Differential equations notes

David Pierce

October 11, 2006; recompiled, November 9, 2007

I originally prepared these notes while teaching Math 253, Introduction to Ordinary Differential Equations, in the fall of 2003. Now I am updating them while teaching Math 219, Introduction to Differential Equations.

The main reference is [1, 7th edition, 2001], though sometimes I consult books like [2] or [3], used at METU years ago.

References

[1] William E. Boyce and Richard C. DiPrima. Elementary differential equations and boundary value problems. John Wiley & Sons Inc., New York, 1965.

[2] Lyman M. Kells. Elementary Differential Equations. McGraw-Hill Book Company, New York, fifth edition, 1960.

[3] Shepley L. Ross. Differential equations. Blaisdell Publishing Company Ginn and Co. New York-Toronto-London, 1964.

[4] Michael Spivak. Calculus. Publish or Perish, Berkeley, California, 2nd edition, 1980.

Contents

0 Introduction 1

1 Examples from natural science 4 1.1 From physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 From ecology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Direction-fields 4

3 First-order ODE’s 5 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 3.2 Integrating factors . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.3 Separable equations . . . . . . . . . . . . . . . . . . . . . . . . . 8 3.4 Linear versus non-linear equations . . . . . . . . . . . . . . . . . 12 3.5 Exact equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Linear equations 18 4.1 Homogeneous with constant coeff. I . . . . . . . . . . . . . . . . 18 4.2 Homogeneous equations in general . . . . . . . . . . . . . . . . . 20 4.3 Homogeneous with constant coeff. II . . . . . . . . . . . . . . . . 24 4.4 Homogeneous with constant coeff. III . . . . . . . . . . . . . . . 26 4.5 Reduction of order . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4.6 Inhomogeneous equations . . . . . . . . . . . . . . . . . . . . . . 28 4.7 Variation of parameters . . . . . . . . . . . . . . . . . . . . . . . 29

5 Laplace transform 29 5.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 5.2 Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5.3 Discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6 Systems of DE’s 35

0 Introduction

Recall the fundamental theorems of calculus:

(1) If f is continuous, then \frac{d}{d x} ∫_{a}^{x} f(t) d t = f(x).

(2) If f = g′, then ∫_{a}^{b} f(x) d x = g(b) − g(a). So anybody who has studied calculus has solved differential equations:

• To find an indefinite integral ∫ f(t) d t is to solve the differential equation

y′ = f(x).  (0.1)

For example, since

∫ 4t ln t d t = 2t^{2} ln t − ∫ 2t d t = 2t^{2} ln t − t^{2} + C,  (0.2)

the solutions of the differential equation

y′ = 4x ln x  (0.3)

are the functions 2x^{2} ln x − x^{2} + C.

• The value of ln x is defined to be ∫_{1}^{x} d t/t. Thus ln is the solution of the

differential equation

y′ = \frac{1}{x}  (0.4)

meeting the initial condition y(1) = 0.

• The inverse of ln is defined to be exp, and we must have

exp′ x = \frac{1}{ln′ exp x} = exp x.  (0.5)

Thus exp is the solution of the differential equation

y′ = y  (0.6)

meeting the initial condition y(0) = 1. Note that equation (0.4) expresses the derivative of a function of x as another function of x; so the equation can be solved by integrating. But in equation (0.6), the derivative of a function y is expressed as a function of that function y; the solution of such equations will require new techniques, which are what this course is all about.

A note on notation

Particular functions can be described in several ways. For example, the squaring function is:

• the function f such that f(x) = x^{2} for all [real] numbers x; • the function x ↦ x^{2} (or t ↦ t^{2}); or

• the function x^{2} (or t^{2}). (See [4, ch. 3, p. 45].) The last notation is ambiguous. The expressions x ↦ x^{2} and t ↦ t^{2} represent the same function; but we cannot indicate this by writing

x^{2} = t^{2}.  (0.7)

Also, what does an expression like

t + t^{2}x  (0.8)

mean? If it indicates a function, then this function could be

t ↦ t + t^{2}x, or x ↦ t + t^{2}x, or possibly (t, x) ↦ t + t^{2}x. (0.9)

The definite integral

∫_{a}^{b} f(t) d t  (0.10)

can also be written

∫_{a}^{b} f.  (0.11)

Of course the latter notation can’t be used if we only know f from a formula: the integral from a to b of the function x ↦ x ln x is

∫_{a}^{b} x ln x d x;  (0.12)

this is a number (depending on a and b), and it is the same number as

∫_{a}^{b} t ln t d t.  (0.13)

The so-called indefinite integral ∫ f(t) d t is usually understood as an instruction or command, namely:

(1) find a function g such that g′ = f;

(2) write down g(t) + C, where C is understood as an arbitrary constant. (See [4, ch. 18, p. 337].) In this case, we would also write

∫ f(s) d s = g(s) + C,  (0.14)

which means that ∫ f(t) d t and ∫ f(s) d s indicate different things. However, this convention is violated at [1, § 2.1, p. 33, equations (21–22)]. There, by integrating

\frac{d}{d t}(e^{at}y) = e^{at}g(t),  (0.15)

the writers conclude

e^{at}y = ∫ e^{as}g(s) d s + c;  (0.16)

they feel the need to distinguish the ‘integration variable,’ now s, from the ‘independent variable,’ t. Yet if the integration can be performed, then, according to the usual convention, an expression h(s) will be obtained. Then one just has to remember to replace s with t here.

To avoid ambiguity, it might be best to let expressions such as ∫ f(t) d t and ∫ f(s) d s and ∫ f indicate the same thing, namely an arbitrary function g such that g′ = f. In other words, these expressions should simply denote a primitive of f. Suppose we want to indicate a value of g, such as g(x). Keeping in mind the fundamental theorem of calculus, we can write, for example,

∫ ^{x} f.  (0.17)

Then the conclusion of (0.2) would be written

∫ ^{x} 4t ln t d t = 2x^{2} ln x − x^{2} + C.  (0.18)

If we are trying to be precise in this way, then it seems that we should write (0.1) in the form

y′ = f.  (0.19)

But this form does not accommodate the example (0.3); this could be written more precisely as

\frac{d y}{d x} = 4x ln x.  (0.20)

One can formally ‘solve’ this as follows:

\frac{d y}{d x} = 4x ln x ⟺ d y = 4x ln x d x  (0.21)
⟺ ∫ d y = ∫ 4x ln x d x  (0.22)
⟺ y = 2x^{2} ln x − x^{2} + C.  (0.23)

The chain rule is that if g is differentiable at a, and f is differentiable at g(a), then f ∘ g is differentiable at a, and

(f ∘ g)′(a) = f′(g(a)) · g′(a).  (0.24)

Hence the equation

(f ∘ g)′ = (f′ ∘ g) · g′  (0.25)

holds on the common domain of the two members. If one thinks of f ∘ g as a function y, and of g as a function x, then one may write

\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t}.  (0.26)

Then d y/ d x is not the derivative y′ of y, but is the quotient of y′ by x′.

1 Examples from natural science

1.1 From physics

In physics, Newton’s second law, F = ma, is a differential equation since a = d v/ d t, that is, a = v̇ (the time-derivative of v).

On a body falling in air, there are two forces:

• mg, due to gravity;

• −γv, due to friction. (Here γ is the ‘drag-coefficient’.)

Then ma = F can be written as mv̇ = mg − γv, that is,

\frac{d v}{d t} = g − \frac{γ}{m}v.  (1.1)

1.2 From ecology

A similar equation arises from consideration of a mouse population of size p attacked by owls. Changes in p are due to:

• the birth-rate r, in units \frac{baby mice}{(mice)(month)};

• the death-rate k, in units \frac{mice}{month} (killed by owls).

(Note how ‘rate’ is used in two different senses.) Hence we have the differential equation

\frac{d p}{d t} = rp − k.  (1.2)

2 Direction-fields

The equations (1.1) and (1.2) have the form

\frac{d y}{d t} = ay − b,  (2.1)

where y is a function of t, and a and b are constant. In turn, the equation (2.1) (along with (0.4) and (0.6)) is a special case of an equation

\frac{d y}{d t} = f(t, y),  (2.2)

where f is a function of two variables. A solution to such an equation can be thought of as a curve in the Cartesian plane whose slope at each point (t, y) has the value f(t, y).

Hence an idea of the solution of (2.2) can be got from a direction-field. Formally, the direction-field associated with (2.2) is just the function f itself. One pictures the direction-field as the plane with a little line of slope f(t, y) at each point (t, y). See figures 1.1.2 and 1.1.4 in the book.

From (1.1) (and the direction-field) one may see that, as t → ∞, so v approaches a ‘terminal velocity’, mg/γ.

From (1.2), one may see that, as t → ∞, so p either drops to 0 or blows up (increases without bound).

3 First-order ordinary differential equations

3.1 Introduction

Equation (2.1) is an ordinary differential equation because y is a function of only one variable, so only ordinary derivatives (and not partial derivatives) appear. Equation (2.1) is also:

• first-order, because only the first derivative of y appears;

• linear, because y and its derivative(s) are not raised to powers or multiplied by each other;

• with constant coefficients, because a and b are constant (that is, the derivatives d a/ d t and d b/ d t are 0).

To solve (2.1), assuming a ≠ 0, first note that one solution is constant: y = b/a. To find all other solutions, we may proceed thus:

\frac{d y}{d t} = ay − b ⟺ \frac{d y}{d t} = a (y − \frac{b}{a})  (3.1)
⟺ \frac{d y/ d t}{y − b/a} = a  (3.2)
⟺ \frac{d}{d t} ln |y − \frac{b}{a}| = a  (3.3)
⟺ ln |y − \frac{b}{a}| = at + C  (3.4)
⟺ |y − \frac{b}{a}| = e^{at}e^{C}  (3.5)
⟺ y = \frac{b}{a} + ce^{at}.  (3.6)

Note that c = ±e^{C}; but what does this constant mean? Suppose y(0) is y_{0}—this is an initial condition. Then y_{0} = b/a + c, so c = y_{0} − b/a. Hence the solution of (2.1) is

y = \frac{b}{a} + (y_{0} − \frac{b}{a}) e^{at}.  (3.7)

For another method of solution, see Exercise 1.2.4.

3.2 Integrating factors

For a third method of solving (2.1), rewrite the equation as

\frac{d y}{d t} − ay = −b.  (3.8)

Now we use a trick: Multiply each side by a function μ(t) (called an integrating factor):

\frac{d y}{d t} · μ(t) − ay · μ(t) = −b · μ(t).  (3.9)

Try to understand the left side as a derivative. We know

\frac{d}{d t}(y · μ(t)) = \frac{d y}{d t} · μ(t) + y · μ′(t).  (3.10)

So let us choose μ so that

−a · μ(t) = μ′(t).  (3.11)

This is a new differential equation. We just need one solution, say μ(t) = e^{−at}. Substituting into (3.9) and integrating, we have

\frac{d y}{d t} = ay − b ⟺ \frac{d y}{d t} · e^{−at} − ay · e^{−at} = −be^{−at}  (3.12)
⟺ \frac{d}{d t}(y · e^{−at}) = −be^{−at}  (3.13)
⟺ y · e^{−at} = \frac{b}{a}e^{−at} + c  (3.14)
⟺ y = \frac{b}{a} + ce^{at},  (3.15)

in agreement with (3.6).

Now: in arriving at (3.13), we had no need to assume that b was constant. Replacing −b with g(t), we have

\frac{d y}{d t} − ay = g(t) ⟺ \frac{d}{d t}(y · e^{−at}) = g(t)e^{−at}  (3.16)
⟺ y · e^{−at} = ∫ ^{t} g(s)e^{−as} d s  (3.17)
⟺ y = e^{at} ∫ ^{t} g(s)e^{−as} d s.  (3.18)

(For the notation, see (0.17).) If g is a polynomial, then we can find ∫ g(t)e^{−at} d t through integration by parts. 3.1 Example. Say g(t) = −(mt + k). We have

∫ (mt + k)e^{−at} d t = −\frac{mt + k}{a} e^{−at} + \frac{m}{a} ∫ e^{−at} d t  (3.19)
= − (\frac{mt + k}{a} + \frac{m}{a^{2}} ) e^{−at} + C.  (3.20)

Therefore

\frac{d y}{d t} − ay = −(mt + k) ⟺ y = \frac{mt + k}{a} + \frac{m}{a^{2}} + Ce^{at}.  (3.21)

Note here that the behavior of y for large t depends on the sign of a:

• if a ⩽ 0, then y ≈ (mt + k)/a + m/a^{2} as t → ∞;

• if a > 0 and C ≠ 0, then y diverges from this value. We can try using integrating factors to solve an arbitrary first-order linear equation, namely

\frac{d y}{d t} + y · p(t) = g(t).  (3.22)

Our integrating factor μ(t) should satisfy

p(t)μ(t) = μ′(t),  (3.23)
\frac{μ′(t)}{μ(t)} = p(t),  (3.24)
\frac{d}{d t} ln μ(t) = p(t).  (3.25)

So we may let

ln μ(t) = ∫ ^{t} p,  (3.26)
μ(t) = exp ∫ ^{t} p.  (3.27)

For this μ, we have

\frac{d y}{d t} + y · p(t) = g(t) ⟺ \frac{d}{d t}(y · μ(t)) = g(t) · μ(t)  (3.28)
⟺ y · μ = ∫ g · μ.  (3.29)

How useful this is depends on whether we can perform the integrations: 3.2 Example. Suppose we want to solve

ty′ + ay = bt^{2},  (3.30)

that is,

y′ + y · \frac{a}{t} = bt.  (3.31)

To find μ as above, we note

∫ ^{t} \frac{a}{s} d s = a ln t + C,  (3.32)

so we may let

μ(t) = exp(a ln t) = t^{a}.  (3.33)

Then we have

y′ + y · \frac{a}{t} = bt ⟺ \frac{d}{d t}(y · t^{a}) = bt^{a+1}  (3.34)
⟺ y · t^{a} = \frac{b}{a + 2}t^{a+2} + C  (3.35)
⟺ y = \frac{b}{a + 2}t^{2} + Ct^{−a}.  (3.36)

We must assume here a ≠ −2.

3.3 Separable equations

We can use the technique of integrating factors to attack an equation

\frac{d y}{d x} = f(x, y),  (3.37)

provided that f(x, y) is linear in y; in this case we solve as we did (3.22). Suppose now that in f(x, y), we can separate the two variables in the sense of putting x in the numerator and y in the denominator, so that (3.37) becomes

\frac{d y}{d x} = \frac{g(x)}{h(y)}.  (3.38)

The notation suggests the following solution:

\frac{d y}{d x} = \frac{g(x)}{h(y)} ⟺ h(y) d y = g(x) d x  (3.39)
⟺ ∫ h(y) d y = ∫ g(x) d x.  (3.40)

If in fact h = H′ and g = G′, then (3.40) is equivalent to

H(y) = G(x) + c.  (3.41)

This equation describes a family of curves, one for each value of c. Each curve determines y implicitly as one or more functions of x. Perhaps we want the curve to contain the point (x_{0}, y_{0}); then we must have

H(y_{0}) = G(x_{0}) + c,  (3.42)

which lets us rewrite (3.41) as

H(y) − H(y_{0}) = G(x) − G(x_{0}),  (3.43)

which in turn means that (3.40) can be written

∫_{y_{0}}^{y} h(t) d t = ∫_{x_{0}}^{x} g(t) d t.  (3.44)

3.3 Example. To solve

\frac{d y}{d x} = \frac{1 − x^{2}}{y^{2}}  (3.45)

so that y(0) = 3, we can rewrite the requirements as

∫_{3}^{y} t^{2} d t = ∫_{0}^{x}(1 − t^{2}) d t,  (3.46)
\frac{y^{3}}{3} − 9 = x − \frac{x^{3}}{3} ,  (3.47)
y^{3} = 27 + 3x − x^{3}.  (3.48)

Since every real number has a unique real cube-root, (3.48) determines y as a unique function of x. One can get a sense for the shape of the function by looking at the direction-field… The technique can be justified more formally as follows. Rewrite (3.38) as

y′ = \frac{g}{h(y)},  (3.49)

where y, g and h are understood to be functions. Equivalently,

h(y)y′ = g,  (3.50)
∫_{x_{0}}^{x} h(y)y′ = ∫_{x_{0}}^{x} g.  (3.51)

If y(x_{0}) = y_{0}, then by so-called u-substitution (here u = y(t)) we can write (3.51) as

∫_{y_{0}}^{y(x)} h = ∫_{x_{0}}^{x} g,  (3.52)

which is (3.44).

Note that linear differential equations with no constant term are separable: 3.4 Example. To solve the equation

y′ = y + y sin x  (3.53)

so that y(0) = 1, we may note

y′ = y + y sin x ⟺ y′ = y(1 + sin x) = \frac{1 + sin x}{1/y}  (3.54)
⟺ ∫_{1}^{y} \frac{d t}{t} = ∫_{0}^{x}(1 + sin t) d t  (3.55)
⟺ ln y = (t − cos t)|_{0}^{x} = x − cos x + 1  (3.56)
⟺ y = exp(x − cos x + 1).  (3.57)

A separable equation with initial condition may not have a unique solution: 3.5 Example. To solve

\frac{d y}{d x} = −\frac{x}{y} and y(1) = 0,  (3.58)

we note that the solutions are just those y satisfying:

∫_{0}^{y} t d t = − ∫_{1}^{x} t d t,  (3.59)
\frac{t^{2}}{2} |_{0}^{y} = −\frac{t^{2}}{2} |_{1}^{t} ,  (3.60)
y^{2} = −x^{2} + 1,  (3.61)
y = ±\sqrt{x^{2} + 1}.  (3.62)

We now have three methods of solution for some equations: 3.6 Example. We can solve

y′ = f(t) · (y + a),  (3.63)

where a is constant, in any of the following ways:

(1) Rewrite as follows:

\frac{y′}{y + a} = f(t),  (3.64)
\frac{d}{d t} ln |y + a| = f(t),  (3.65)
ln |y + a| = ∫ f + C,  (3.66)
|y + a| = exp C exp ∫ f,  (3.67)
y + a = c exp ∫ f,  (3.68)
y = c exp ∫ f − a.  (3.69)

In particular,

y_{0} = c exp ∫_{t_{0}}^{t_{0}} f − a = c − a,  (3.70)

so the solution is

y = (a + y_{0}) exp ∫_{t_{0}}^{t} f − a.  (3.71)

(2) The equation is separable:

\frac{d y}{d t} = \frac{f(t)}{(y + a)^{−1}} ,  (3.72)
∫_{y_{0}}^{y} \frac{d x}{x + a} = ∫_{t_{0}}^{t} f(x) d x,  (3.73)
ln ±(y + a) − ln ±(y_{0} + a) = ∫_{t_{0}}^{t} f,  (3.74)

which is equivalent to (3.71).

(3) We can use integrating factors; but watch out: We want to solve

\frac{d y}{d t} · μ(t) − y · f(t) · μ(t) = a · f(t) · μ(t),  (3.75)

where the left side is

\frac{d}{d t}(y · μ(t)),  (3.76)

so we want

μ′(t) = −f(t) · μ(t),  (3.77)
\frac{μ′(t)}{μ(t)} = −f(t),  (3.78)
\frac{d}{d t} ln |μ(t)| = −f(t),  (3.79)
ln |μ(t)| = − ∫ ^{t} f.  (3.80)

Any μ satisfying this will work, so let

μ(t) = exp ∫ ^{t} f.  (3.81)

Then we have

\frac{d}{d t}(y · μ(t)) = a · f(t) · μ(t) = −a · μ′(t),  (3.82)
∫_{t_{0}}^{t} \frac{d}{d u}(y · μ(u)) d u = ∫_{t_{0}}^{t} μ′(u) d u.  (3.83)

If we assume μ(t_{0}) = 0, then we get

y · μ(t) = −a · μ(t),  (3.84)
y = −a,  (3.85)

which is only a special case of the earlier solutions. But because of its definition here, μ(t) can never be 0. We may specify

μ(t) = exp ∫_{t_{0}}^{t} f,  (3.86)

so μ(t_{0}) = 1. Then (3.84) gives us

t · μ(t) − y_{0} = −a(μ(t) − 1),  (3.87)
y = \frac{a + y_{0}}{μ(t)} − a = (a + y_{0}) exp ∫_{t_{0}}^{t} f − a,  (3.88)

as before. 3.7 Example. We find an expression for the velocity v of a projectile fired vertically from an airless planet. We may let x be the distance of the projectile from the center of the planet. From Newton we have

\frac{d v}{d t} = −\frac{k}{x^{2}}  (3.89)

for some positive constant k that is proportional to the planet’s mass.

Either the projectile eventually returns to the planet, or not.

During certain time-intervals, ‘v is a function of x,’ that is, v = f ∘ x for some function f; so by the Chain Rule (see p. 3),

\frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v · \frac{d v}{d x}.  (3.90)

Thus we get a separable equation:

v · \frac{d v}{d x} = −\frac{k}{x^{2}} ,  (3.91)
∫_{v_{0}}^{v} u d u = −k ∫_{x_{0}}^{x} \frac{d u}{u^{2}},  (3.92)
v^{2} − v_{0}^{2} = 2k (\frac{1}{x} − \frac{1}{x_{0}} ) ,  (3.93)
v^{2} = \frac{2k}{x} + v_{0}^{2} − \frac{2k}{x_{0}} .  (3.94)

The limiting case is v_{0}^{2} = 2k/x_{0}; the projectile returns if and only if

v_{0} ⩽ \sqrt{\frac{2k}{x_{0}}} .  (3.95)

3.8 Example. [I didn’t do this in class.] Say a container (a bucket, a pond) of volume V is fed at a constant rate a (say in liters/minute) with a solution of some chemical compound with a variable concentration of f (in grams/liter, say). The same container is drained at the same rate. Assume that at any moment the concentration of compound in the container is uniform. Let the total mass of the compound in the container be Q. Then we get

Q′ = af(t) − a\frac{Q}{V},  (3.96)

a linear equation soluble by means of the integrating factor exp(at/V):

a · f(t) · exp \frac{at}{V} = Q′(t) · exp \frac{at}{V} + \frac{a}{V} Q(t) exp \frac{at}{V} = \frac{d}{d t} (Q′(t) · exp \frac{at}{V}) ,

(3.97)

a ∫ f(t) exp \frac{at}{V} d t + C.  (3.98)

Say t_{0} = 0 and Q(t_{0}) = Q_{0}. Then

Q(t) = \frac{∫_{0}^{t} f(u) exp \frac{au}{V} d u}{exp \frac{at}{V}} + Q_{0}.  (3.99)

3.4 Linear versus non-linear equations

Here is precise formulation of the method of integrating factors as applied to linear differential equations: 3.9 Theorem. Suppose p and g are continuous functions on the open interval (α, β), and α < t_{0} < β. Then for any real number y_{0} there is a unique function y that is differentiable on (α, β) and solves the initial-value problem

y′ + y · p = g,  (3.100)
y(t_{0}) = y_{0}.  (3.101)

Proof. Let μ be the function on (α, β) given by

μ(t) = exp ∫_{t_{0}}^{t} p.  (3.102)

Note that μ is well-defined on (α, β) since p is continuous. Also,

μ′ = p · μ,  (3.103)

μ(t_{0}) = 1 and μ(t) is never 0. Suppose (3.100) and (3.101) hold. Then we have

y′ · μ + y · p · μ = g · μ,  (3.104)
(y · μ)′ = g · μ,  (3.105)
∫_{t_{0}}^{t}(y · μ)′ = ∫_{t_{0}}^{t} g · μ,  (3.106)
y(t)μ(t) − y_{0} = ∫_{t_{0}}^{t} g · μ,  (3.107)

y(t) = \frac{∫_{t_{0}}^{t} g · μ}{μ(t)} + \frac{y_{0}}{μ(t)}  (3.108)

for all t in (α, β). Thus y is unique. Multiplying by μ(t) and differentiating, one can check that the value of y given by (3.108) does satisfy (3.100) and (3.101).

3.10 Example. In Example 3.2, we consider the equation

ty′ + ay = bt^{2}.  (3.109)

The corresponding linear equation is

y′ + y · \frac{a}{t} = bt.  (3.110)

Then Theorem 3.9 tells that, if (t_{0}, y_{0}) is a point in the plane such that t_{0} =≠ 0, then (3.109) has a unique solution such that y(t_{0}) = y_{0}, and y is defined on: (0, ∞), if t_{0} > 0, and on (−∞, 0), if t_{0} < 0. If a ≠ −2, we found

y = \frac{b}{a + 2}t^{2} + Ct^{−a},  (3.111)

from which we get

y_{0} = \frac{b}{a + 2}t_{0}^{2} + Ct_{0}^{−a},  (3.112)
C = y_{0}t_{0}^{a} − \frac{b}{a + 2}t_{0}^{a+2},  (3.113)
y = \frac{b}{a + 2}t^{2} + y_{0} (\frac{t_{0}}{t} )^{a} − \frac{b}{a + 2} (\frac{t_{0}}{t} )^{a} t_{0}^{2};  (3.114)

this is the unique solution. But say a = −1; then

y = bt^{2} + \frac{y_{0}}{t_{0}} t − bt_{0}t.  (3.115)

Thus y is well-defined on ℝ (and is a solution of (3.109)).

Does a theorem like Theorem 3.9 hold for an arbitrary first-order equation,

y′(t) = f(t, y(t))?  (3.116)

In time we shall give some justification for the following assertion:

Suppose a rectangle (α, β)×(γ, δ) contains a point (t_{0}, y_{0}), and on this rectangle, a function f is continuous and has a continuous partial derivative in the second coordinate. Then the initial-value problem

y′(t) = f(t, y(t)),  (3.117)
y(t_{0}) = y_{0}  (3.118)

has a unique solution on some interval containing t_{0}.

To see why this might be true, consider the direction-field for the equation. Note how continuity matters: For example, in Example 3.5, we have an initial-value problem d y/ d x = −x/y and y(1) = 0 with two solutions; but the function

(x, y) ↦ −\frac{x}{y}  (3.119)

is not defined (much less continuous) at (0, 1).

We also have the following: 3.11 Example. Is there an initial-value problem whose solutions are the curves y = at^{2}? These pass through (0, 0), and on such curves we have

y′ = 2at,  (3.120)
a = \frac{y}{t^{2}} ,  (3.121)
y′ = \frac{2y}{t} ,  (3.122)
y′t = 2y.  (3.123)

The direction-field is undefined at (0, 0), so the proposed theorem does not apply. We can still solve the equation:

\frac{d y}{2y} = \frac{d t}{t},  (3.124)
\frac{1}{2}(ln y − ln y_{0} = ln t − ln t_{0},  (3.125)
ln y + 2 ln t_{0} = 2 ln t + ln y_{0},  (3.126)
yt_{0}^{2} = t^{2}y_{0}.  (3.127)

3.12 Example. Assume y_{0} ≠ 0. If we solve

y′ = y^{2}  (3.128)

as a separable equation, we get

t − t_{0} = ∫_{y_{0}}^{y} u^{−2} d u = \frac{1}{y_{0}} − \frac{1}{y} ,  (3.129)
y = \frac{y_{0}}{1 − (t − t_{0})y_{0}} .  (3.130)

Then y is undefined when t = t_{0} +1/y_{0}, something not evident from the original equation (which meets the criteria of the proposed theorem on all rectangles). The solution should be understood as defined on: (−∞, t_{0} + 1/y_{0}), if y_{0} > 0; and on (t_{0} + 1/y_{0}, ∞), if y_{0} < 0. If we require y_{0} = 0, then, by inspection, the equation has the solution y = 0, defined on (−∞, ∞); by the proposed theorem, the solution is unique.

3.5 Exact equations

We ask which first-order equations y′ = f(x, y) have an implicit solution

ψ(x, y) = 0.  (3.131)

Differentiating the solution, we get

ψ_{x}(x, y) + ψ_{y}(x, y) · y′ = 0.  (3.132)

(Note that ψ_{x} is just ∂ψ/∂x, &c.) Such an equation is called exact, and its solutions have the form

ψ(x, y) = C.  (3.133)

Note that a multiple

μ(x, y) · ψ_{x}(x, y) + μ(x, y) · ψ_{y}(x, y) · y′ = 0  (3.134)

of an exact equation will not generally be exact, although it is solved by all solutions of the exact equation. 3.13 Example. The separable equation

f(x) + g(y) · y′ = 0  (3.135)

is exact, since it has the form (3.132) with

ψ(x, y) = ∫ f(x) d x + ∫ g(y) d y.  (3.136)

3.14 Example. The linear equation

y′ + y · g(x) = p(x)  (3.137)

is not generally exact, but if μ′(x) = g(x)·μ(x), then we can rewrite the equation as

y · μ′(x) − p(x) · μ(x) + μ(x) · y′ = 0,  (3.138)

which has the form (3.132) with

ψ(x, y) = y · μ(x) − ∫ p(x) · μ(x) d x.  (3.139)

3.15 Example. The equation

2x + y^{2} + 2xy · y′ = 0  (3.140)

has the form of (3.132), where

ψ(x, y) = x^{2} + xy^{2}.  (3.141)

How can you tell whether an equation of the form

M(x, y) + N(x, y) · y′ = 0  (3.142)

is exact?

3.16 Lemma. Let M and N be continuous functions of (x, y) on an open rectangle R, and suppose M_{y} and N_{x} exist and are continuous on that rectangle. The following are equivalent:

(1) There is a function ψ such that ψ_{x} = M and ψ_{y} = N on R.

(2) M_{y} = N_{x} on R.

Proof. If ψ exists, then ψ_{xy} = M_{y} and ψ_{yx} = N_{x}; since these derivatives are continuous, they are equal.

If M_{y} = N_{x}, we recover ψ by integrating: Say

ψ(x, y) = ∫ M(x, y) d x + h(y)  (3.143)

for some function h. Then

ψ_{x} = M.  (3.144)

We want to choose h so that ψ_{y} = N. We have

ψ_{y} = ∫ M_{y}(x, y) d x + h′(y),  (3.145)
h′(y) = ψ_{y} − ∫ M_{y}(x, y) d x.  (3.146)

We want the right member of this equation to be N − ∫ M_{y}(x, y) d x. It is enough if this function is constant in x, as h′(y) should be: that is, it is enough if

0 = \frac{∂}{∂x} (N − ∫ M_{y}(x, y) d x) = N_{x} − M_{y},  (3.147)

which is true by assumption. Hence we may choose

h = ∫ (N − ∫ M_{y}(x, y) d x) d y.  (3.148)

3.17 Example. To solve

2xe^{y} − y sin x + (x^{2}e^{y} + cos x − 1)y′ = 0,  (3.149)

we first check for exactness:

\frac{∂}{∂y} (2xe^{y} − y sin x) = 2xe^{y} − sin x,  (3.150)
\frac{∂}{∂x}(x^{2}e^{y} + cos x − 1) = 2xe^{y} − sin x,  (3.151)

equal, so the equation is exact. We now look for ψ so that

ψ_{x} = 2xe^{y} − y sin x,  (3.152)
ψ_{y} = x^{2}e^{y} + cos x − 1.  (3.153)

One may be able to see that the solution is

x^{2}e^{y} + y cos x − y = C.  (3.154)

Otherwise, find ψ by integrating:

ψ = ∫ ψ_{x} d x  (3.155)
= ∫ (2xe^{y} − y sin x) d x  (3.156)
= x^{2}e^{y} + y cos x + h(y);  (3.157)

differentiating with respect to y gives

x^{2}e^{y} + cos x − 1 = ψ_{y} = x^{2}e^{y} + cos x + h′(y),  (3.158)
h′(y) = −1,  (3.159)
h(y) = −y + C.  (3.160)

3.18 Example. The equation

y^{2} + xy + (x^{2} + 3xy)y′ = 0  (3.161)

is not exact. Perhaps we can find an integrating factor: some μ(x, y) such that the equation

μ(x, y) · (y^{2} + xy) + μ(x, y) · (x^{2} + 3xy)y′ = 0  (3.162)

is exact. Then we should have

μ(x, y) · (2y + x) + μ_{y}(x, y) · (y^{2} + xy) = μ(x, y) · (2x + 3y) + μ_{x}(x, y) · (x^{2} + 3xy),

(3.163)

a partial differential equation, which we are not prepared to solve in general. But perhaps there is a solution μ(x); that is, perhaps we can solve

μ(x) · (2y + x) = μ(x) · (2x + 3y) + μ′(x) · (x^{2} + 3xy),  (3.164)
μ′(x) = μ(x)\frac{2y + x − (2x + 3y)}{x^{2} + 3xy},  (3.165)
μ′(x) = −μ(x)\frac{x + y}{x(x + 3y)},  (3.166)

not soluble by current techniques. Try again: perhaps (3.163) has a solution μ(y), that is, we may be able to solve

μ(y) · (2y + x) + μ′(y) · (y^{2} + xy) = μ(y) · (2x + 3y),  (3.167)
μ′(y) = μ(y)\frac{2x + 3y − (2y + x)}{y^{2} + xy} = μ(y)\frac{x + y}{y(x + y)} = \frac{μ(y)}{y},  (3.168)

so we may let μ(y) = y. We now have to solve the exact equation

y^{3} + xy^{2} + (x^{2}y + 3xy^{2}) · y′ = 0.  (3.169)

The solution is ψ(x, y) = C, where

ψ = ∫ ψ_{x}(x, y) d x = ∫ (y^{3} + xy^{2}) d x = xy^{3} + \frac{1}{2}x^{2}y^{2} + h(y),  (3.170)
x^{2}y + 3xy^{2} = ψ_{y} = 3xy^{2} + x^{2}y + h′(y),  (3.171)
h′(y) = 0,  (3.172)

so h is constant and the solution is

xy^{3} + \frac{1}{2}x^{2}y^{2} = C.  (3.173)

4 Linear equations

An arbitrary linear ordinary differential equation can be written

y · f_{0}(t) + y′ · f_{1}(t) + y′′ · f_{2}(t) + ⋯ + y^{(n)} · f_{n}(t) = g(t)  (4.1)

or, more neatly,

∑_{i=0}^{n} y^{(i)} · f_{i}(t) = g(t);  (4.2)

if f_{n} ≠ 0, then this equation is of order n, and one can rewrite the equation as

∑_{i=0}^{n−1} y^{(i)} · \frac{f_{i}(t)}{f_{n}(t)} + y^{(n)} = \frac{g(t)}{f_{n}(t)}.  (4.3)

4.1 Homogeneous equations with constant coefficients, I

The equation (4.2) is homogeneous if g = 0. An arbitrary second-order linear homogeneous equation with constant coefficients is

y′′ + ay′ + b = 0.  (4.4)

4.1 Example. One may notice that the equation

y′′ − y = 0  (4.5)

has solutions e^{t} and e^{−t}. Then, because of the sum- and constant-rules of calculus, every function

c_{0}e^{t} + c_{1}e^{−t}  (4.6)

is a solution (where the c_{i} are constants). An initial condition such as

y(0) = 3  (4.7)

gives us only

c_{0} + c_{1} = 3;  (4.8)

to have a unique solution, we can impose an initial condition on y′, such as

y′(0) = 2;  (4.9)

then we have also

c_{0} − c_{1} = 2.  (4.10)

The system of equations (4.8) and (4.10) yields

y = \frac{5}{2}e^{t} + \frac{1}{2}e^{−t}  (4.11)

as the unique solution of the form (4.6) to the initial-value problem.

We may hypothesize that (4.4) has a solution e^{rt} for some r. Then we can check:

\frac{d^{2}}{d t^{2}} e^{rt} + a\frac{d}{d t}e^{rt} + be^{rt} = 0 ⟺  (4.12)
(r^{2} + ar + b)e^{rt} = 0 ⟺  (4.13)
r^{2} + ar + b = 0.  (4.14)

Equation (4.14) is called the characteristic equation of (4.4); it is a quadratic equation, with two or one real root, or else two or one complex, non-real root.

Suppose first that (4.14) has the two distinct real roots r_{0} and r_{1}. Then, just as in Example 4.1, Equation (4.4) has the solutions

y = c_{0}e^{r_{0}t} + c_{1}e^{r_{1}t}.  (4.15)

This is called the general solution, since it will turn out that all solutions have this form. The initial conditions

y(t_{0}) = y_{0} and y′(t_{0}) = y′_{0}  (4.16)

yield the linear system

\begin{cases}y_{0} = c_{0}e^{r_{0}t_{0}} + c_{1}e^{r_{1}t_{0}}, \\ y′_{0} = r_{0}c_{0}e^{r_{0}t_{0}} + r_{1}c_{1}e^{r_{1}t_{0}}, \end{cases}  (4.17)

which has a unique solution (c_{0}, c_{1}): We can check this just by solving the system, or by using theory:

If a, b, c and d are numbers, we write

|\begin{matrix}a & b \\ c & d \end{matrix}|  (4.18)

for the number ad − bc; this is the determinant of the 2 × 2 matrix

(\begin{matrix}a & b \\ c & d \end{matrix}) .  (4.19)

If this is not 0, then the linear system

\begin{cases}ax + by = e, \\ cx + dy = f \end{cases}  (4.20)

20  4 LINEAR EQUATIONS  November 9, 2007

has the solution

x = \begin{matrix}| & \begin{matrix}e & b \\ f & d \end{matrix} & | \\ | & \begin{matrix}a & b \\ c & d \end{matrix} & | \end{matrix} = \frac{ed − bf}{ad − bc} , y = \begin{matrix}| & \begin{matrix}a & e \\ c & f \end{matrix} & | \\ | & \begin{matrix}a & b \\ c & d \end{matrix} & | \end{matrix} = \frac{af − ec}{ad − bc},  (4.21)

as one can check by computing. So (4.17) has the solution

c_{0} = \frac{y_{0}r_{1} − y′_{0}}{r_{1} − r_{0}} e^{−r_{0}t_{0}}, c_{1} = \frac{y′_{0} − y_{0}r_{0}}{r_{1} − r_{0}} e^{−r_{1}t_{0}}.  (4.22)

4.2 Example. For the general solution of

y′′ + y′ − 12y = 0,  (4.23)

form the characteristic equation,

r^{2} + r − 12 = 0,  (4.24)

whose roots are −4 and 3; so the general solution is

y = c_{0}e^{−4t} + c_{1}e^{3t}.  (4.25)

Under the initial conditions

y(0) = 5 and y′(0) = −1,  (4.26)

we get the system

\begin{cases}5 = c_{0} + c_{1}, \\ −1 = −4c_{0} + 3c_{1}, \end{cases}  (4.27)

with solution (16/7, 19/7), whence

y = \frac{16}{7}e^{−4t} + \frac{19}{7}e^{3t}.  (4.28)

Note that the graph has a minimum point (t_{c}, y_{c}), where

0 = y′_{c} = −\frac{64}{7}e^{−4t_{c}} + \frac{57}{7}e^{3t_{c}},  (4.29)
e^{−7t_{c}} = \frac{57}{64},  (4.30)
t_{c} = \frac{1}{7} ln \frac{64}{57}.  (4.31)

Note how the long-term behavior of (4.15) depends on the signs of r_{0} and r_{1}.

4.2 Homogeneous equations in general

An arbitrary second-order linear differential equation can be written

y′′ + p(t)y′ + q(t)y = g(t).  (4.32)

It can be proved that, if the functions p, q and g are continuous on an open interval, then the equation has a unique solution on the whole interval satisfying initial conditions

y_{0} = y(t_{0}) and y′_{0} = y′(t_{0}).  (4.33)

We are interested now in the homogeneous case, where g = 0. 4.3 Theorem (principle of superposition). If ϕ_{0} and ϕ_{1} are solutions of

y′′ + p(t)y′ + q(t)y = 0,  (4.34)

and c_{0} and c_{1} are constants, then

c_{0}ϕ_{0} + c_{1}ϕ_{1}  (4.35)

is also a solution.

As in Example 4.1, the proof is by the sum- and constant-rules of differential calculus. Since the 0-function is a solution of (4.34), another way to state the conclusion of the theorem is that the set of solutions of (4.34) is a vector-space.

Suppose one does know two solutions as in the theorem. Then a solution satisfying (4.33) can be found by solving

\begin{cases}y_{0} = c_{0}ϕ_{0}(t_{0}) + c_{1}ϕ_{1}(t_{0}), \\ y′_{0} = c_{0}ϕ′_{0}(t_{0}) + c_{1}ϕ′_{1}(t_{0}). \end{cases}  (4.36)

This system has a solution if the determinant

|\begin{matrix}ϕ_{0}(t_{0}) & ϕ_{1}(t_{0}) \\ ϕ′_{0}(t_{0}) & ϕ′_{1}(t_{0}) \end{matrix}|  (4.37)

is not 0. This determinant is called the Wronskian of ϕ_{0} and ϕ_{1} at t_{0} and can be denoted

W (ϕ_{0}, ϕ_{1})(t_{0}).  (4.38)

Suppose this is not 0; then:

• for the equation (4.34) with initial conditions (4.33) there is a solution of the form (4.35);

• the functions ϕ_{0} and ϕ_{1} compose a fundamental set of solutions to (4.34);

• equation (4.35) can be called a general solution to (4.34);

• the space of solutions to (4.34) is 2-dimensional. 4.4 Example. Suppose the characteristic equation of (4.4) has distinct real roots r_{0} and r_{1}. Let ϕ_{i}(t) = e^{r_{i}t}. Then

W (ϕ_{0}, ϕ_{1})(t) = |\begin{matrix}e^{r_{0}t} & e^{r_{1}t} \\ r_{0}e^{r_{0}t} & r_{1}e^{r_{1}t} \end{matrix}| = (r_{1} − r_{0})e^{(r_{0}+r_{1})t} ≠ 0,  (4.39)

so ϕ_{0} and ϕ_{1} compose a pair of fundamental solutions of (4.4).

4.5 Example. The equation

2t^{2}y′′ + 3ty′ − 1 = 0  (4.40)

can be written

2y′′ + 3t^{−1}y′ − t^{−2} = 0,  (4.41)

so by the claim at the beginning of this sub-section, the equation has solutions on (0, ∞). One can check that two such solutions are t^{1/2} and t^{−1}. These are fundamental solutions since

W (t^{1/2}, t^{−1}) = |\begin{matrix}t^{1/2} & t^{−1} \\ t^{−1/2}/2 & −t^{−2} \end{matrix}| = \frac{−3t^{−3/2}}{2} ≠ 0  (4.42)

when t > 0. 4.6 Example. Suppose ϕ_{i} are solutions of (4.34) satisfying the initial conditions

ϕ_{0}(t_{0}) = 1, ϕ′_{0}(t_{0}) = 0,  (4.43)
ϕ_{1}(t_{0}) = 0, ϕ′_{1}(t_{0}) = 1.  (4.44)

Then the ϕ_{i} compose a fundamental pair of solutions, since their Wronskian is 1.

The last example shows that, by the claim at the beginning of the sub-section, fundamental solutions always exist. More than one pair exists for a given equation: 4.7 Example. Equation (4.5) has fundamental solutions e^{t} and e^{−t}. If we solve (4.5) for the fundamental solutions in the last example, we get (e^{t} + e^{−t})/2 and (e^{t} − e^{−t})/2, that is, cosh t and sinh t. So the general solution can be written

k_{0} cosh t + k_{1} sinh t.  (4.45)

Thus, (4.5) has two neat pairs of fundamental solutions, namely, {e^{t}, e^{−t}} and {cosh, sinh}.

If W (ϕ_{0}, ϕ_{1})(t_{0}) ≠ 0 for some t_{0}, then the homogeneous linear system

\begin{cases}c_{0}ϕ_{0}(t_{0}) + c_{1}ϕ_{1}(t_{0}) = 0, \\ c_{0}ϕ′_{0}(t_{0}) + c_{1}ϕ′_{1}(t_{0}) = 0. \end{cases}  (4.46)

has the unique solution (0, 0). This means that the only solution to the equation

c_{0}ϕ_{0} + c_{1}ϕ_{1} = 0  (4.47)

is (0, 0). By definition then, the functions ϕ_{0} and ϕ_{1} are linearly independent. We have shown: 4.8 Theorem. If the Wronskian of two functions is non-zero at one point, then the two functions are linearly independent.

The converse is false: that is, there are linearly independent functions whose Wronskian is 0 at all points:

4.9 Example. On the interval (−2, 2), let ϕ_{0}(t) = t^{3} and let ϕ_{1}(t) = |t^{3}| = t^{2} |t|. Then ϕ′_{0}(t) = 3t^{2}, while ϕ′_{1}(t) = 3t |t|. Hence

W (ϕ_{0}, ϕ_{1})(t) = |\begin{matrix}t^{3} & t^{2} |t| \\ 3t^{2} & 3t |t| \end{matrix}| = 0  (4.48)

for all t. However, suppose c_{0}ϕ_{0} + c_{1}ϕ_{1} = 0 on (−2, 2). Then at 1 we have c_{0} + c_{1} = 0, while at −1 we have −c_{0} + c_{1} = 0; solved simultaneously, these equations have the unique solution (0, 0). Therefore ϕ_{0} and ϕ_{1} are linearly independent. 4.10 Example. If r_{0} ≠ r_{1}, then the functions e^{r_{0}t} and e^{r_{1}t} are linearly independent by Example 4.4.

If the ϕ_{i} are solutions to a second-order linear homogeneous differential equation, then examples like 4.9 are not possible: 4.11 Theorem. Suppose ϕ_{0} and ϕ_{1} are defined on an open interval I and are solutions of

y′′ + p(t)y′ + q(t)y = 0.  (4.49)

If their Wronskian is 0 at one point of I, then the functions are linearly dependent.

Proof. Suppose W (ϕ_{0}, ϕ_{1})(t_{0}) = 0. Then the system (4.46) has a non-zero solution (c_{0}, c_{1}). Let ψ be the function c_{0}ϕ_{0} + c_{1}ϕ_{1}. Then, by (4.46), we have ψ(t_{0}) = 0 and ψ′(t_{0}) = 0. But ψ is also a solution of (4.49), and there is only one solution meeting the initial conditions y(t_{0}) = 0 and y′(t_{0}) = 0, namely, y = 0. Hence ψ = 0, which means that the ϕ_{i} are linearly dependent.

We now know that the following are equivalent conditions on a pair {ϕ_{0}, ϕ_{1}} of solutions of (4.49):

(1) W (ϕ_{0}, ϕ_{1})(t_{0}) ≠ 0 for some t_{0}.

(2) {ϕ_{0}, ϕ_{1}} is a fundamental set of solutions of (4.49).

(3) The ϕ_{i} are linearly independent.

(4) W (ϕ_{0}, ϕ_{1})(t_{0}) ≠ 0 for all t_{0} (in the domain of the ϕ_{i}).

In particular, we can say something about W (ϕ_{0}, ϕ_{1}) without knowing the ϕ_{i}: it is 0 either everywhere or nowhere. We can say more: 4.12 Theorem (Abel). If ϕ_{0} and ϕ_{1} are solutions of (4.49), then

W (ϕ_{0}, ϕ_{1})(t) = c exp (− ∫ ^{t} p) ,  (4.50)

where c is constant (that is, c is independent of t, or d c/ d t = 0; the value of c does depend on ϕ_{0} and ϕ_{1}).

Proof. Write the Wronskian here as W. We have

W = ϕ_{0}ϕ′_{1} − ϕ_{1}ϕ′_{0},  (4.51)
W′ = ϕ_{0}ϕ′′_{1} − ϕ_{1}ϕ′′_{0}.  (4.52)

24  4 LINEAR EQUATIONS  November 9, 2007

From (4.49), we also have

ϕ_{1}ϕ′′_{0} + pϕ_{1}ϕ′_{0} + qϕ_{1}ϕ_{0} = 0,  (4.53)
ϕ_{0}ϕ′′_{1} + pϕ_{0}ϕ′_{1} + qϕ_{0}ϕ_{1} = 0,  (4.54)

which by subtraction yields

W′ + pW = 0.  (4.55)

This is a first-order differential equation we can solve, and its solution is (4.50).

4.3 Homogeneous equations with constant coefficients, II

By observation, the equation

y′′ + y = 0  (4.56)

has a fundamental set {cos, sin} of solutions. The characteristic equation

r^{2} + 1 = 0  (4.57)

has roots ±i. Suppose the equation

r^{2} + ar + b = 0  (4.58)

has roots λ ± iμ, where λ and μ are real, and μ ≠ 0 (so that the two roots are distinct). One may guess that a fundamental set of solutions of

y′′ + ay′ + by = 0  (4.59)

is {e^{λt} cos μt, e^{λt} sin μt}. The guess is correct, as we can verify: For example, let

ϕ(t) = e^{λt} cos μt;  (4.60)

then

ϕ′(t) = λe^{λt} cos μt − μe^{λt} sin μt,  (4.61)
ϕ′′(t) = λ^{2}e^{λt} cos μt − 2λμe^{λt} sin μt − μ^{2}e^{λt} cos μt,  (4.62)

and therefore

(ϕ′′ + aϕ′ + bϕ)(t) =

(λ^{2} − μ^{2} + aλ + b)e^{λt} cos μt − μ(2λ + a)e^{λt} sin μt. (4.63)

By assumption, we have

r^{2} + ar + b = (r − λ − iμ)(r − λ + iμ)  (4.64)
= (r − λ)^{2} − (iμ)^{2}  (4.65)
= r^{2} − 2rλ + λ^{2} + μ^{2},  (4.66)

whence

a = −2λ,  (4.67)
b = λ^{2} + μ^{2}  (4.68)

and therefore

2λ + a = 0,  (4.69)
λ^{2} − μ^{2} + aλ + b = λ^{2} − μ^{2} − 2λ^{2} + λ^{2} + μ^{2} = 0,  (4.70)
ϕ′′ + aϕ′ + bϕ = 0.  (4.71)

Likewise if ϕ(t) = e^{λt} sin μt. Therefore e^{λt} cos μt and e^{λt} sin μt are solutions of (4.59). They comprise a fundamental set since their Wronskian is non-zero:

|\begin{matrix}e^{λt} cos μt & e^{λt} sin μt \\ e^{λt}(λ cos μt − μ sin μt) & e^{λt}(λ sin μt + μ cos μt) \end{matrix}| =

e^{2λt} (λ |\begin{matrix}cos μt & sin μt \\ cos μt & sin μt \end{matrix}| + μ |\begin{matrix}cos μt & sin μt \\ − sin μt & cos μt) \end{matrix}|) = μe^{2λt}. (4.72)

Why does this approach work? We might expect complex exponentiation to be defined so that (4.59) has solutions e^{(λ±iμ)t}. In particular, if ϕ(t) = e^{it}, then we should have

ϕ′(t) = iϕ(t),  (4.73)
ϕ′′(t) = −ϕ(t),  (4.74)

so that ϕ solves the initial-value problem

y′′ + y = 0, y(0) = 1, y′(0) = i.  (4.75)

But this problem should have a unique solution c_{0} cos +c_{1} sin, where—since (c_{0} cos +c_{1} sin)′ = −c_{0} sin +c_{1} cos—we have

1 = c_{0} cos 0 + c_{1} sin 0 = c_{0},  (4.76)
i = −c_{0} sin 0 + c_{1} cos 0 = c_{1}.  (4.77)

Therefore we should define

e^{it} = cos t + i sin t;  (4.78)

this is known as Euler’s formula. From it we have

e^{−it} = cos t − i sin t,  (4.79)

whence

e^{(λ±iμ)t} = e^{λt}e^{±iμt} = e^{λt}(cos μt ± i sin μt).  (4.80)

Hence, by Euler’s formula, e^{(λ±iμ)t} are solutions of (4.59). Moreover, the solutions we found originally are given by

e^{λt} cos μt = \frac{1}{2}(e^{(λ+iμ)t} + e^{(λ−iμ)t}),  (4.81)
e^{λt} sin μt = \frac{1}{2i}(e^{(λ+iμ)t} − e^{(λ−iμ)t}).  (4.82)

(4.83)

Thus {e^{(λ±iμ)t}} is a fundamental set of solutions of (4.59).

Note in particular

cos t = \frac{1}{2}(e^{it} + e^{it}),  (4.84)
sin t = \frac{1}{2i}(e^{it} − e^{it}).  (4.85)

4.13 Example. The equation

y′′ − y′ + y = 0  (4.86)

has characteristic equation

r^{2} − r + 1 = 0,  (4.87)

whence

r = \frac{1 ± i\sqrt{3}}{2},  (4.88)

so the general solution is

e^{t/2} (c_{0} cos \frac{t\sqrt{3}}{2} + c_{1} sin \frac{t\sqrt{3}}{2}) .  (4.89)

4.14 Remark. The product

e^{λt}(c_{0} cos μt + i sin μt)  (4.90)

can also be written

Ae^{λt} sin(μt + θ)  (4.91)

for some A and θ. One can verify this by angle-addition formulas. Alternatively, A sin(t + θ) is a solution of (4.56), so it is c_{0} cos t + c_{1} sin t for some c_{0} and c_{1}; moreover,

A sin θ = c_{0},  (4.92)
A cos θ = c_{1}.  (4.93)

From this, A^{2} = c_{0}^{2}+c_{1}^{2}, and then θ is uniquely determined in the interval [0, 2π).

4.4 Homogeneous equations with constant coefficients, III

Suppose the characteristic equation (4.58) has a double root—which must therefore be −a/2. Then (4.59) has the solution e^{−at/2}. Call this ϕ(t). Then tϕ(t) is also a solution. Indeed, we have

\frac{d}{d t}(tϕ(t)) = ϕ(t) − \frac{a}{2}tϕ(t),  (4.94)
\frac{d^{2}}{d t^{2}} (tϕ(t)) = −aϕ(t) + \frac{a^{2}}{4} tϕ(t)  (4.95)

and hence

\frac{d^{2}}{d t^{2}} (tϕ(t)) + a\frac{d}{d t}(tϕ(t)) + btϕ(t) = (\frac{a^{2}}{4} − \frac{a^{2}}{2} + b) tϕ(t) = 0  (4.96)

since 4b = a^{2}. Moreover, ϕ(t) and tϕ(t) are linearly independent, since

|\begin{matrix}ϕ(t) & tϕ(t) \\ ϕ′(t) & ϕ(t) + tϕ′(t) \end{matrix}| = |\begin{matrix}ϕ(t) & tϕ(t) \\ 0 & ϕ(t) \end{matrix}| + |\begin{matrix}ϕ(t) & tϕ(t) \\ ϕ′(t) & tϕ′(t) \end{matrix}| = ϕ(t)^{2}.  (4.97)

Thus {e^{−at/2}, te^{−at/2}} is a fundamental set of solutions of (4.59) when the characteristic equation has a double root. 4.15 Example. The equation

y′′ − 2y + y = 0  (4.98)

has characteristic equation

0 = r^{2} − 2r + 1 = (r − 1)^{2},  (4.99)

so {e^{t}, te^{t}} is a fundamental set of solutions. Let ϕ(t) = c_{0}e^{t} + c_{1}te^{t}, so that ϕ′(t) = (c_{0} + c_{1})e^{t} + c_{1}te^{t}. The initial conditions

y(0) = 1 and y′(0) = 0  (4.100)

give us

\begin{cases}1 = c_{0}, \\ 0 = c_{0} + c_{1}, \end{cases}  (4.101)

so the solution is e^{t} − te^{t}, that is, (1 − t)e^{t}.

4.5 Reduction of order

The solution te^{−at/2} in the previous subsection can be derived from the solution e^{−at/2} by reduction of order, that is, by reducing the original second-order equation (4.59) to a first-order equation. The general procedure is as follows.

Suppose ϕ is a solution of of (4.34). We try to find a function v so that vϕ is also a solution. We have

(vϕ)′ = vϕ′ + v′ϕ,  (4.102)
(vϕ)′′ = vϕ′′ + 2v′ϕ′ + v′′ϕ,  (4.103)

whence

(vϕ)′′ + p(vϕ)′ + qvϕ = 2v′ϕ′ + v′′ϕ + pv′ϕ.  (4.104)

Thus v should be a solution of the equation

2v′ϕ′ + v′′ϕ + pv′ϕ = 0,  (4.105)

which means v′ should be a solution of

ϕ(v′)′ + (2ϕ′ + pϕ)v′ = 0,  (4.106)

a first-order equation—in fact, linear-homogeneous (hence separable). 4.16 Example. If our equation is (4.98), with solution e^{t}, then (4.106) becomes

e^{t}v′′(t) = 0,  (4.107)

so v′ is constant, and we may let v(t) = t.

4.17 Example. The equation

2t^{2}y′′ + 3ty′ − y = 0  (4.108)

has the solution 1/t. We want another solution, v(t)/t. We have

\frac{d}{d t} (\frac{v(t)}{t}) = \frac{tv′(t) − v(t)}{t^{2}} = \frac{v′(t)}{t} − \frac{v(t)}{t^{2}},  (4.109)
\frac{d^{2}}{d t^{2}} (\frac{v(t)}{t}) = \frac{tv′′(t) − v′(t)}{t^{2}} − \frac{t^{2}v′(t) − 2tv(t)}{t^{4}}  (4.110)
= \frac{t^{2}v′′(t) − 2tv′(t) + 2v(t)}{t^{3}},  (4.111)

so replacing y with v(t)/t in (4.108) gives

2(t^{2}v′′(t) − 2tv′(t) + 2v(t)) + 3(tv′(t) − v(t)) − v(t) = 0,  (4.112)
2t^{2}v′′(t) − tv′(t) = 0,  (4.113)
\frac{v′′(t)}{v′(t)} = \frac{1}{2t},  (4.114)
ln v′(t) = \frac{1}{2} ln t + C;  (4.115)

letting C be 0, we get

v′(t) = t^{1/2},  (4.116)
v = ∫ t^{1/2} d t = \frac{2}{3}t^{3/2} + C,  (4.117)

where again we may let C be 0.

4.6 Inhomogeneous equations

We know that the set of solutions of (4.34) is a two-dimensional vector-space V. If ψ is a solution of

y′′ + p(t) · y′ + q(t) · y = g(t),  (4.118)

then the set of solutions is V + ψ, the translation of V by ψ. This means two things:

• If ψ_{0} and ψ_{1} solve (4.118), then ψ_{0} − ψ_{1} solves (4.34).

• If ψ solves (4.118), and ϕ solves (4.34), then ψ + ϕ solves (4.118).

So if we know the general solution to (4.34), then we know every solution to (4.118), provided we know just one solution. To find one solution, we have two methods: 4.18 Example. To solve

y′′ + 3y′ − 4y = e^{2t},  (4.119)

we note first that the general solution of the corresponding homogeneous equation

y′′ + 3y′ − 4y = e^{2t}  (4.120)

is c_{0}e^{t} + c_{1}e^{−4t}. We guess that the inhomogeneous equation has solution Ae^{2t}, where A is an undetermined coefficient. If ϕ(t) = Ae^{2t}, then ϕ′ = 2ϕ and ϕ′′ = 4ϕ, so we want

(ϕ′′ + 3ϕ′ − 4ϕ)(t) = e^{2t},  (4.121)
(4 + 3 · 2 − 4)Ae^{2t} = e^{2t},  (4.122)
6A = 1.  (4.123)

So the inhomogeneous equation has general solution

c_{0}e^{t} + c_{1}e^{−4t} + \frac{1}{6}e^{2t}.  (4.124)

If ϕ_{i} solves y′′ + p · y′ + q · y = g_{i}, then ϕ_{0} + ϕ_{1} solves

y′′ + p · y′ + q · y = g_{0} + g_{1}.  (4.125)

So the method of undetermined coefficients has a broader application: 4.19 Example. To find a particular solution of

y′′ + 3y′ − 4y = e^{2t} − e^{6t},  (4.126)

it is enough to take e^{2t}/6 from the previous example and add it to a solution of y′′ + 3y′ − 4y = −e^{6t}, such as −e^{6t}/50. 4.20 Example. To solve

y′′ + 3y′ − 4y = 2 sin t,  (4.127)

remember that 2i sin t = e^{it} − e^{−it}; so we can look for a particular solution of the form

Ce^{it} + De^{−it};  (4.128)

but since also 2 cos t = e^{it} + e^{−it}, we can also look for a solution

A cos t + B sin t.  (4.129)

4.7 Variation of parameters

[Not typed up yet.]

5 Laplace transform

5.1 Basics

Suppose f is a function defined on [0, ∞). The function f (if it exists) defined on an interval [p, ∞) by

F (s) = ∫_{0}^{∞} e^{−st}f(t) d t  (5.1)

is called the Laplace transform of f and is denoted

ℒ {f} .  (5.2)

By convention here, f is treated as f(t), and F as F (s). 5.1 Example. By integration by parts, we have

ℒ {t} = ∫_{0}^{∞} te^{−st} d t  (5.3)
= lim_{x→∞} ∫_{0}^{x} te^{−st} d t  (5.4)
= lim_{x→∞} ([\frac{te^{−st}}{−s} ]_{0}^{x} + \frac{1}{s} ∫_{0}^{x} e^{−st} d t)  (5.5)
= lim_{x→∞} [\frac{e^{−st}}{−s^{2}} ]_{0}^{x}  (5.6)
= \frac{1}{s^{2}} .  (5.7)

One important property of ℒ {} is easily proved: 5.2 Theorem. ℒ {} is linear, that is,

ℒ {af + bg} = aℒ {f} + bℒ {g} ,  (5.8)

where f and g are functions, and a and b are constants.

To use ℒ {} in solving differential equations, we can relate ℒ {f} and ℒ {f′}. First we identify a condition on f guaranteeing that ℒ {f} exists: 5.3 Theorem. If f is defined on [0, ∞) and is integrable on each interval [0, t], and if lim_{t→∞} e^{−pt}f(t) = 0 for some (positive) p, then ℒ {f} exists.

Proof. By hypothesis, there is M such that, if M ⩽ t, then |e^{−pt}f(t)| ⩽ 1. Suppose p < s. Then |e^{−st}f(t)| ⩽ e^{(p−s)t} |e^{−pt}f(t)| ⩽ e^{(p−s)t} when M ⩽ t. Since ∫_{0}^{∞} e^{(p−s)t} exists, ℒ {f} must exist.

If f meets the condition that lim_{t→∞} e^{−pt}f(t) = 0 for some (positive) p, then f may be said to have exponential order. 5.4 Theorem. If f is continuously differentiable and has exponential order, then ℒ {f′} exists and is given by

ℒ {f′} = s · ℒ {f} − f(0).  (5.9)

Proof. Using integration by parts, we have

ℒ {f′} = ∫_{0}^{∞} e^{−st}f′(t) d t  (5.10)
= [e^{−st}f(t)]_{0}^{∞} + s ∫_{0}^{∞} e^{−st}f(t) d t  (5.11)
= [e^{−st}f(t)]_{0}^{∞} + sℒ {f(t)} .  (5.12)

Since f has exponential order, we must have

lim_{x→∞} e^{−sx}f(x) = 0  (5.13)

if s is large enough. The claim now follows.

In fact the last theorem is true if f′ is merely ‘piecewise’ continuous. A third fact we shall not prove: 5.5 Theorem. If ℒ {f} = ℒ {g}, then f = g.

We can now solve some initial-value problems: 5.6 Example. To solve

y′ − 2y = e^{4t}, y(0) = 3,  (5.14)

suppose ϕ is a solution. Then

ℒ {ϕ′ − 2ϕ} = ℒ {e^{4t}} ,  (5.15)
ℒ {ϕ′} − 2 · ℒ {ϕ} = ∫_{0}^{∞} e^{−st}e^{4t} d t,  (5.16)
s · ℒ \{ϕ\} − ϕ(0) − 2 · ℒ \{ϕ\} = \frac{1}{s − 4},  (5.17)
(s − 2) · ℒ \{ϕ\} − 3 = \frac{1}{s − 4},  (5.18)
(s − 2) · ℒ \{ϕ\} = 3 + \frac{1}{s − 4} = \frac{3s − 11}{s − 4},  (5.19)
ℒ \{ϕ\} = \frac{3s − 11}{(s − 2)(s − 4)}.  (5.20)

Now decompose the right side into ‘partial fractions’:

\frac{3s − 11}{(s − 2)(s − 4)} = \frac{A}{s − 2} + \frac{B}{s − 4},  (5.21)

where

3s − 11 = A(s − 4) + B(s − 2) = (A + B)s − 4A − 2B,  (5.22)

whence

\begin{cases}A + B = 3, \\ 4A + 2B = 11, \end{cases}  (5.23)

so that

A = \begin{matrix}| & 3 1 11 2 & | \\ | & 1 1 4 2 & | \end{matrix} = \frac{6 − 11}{2 − 4} = \frac{5}{2}, B = 3 − \frac{5}{2} = \frac{1}{2}.  (5.24)

We now have

ℒ \{ϕ\} = \frac{5}{2} (\frac{1}{s − 2}) + \frac{1}{2} (\frac{1}{s − 4}) .  (5.25)

Our calculation of ℒ {e^{4t}} can be adapted to show

ℒ \{e^{at}\} = \frac{1}{s − a}.  (5.26)

So we have

ℒ \{ϕ\} = \frac{5}{2}ℒ \{e^{2t}\} + \frac{1}{2}ℒ \{e^{4t}\}  (5.27)
= ℒ \{\frac{5}{2}e^{2t} + \frac{1}{2}e^{4t}\} ,  (5.28)

and therefore

ϕ = \frac{5}{2}e^{2t} + \frac{1}{2}e^{4t}.  (5.29)

To solve higher-order differential equations, we can use a generalization of Theorem 5.4: 5.7 Theorem. If f^{(n−1)} is continuously differentiable and has exponential order, then ℒ {f^{(n)}} exists and is given by

ℒ {f^{(n)}} = s^{n} · ℒ {f} − ∑_{k=0}^{n−1} s^{k}f^{(n−1−k)}(0).  (5.30)

Proof. Induction.

5.8 Example. To solve

y′′ + y = e^{−2t} sin t, y(0) = 0, y′(0) = 0,  (5.31)

apply ℒ {} to the differential equation, and use the initial values:

ℒ {y′′} + ℒ {y} = ℒ {e^{−2t} sin t} ,  (5.32)
s^{2} · ℒ {y} − s · y(0) − y′(0) + ℒ {y} = ℒ {e^{−2t} sin t} ,  (5.33)
(s^{2} + 1) · Y = ℒ {e^{−2t} sin t} ,  (5.34)

where Y = ℒ {y}.

To continue with the solution, we need more facts about Laplace transforms: 5.9 Theorem. If ℒ {f(t)} = F (s), then F (s + a) = ℒ {e^{−at}f(t)}.

So ℒ {e^{−2t} sin t} = F (s + 2), where F (s) = ℒ {sin t}. Now,

ℒ {sin} = ∫_{0}^{∞} e^{−st} sin t d t  (5.35)
= [−e^{−st} cos t]_{0}^{∞} − s ∫_{0}^{∞} e^{−st} cos t d t  (5.36)
= 1 − s · ℒ {cos} ,  (5.37)

and similarly

ℒ {cos} = s · ℒ {sin}  (5.38)
= s − s^{2} · ℒ {cos} ,  (5.39)

so that

ℒ \{cos\} = \frac{s}{s^{2} + 1}, ℒ \{sin\} = \frac{1}{s^{2} + 1}.  (5.40)

Alternatively, one can use that

ℒ {cos} + i · ℒ {sin} = ∫_{0}^{∞} e^{−st}(cos t + i sin t) d t  (5.41)
= ∫_{0}^{∞} e^{−st}e^{it} d t = [\frac{1}{−s + i}e^{(−s+i)t}]_{0}^{∞}  (5.42)
= \frac{1}{s − i}  (5.43)
= \frac{s + i}{s^{2} + 1}.  (5.44)

Continuing with Example 5.8, we have

(s^{2} + 1) · Y = \frac{1}{(s + 2)^{2} + 1},  (5.45)
Y = \frac{1}{(s^{2} + 1)(s^{2} + 4s + 5)},  (5.46)
Y = \frac{As + B}{s^{2} + 1} + \frac{Cs + D}{s^{2} + 4s + 5},  (5.47)

where one should calculate that

A = −\frac{1}{8}, B = \frac{1}{8}, C = \frac{1}{8}, D = \frac{3}{8}.  (5.48)

In any case, we now have

Y = A\frac{s}{s^{2} + 1} + B \frac{1}{s^{2} + 1} + C \frac{s}{(s + 2)^{2} + 1} + D \frac{1}{(s + 2)^{2} + 1}  (5.49)
= Aℒ {cos} + Bℒ {sin} + Cℒ {e^{−2t} cos t} + Dℒ {e^{−2t} sin t} ,  (5.50)

so that

y = A cos t + B sin t + Ce^{−2t} cos t + De^{−2t} sin t.  (5.51)

In the example, if sin t had been replaced with sin 5t, we could have used the following: 5.10 Theorem. If ℒ {f} = F, then ℒ {f(at)} = F (s/a)/a.

Proof. By the substitution

u = at, d u = a d t,  (5.52)

we have

ℒ {f(at)} = ∫_{0}^{∞} e^{−st}f(at) d t  (5.53)
= \frac{1}{a} ∫_{0}^{∞} e^{−su/a}f(u) d u  (5.54)
= \frac{1}{a} · F (\frac{s}{a}) ,  (5.55)

where F (s) = ℒ {f}.

Hence for example

ℒ \{cos at\} = \frac{s}{s^{2} + a^{2}} , ℒ \{sin at\} = \frac{a}{s^{2} + a^{2}} .  (5.56)

5.2 Springs

Suppose an object is subject to the force of a spring (and gravity). Let u(t) be the displacement of the object from its resting position. Then the force on the spring is mu′′(t). The force may have three components:

(1) −ku(t), by Hooke’s Law;

(2) −γu′(t), due to friction (damping);

(3) some other force F (t). Thus arises the 2nd-order linear differential equation

mu′′ + γu′ + ku = F (t),  (5.57)

where the coefficients m, γ and k are non-negative constants. Laplace transforms will be particularly useful in case F has discontinuities.

5.3 Discontinuities

5.11 Example. To solve

y′′ + 2y′ + 5y = F (t), y(0) = 1, y′(0) = 0,  (5.58)

where

F (t) = \begin{cases}1, & if 0 ⩽ t < π; \\ 0, & if π ⩽ t, \end{cases}  (5.59)

note that

ℒ {F } = ∫_{0}^{∞} e^{−st}F (t) d t  (5.60)
= ∫_{0}^{π} e^{−st} d t  (5.61)
= [−\frac{e^{−st}}{s} ]_{0}^{π}  (5.62)
= \frac{1 − e^{sπ}}{s} .  (5.63)

So we have

ℒ \{y′′\} + 2ℒ \{y′\} + 5ℒ \{y\} = \frac{1 − e^{sπ}}{s} ,  (5.64)
s^{2}Y − s + 2sY − 2 + 5Y = \frac{1 − e^{sπ}}{s} ,  (5.65)
(s^{2} + 2s + 5)Y = \frac{1 − e^{sπ}}{s} + s + 2,  (5.66)
(s^{2} + 2s + 5)Y = \frac{s^{2} + 2s + 1}{s} − \frac{e^{sπ}}{s} ,  (5.67)
Y = \frac{A}{s} + \frac{Bs + C}{s^{2} + 2s + 5} − e^{sπ} (\frac{D}{s} + \frac{Es + F}{s^{2} + 2s + 5}) ,  (5.68)
Y = \frac{A}{s} + \frac{Bs + C}{(s + 1)^{2} + 4} − e^{sπ} (\frac{D}{s} + \frac{Es + F}{(s + 1)^{2} + 4}) .  (5.69)

To complete the solution, we need the following:…

6 Systems of differential equations

We want to solve systems of the form

\begin{cases}x′_{1} = a_{1 1}x_{1} + a_{1 2}x_{2} + ⋯ + a_{1 n}x_{n} \\ x′_{2} = a_{2 1}x_{1} + a_{2 2}x_{2} + ⋯ + a_{2 n}x_{n} \\ … \\ x′_{n} = a_{n 1}x_{1} + a_{n 2}x_{2} + ⋯ + a_{n n}x_{n} \end{cases}  (6.1)

We can write this as the matrix equation

x′ = Ax,  (6.2)

where

x = [x_{1} ; ⋮ ; x_{n}] and A = [\begin{matrix}a_{1 1} & … & a_{1 n} \\ ⋮ & & ⋮ \\ a_{n 1} & … & a_{n n} \end{matrix}] .  (6.3)

In case n = 1, we have x′ = ax, with general solution ce^{at}; that is, all solutions are linear combinations of e^{at}. In the general case, we look for solutions like e^{at}. They should be vectors—say ξe^{rt}, that is,

[ξ_{1}e^{rt} ; ⋮ ; ξ_{n}e^{rt}] .  (6.4)

Plugging into (6.2) gives

rξe^{rt} = Aξe^{rt},  (6.5)
rξ = Aξ,  (6.6)
rIξ = Aξ,  (6.7)
(rI − A)ξ = 0,  (6.8)

where

I = [\begin{matrix}1 & 0 & … & 0 \\ 0 & 1 & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & 1 \end{matrix}] .  (6.9)

Here r is an eigenvalue of A, and ξ is a corresponding eigenvector. This means ξ ≠ 0, and:

(1) Multiplying ξ by the matrix A is the same as multiplying by a scalar (number);

(2) this scalar is r.

We find r first by solving

det(λI − A) = 0,  (6.10)

the characteristic equation of A. 6.1 Example. Let A = [\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix}]; the characteristic equation is

|\begin{matrix}λ − 1 & 0 \\ 0 & λ + 1 \end{matrix}| = 0,  (6.11)
λ^{2} − 1 = 0,  (6.12)
λ = ±1.  (6.13)

That is, A has eigenvalues ±1. For the eigenvectors corresponding to 1, we solve:

(A − I) [\begin{matrix}x \\ y \end{matrix}] = 0,  (6.14)
[\begin{matrix}0 & 0 \\ 0 & −2 \end{matrix}] [\begin{matrix}x \\ y \end{matrix}] = 0,  (6.15)
−2y = 0.  (6.16)

So the eigenvectors are the non-zero multiples of [\begin{matrix}1 \\ 0 \end{matrix}]. For the eigenvectors corresponding to −1, we solve:

(A + I) [\begin{matrix}x \\ y \end{matrix}] = 0,  (6.17)
[\begin{matrix}2 & 0 \\ 0 & 0 \end{matrix}] [\begin{matrix}x \\ y \end{matrix}] = 0,  (6.18)
2x = 0.  (6.19)

So the eigenvectors are the non-zero multiples of [\begin{matrix}0 \\ 1 \end{matrix}].

Hence the equation (6.2) has, in this case, the general solution

c_{0} [\begin{matrix}1 \\ 0 \end{matrix}] e^{t} + c_{1} [\begin{matrix}0 \\ 1 \end{matrix}] e^{−t}.  (6.20)

Consider the direction-field… 6.2 Example. A = [\begin{matrix}3 & −2 \\ 4 & −3 \end{matrix}]. Eigenvalues: ±1 again. Eigenvectors: (1, 1) and (1, 2).