution of the differential equation is

y = c1e(1+i)x + c2e(1-i)x = ex(c1eix + c2e-ix)

Using Euler's formula,

e±ix = cos x ± i sin x

Thus, we can write

y = ex(c1 cos x + c2 sin x)

(d) d4y/dx4 - 4 d3y/dx3 + 7 d2y/dx2 - 6 dy/dx + 2y = 0

Proceeding as above, the characteristic equation is

r4 - 4r3 + 7r2 - 6r + 2 = 0

The constant term of the polynomial is 2, while the leading coefficient is 1. Thus, the possible rational roots of the characteristic equation are ± 2 and ± 1.

Assume that r = 1 is a root, and divide the polynomial by r - 1 using synthetic division.

1  -4  7  -6  2 | 1
    1 -3   4 -2 +--
_______________
1  -3  4  -2  0


Thus, r = 1 is a root, and

(r - 1)(r3 - 3r2 + 4r - 2) = 0

Assume that r = 1 is a root of r3 - 3r2 + 4r - 2, and divide this polynomial by r - 1 using synthetic division.

1  -3  4  -2 | 1
    1 -2   2 +--
____________
1  -2  2   0


So r is a double root of the original polynomial, and

(r - 1)2(r2 - 2r + 2) = 0

The remaining roots can be obtained by setting

r2 - 2r + 2 = 0
=> r = {2 ± sqrt[(-2)2 - (4)(1)(2)]} / [(2)(1)] = [2 ± sqrt(-4)] / 2 = (2 ± 2i) / 2 = 1 ± i

So the four roots of the characteristic equation are r = 1, 1, and 1 ± i, and the general solution of the differential equation is

y = (c1 + c2x)ex + c3e(1+i)x + c4e(1-i)x = (c1 + c2x)ex + (c3eix + c4e-ix)ex

which can be rewritten in the form

y = ex(c1 + c2x + c3 cos x + c4 sin x)

(e) d3y/dx3 - y = 0

Proceeding as above, the characteristic equation is

r3 - 1 = 0

The possible rational roots are r = ± 1. Assume that r = 1 is a root and divide r3 - 1 by r - 1 using synthetic division.

1  0  0  -1 | 1
   1  1   1 +--
___________
1  1  1   0


Thus, r = 1 is a root, and

(r - 1)(r2 + r + 1) = 0

The remaining roots can be determined by setting

r2 + r + 1 = 0
=> r = {- 1 ± sqrt[12 - (4)(1)(1)]} / [(2)(1)] = [- 1 ± sqrt(-3)] / 2 = [- 1 ± i sqrt(3)] / 2
= - 1/2 ± [sqrt(3)/2]i

The general solution of the differential equation is

y = c1ex + e-x/2{c2e[sqrt(3)x/2]i + c3e-[sqrt(3)x/2]i}

This can be rewritten as

y = c1ex + e-x/2{c2 cos[sqrt(3)x/2] + c3 sin[sqrt(3)x/2]}

(f) d2y/dx2 - 2iy = 0 (i2 = - 1)

Proceeding as above, the characteristic equation is

r2 - 2i = 0 => r = ± sqrt(2i) (1)

A complex number of the form

z = a + bi

where a and b are real, can be written in the form reiq, where

z = a + bi = reiq = r cos q + ir sin q

Equating the real and imaginary parts, we get

a = r cos q (2)
b = r sin q (3)

Squaring (2) and (3), and adding the results yields

a2 + b2 = r2 cos2 q + r2 sin2 q = r2(cos2 q + sin2 q) = r2 => r = sqrt(a2 + b2)

Dividing (3) by (2),

b/a = tan q => q = arctan(b/a)

The complex number

z = 2i

can be written in the form reiq where r = 2 and q = arctan(2/0) = arctan(∞) = p/2 or 3p/2. We reject the solution q = 3p/2 because this yields reiq = r cos q + ir sin q = 2 cos(3p/2) + 2i sin(3p/2) = - 2i. Thus,

z = 2i = 2ei(p/2)
=> sqrt(z) = sqrt(2i) = [sqrt(2)]ei(p/4) = [sqrt(2)][cos(p/4) + i sin(p/4)]
= [sqrt(2)][sqrt(2)/2 + i sqrt(2)/2] = 1 + i

Making use of this result and returning to (1), we conclude that the roots of the characteristic equation are

r = ± sqrt(2i) = ± (1 + i)

The general solution of the differential equation is

y = c1e(1+i)x + c2e-(1+i)x

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