Linear Differential Equations with Constant Coefficients

Solve the following differential equations:

(a) d

(b) d

(c) d

(d) d

(e) d

(f) d

[from Hildebrand, Francis B. 1976,

SOLUTION

(a) d

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

y = e

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

dy/dx = re

d

Substituting these into the differential equation, we get

r

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

y = c

(b) d

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

r

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 r

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 r

r

The general solution of the differential equation is

y = (c

where the first term, (c

(c) d

Proceeding as above, the characteristic equation is

r

The general solution of the differential equation is

y = c

Using Euler's formula,

e

Thus, we can write

y = e

(d) d

Proceeding as above, the characteristic equation is

r

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)(r

Assume that r = 1 is a root of r

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)

The remaining roots can be obtained by setting

r

=> r = {2 ± sqrt[(-2)

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 = (c

which can be rewritten in the form

y = e

(e) d

Proceeding as above, the characteristic equation is

r

The possible rational roots are r = ± 1. Assume that r = 1 is a root and divide r

1 0 0 -1 | 1

1 1 1 +--

___________

1 1 1 0

Thus, r = 1 is a root, and

(r - 1)(r

The remaining roots can be determined by setting

r

=> r = {- 1 ± sqrt[1

= - 1/2 ± [sqrt(3)/2]i

The general solution of the differential equation is

y = c

This can be rewritten as

y = c

(f) d

Proceeding as above, the characteristic equation is

r

A complex number of the form

z = a + bi

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

z = a + bi = re

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

a

Dividing (3) by (2),

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

The complex number

z = 2i

can be written in the form re

z = 2i = 2e

=> sqrt(z) = sqrt(2i) = [sqrt(2)]e

= [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 = c