Chapter 4 · Section 4.1

4.1 Basic Theory of Linear Differential Equations

So far we have solved first-order equations one type at a time. From now on the main characters are equations of order two and higher, and almost all of them will be linear. Linear equations are worth a whole theory of their own for two reasons. First, they appear everywhere in engineering: a mass on a spring, the charge in an RLC circuit, the deflection of a beam and the temperature in a cooling fin are all described by linear equations. Second, linear equations have a beautifully simple structure. Once you understand that structure, the solving methods of the next sections become almost obvious.

This section contains very little "solving". Its job is to answer three questions. When does a linear initial-value problem have exactly one solution? How many solutions do we need to build every solution? And how do we check that the solutions we found are really different from each other? The answers involve continuity, superposition and a determinant called the Wronskian.

The nth-order linear equation

Definition 4.1.1 (Linear equation of order n)

A linear differential equation of order $n$ in the unknown $y (x)$ is an equation that can be written as

a_{0} (x) y^{(n)} + a_{1} (x) y^{(n - 1)} + \dots + a_{n - 1} (x) y^{'} + a_{n} (x) y = F (x),

where the coefficients $a_{0}, a_{1}, \dots, a_{n}$ and the right-hand side $F$ depend on $x$ only, and $a_{0}$ is not identically zero. If $F (x) \equiv 0$ the equation is called homogeneous; otherwise it is nonhomogeneous.

"Linear" means that $y$ and its derivatives appear only to the first power, are never multiplied by each other, and never sit inside a function such as $sin y$ or $e^{y^{'}}$ . The coefficients, however, may be any functions of $x$ . For example,

x^{2} y^{''} - 3 x y^{'} - 5 y = 0 and y^{'''} + e^{x} y^{'} + (cos x) y = x^{4}

are linear, while $y^{''} + y^{2} = 0$ and $y y^{''} + y^{'} = 1$ are not.

A familiar engineering example is the series RLC circuit. If $q (t)$ is the charge on the capacitor, $L$ the inductance, $R$ the resistance, $C$ the capacitance and $E (t)$ the applied voltage, then

L q^{''} + R q^{'} + \frac{1}{C} q = E (t),

a second-order linear equation with constant coefficients. We will meet it again in Chapter 5.

Normalized form

On any interval where $a_{0} (x) \neq = 0$ we may divide the whole equation by $a_{0} (x)$ . The result has leading coefficient $1$ :

y^{(n)} + p_{1} (x) y^{(n - 1)} + \dots + p_{n - 1} (x) y^{'} + p_{n} (x) y = g (x), p_{k} = \frac{a _{k}}{a _{0}}, g = \frac{F}{a _{0}} .

We call this the normalized form. Several formulas in this chapter (reduction of order below, variation of parameters in Section 4.4) are written for the normalized form, so always divide by the leading coefficient before using them. Points where $a_{0} (x) = 0$ are called singular points of the equation; something unusual may happen there, and we will keep away from them for now.

Operator notation

Writing long equations again and again is tiring, so we introduce a shorthand. Let $D$ stand for "differentiate with respect to $x$ ":

Dy = y^{'}, D^{2} y = y^{''}, D^{k} y = y^{(k)} .

Then the left-hand side of the linear equation is the result of applying the operator

L = a_{0} (x) D^{n} + a_{1} (x) D^{n - 1} + \dots + a_{n - 1} (x) D + a_{n} (x)

to $y$ , and the equation is simply $L [y] = F (x)$ . For instance, $y^{''} - 4 y^{'} + 7 y$ can be written $(D^{2} - 4 D + 7) y$ . The most important property of $L$ is that it is linear: for any (sufficiently differentiable) functions $f, g$ and constants $c_{1}, c_{2}$ ,

L [c_{1} f + c_{2} g] = c_{1} L [f] + c_{2} L [g] .

This follows because the derivative of a sum is the sum of the derivatives and constants can be pulled out of a derivative. Almost every result in this section is a consequence of this one line. You will see the $D$ -notation again in Section 4.2 and in Chapter 7 for systems.

Existence and uniqueness

An $n$ th-order equation needs $n$ pieces of initial information. Think of a mass on a spring: to predict the motion you must know where the mass starts and how fast it is moving. For order $n$ we prescribe the value of $y$ and of its first $n - 1$ derivatives at one point $x_{0}$ .

Theorem 4.1.1 (Existence and uniqueness for linear IVPs)

Consider

a_{0} (x) y^{(n)} + a_{1} (x) y^{(n - 1)} + \dots + a_{n} (x) y = F (x) .

Suppose that on an interval $I$

the functions $a_{0}, a_{1}, \dots, a_{n}$ and $F$ are continuous, and
$a_{0} (x) \neq = 0$ for every $x$ in $I$ .

Let $x_{0}$ be any point of $I$ and let $c_{0}, c_{1}, \dots, c_{n - 1}$ be any real numbers. Then there is exactly one solution $y$ of the equation that satisfies

y (x_{0}) = c_{0}, y^{'} (x_{0}) = c_{1}, \dots, y^{(n - 1)} (x_{0}) = c_{n - 1},

and this solution is defined on the whole interval $I$ .

What is special here

For nonlinear first-order equations (Chapter 1) the existence theorem only promised a solution on some small interval around $x_{0}$ , and the solution could blow up nearby. For linear equations the guarantee is much stronger: the solution lives on the entire interval on which the coefficients are continuous and $a_{0}$ does not vanish. To find that interval, list the points where some coefficient fails to be continuous or where $a_{0} = 0$ , and take the largest open interval that contains $x_{0}$ and avoids all of them.

Example 4.1.1

Find the largest interval on which Theorem 4.1.1 guarantees a unique solution of

(x^{2} - 9) y^{''} + \frac{1}{x + 5} y^{'} + (ln x) y = cos x, y (2) = 1, y^{'} (2) = 0.

What changes if the initial conditions are given at $x_{0} = 4$ instead?

Solution

Coefficients. Here $a_{0} (x) = x^{2} - 9$ , $a_{1} (x) = \frac{1}{x + 5}$ , $a_{2} (x) = ln x$ and $F (x) = cos x$ .
Where are they continuous? $a_{0}$ and $F$ are continuous everywhere. $a_{1}$ is continuous except at $x = - 5$ . $a_{2} = ln x$ is defined and continuous only for $x > 0$ . So all four are continuous on $(0, \infty)$ .
Where is $a_{0} = 0$ ? $x^{2} - 9 = 0$ gives $x = 3$ or $x = - 3$ . Inside $(0, \infty)$ we must remove $x = 3$ .
Candidate intervals. The hypotheses hold on $(0, 3)$ and on $(3, \infty)$ .
Pick the one containing $x_{0}$ . Since $x_{0} = 2$ lies in $(0, 3)$ , the theorem guarantees exactly one solution on $(0, 3)$ .
Other starting point. If $x_{0} = 4$ , the relevant interval is $(3, \infty)$ .

Note that the theorem gives a guarantee on $(0, 3)$ . It does not claim that the solution must misbehave at $x = 3$ ; it simply says nothing there.

A special case of the theorem is used so often that it deserves its own name.

Theorem 4.1.2 (Only the zero solution starts at rest)

Let the hypotheses of Theorem 4.1.1 hold for the homogeneous equation $L [y] = 0$ on $I$ , and let $x_{0} \in I$ . If $y$ is a solution with

y (x_{0}) = y^{'} (x_{0}) = \dots = y^{(n - 1)} (x_{0}) = 0,

then $y (x) = 0$ for every $x$ in $I$ .

Why: the function $y \equiv 0$ obviously solves the homogeneous equation and has all these zero initial values. By uniqueness, it is the only such solution. Physically: a spring that starts at its rest position with zero velocity and is not pushed by any force simply stays at rest.

The superposition principle

For homogeneous linear equations, solutions can be added and scaled, and the result is again a solution.

Theorem 4.1.3 (Superposition for homogeneous equations)

If $y_{1}, y_{2}, \dots, y_{m}$ are solutions of the homogeneous linear equation $L [y] = 0$ on an interval $I$ , then every linear combination

y = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{m} y_{m} (c_{1}, \dots, c_{m} constants)

is also a solution on $I$ .

Proof. By linearity of $L$ ,

L [c_{1} y_{1} + \dots + c_{m} y_{m}] = c_{1} L [y_{1}] + \dots + c_{m} L [y_{m}] = c_{1} \cdot 0 + \dots + c_{m} \cdot 0 = 0.

Common mistake

Superposition in this form works only for homogeneous linear equations. If $y_{1}$ and $y_{2}$ both solve $L [y] = F$ with $F \neq = 0$ , then $L [y_{1} + y_{2}] = 2 F$ , so $y_{1} + y_{2}$ is not a solution of the same equation. And for a nonlinear equation such as $y^{''} = - y^{2}$ the sum of two solutions is in general not a solution at all.

Superposition tells us that solutions form a "space" closed under addition and scaling. The natural next question is: how many solutions do we need so that their combinations give all solutions? To answer it we need the idea of independence.

Linear dependence and independence

Definition 4.1.2 (Linear dependence and independence)

Functions $f_{1}, f_{2}, \dots, f_{n}$ are linearly dependent on an interval $I$ if there exist constants $c_{1}, c_{2}, \dots, c_{n}$ , not all zero, such that

c_{1} f_{1} (x) + c_{2} f_{2} (x) + \dots + c_{n} f_{n} (x) = 0 for every x \in I .

If the only constants that make this true for every $x \in I$ are $c_{1} = c_{2} = \dots = c_{n} = 0$ , the functions are linearly independent on $I$ .

Dependent means "at least one of the functions is redundant": it can be written as a combination of the others, so it adds nothing new. For two functions the test is especially simple: $f_{1}, f_{2}$ are dependent on $I$ exactly when one of them is a constant multiple of the other there. For instance, $sin 2 x$ and $sin x cos x$ are dependent (the first is twice the second), while $e^{x}$ and $e^{- x}$ are independent (their ratio $e^{2 x}$ is not constant).

Figure 4.1.1. Left:

sin 2 x

is exactly twice

sin x cos x

, so their ratio is the constant

2

and the two functions are dependent. Right: the ratio of

e^{x}

e^{- x}

e^{2 x}

, which changes with

x

. No constant multiple of one function gives the other, so they are independent.

Example 4.1.2

Show that $f_{1} (x) = 3 x^{2} - x$ , $f_{2} (x) = x + 4$ and $f_{3} (x) = 6 x^{2} + 8$ are linearly dependent on $(- \infty, \infty)$ .

Solution

Set up the condition. We look for constants, not all zero, with $c_{1} (3 x^{2} - x) + c_{2} (x + 4) + c_{3} (6 x^{2} + 8) = 0$ for all $x$ .
Collect powers of $x$ . $(3 c_{1} + 6 c_{3}) x^{2} + (- c_{1} + c_{2}) x + (4 c_{2} + 8 c_{3}) = 0.$ A polynomial is zero for every $x$ only if all its coefficients are zero.
Solve. From the $x$ term, $c_{2} = c_{1}$ . From the $x^{2}$ term, $c_{3} = - c_{1} /2$ . The constant term then gives $4 c_{1} + 8 (- c_{1} /2) = 0$ , which holds for every $c_{1}$ .
Choose a nonzero solution. Taking $c_{1} = 2$ gives $c_{2} = 2$ , $c_{3} = - 1$ . Check: $2 (3 x^{2} - x) + 2 (x + 4) - (6 x^{2} + 8) = 6 x^{2} - 2 x + 2 x + 8 - 6 x^{2} - 8 = 0.$ Since not all constants are zero, the functions are dependent. Equivalently, $f_{3} = 2 f_{1} + 2 f_{2}$ .

The Wronskian

Checking independence directly from the definition can be tedious. A determinant built from the functions and their derivatives often does the job in one line.

Definition 4.1.3 (Wronskian)

Let $f_{1}, \dots, f_{n}$ each have at least $n - 1$ derivatives on $I$ . Their Wronskian is the function

W (f_{1}, \dots, f_{n}) (x) = f_{1} f_{1}^{'} ⋮ f_{1}^{(n - 1)} f_{2} f_{2}^{'} ⋮ f_{2}^{(n - 1)} \dots \dots \dots f_{n} f_{n}^{'} ⋮ f_{n}^{(n - 1)} .

For two functions this is simply $W (f_{1}, f_{2}) = f_{1} f_{2}^{'} - f_{2} f_{1}^{'}$ .

Why a determinant? Suppose $c_{1} f_{1} + c_{2} f_{2} = 0$ on $I$ . Differentiating gives a second equation, $c_{1} f_{1}^{'} + c_{2} f_{2}^{'} = 0$ . At each $x$ this is a $2 \times 2$ linear system for $c_{1}, c_{2}$ :

(f_{1} (x) f_{1}^{'} (x) f_{2} (x) f_{2}^{'} (x)) (c_{1} c_{2}) = (00) .

If the determinant of this matrix is nonzero at even one point $x_{0}$ , the only solution at that point is $c_{1} = c_{2} = 0$ . So the functions must be independent. The same reasoning works for $n$ functions (differentiate $n - 1$ times). This proves the first test.

Theorem 4.1.4 (Wronskian test for arbitrary functions)

Let $f_{1}, \dots, f_{n}$ have $n - 1$ derivatives on $I$ . If $W (f_{1}, \dots, f_{n}) (x_{0}) \neq = 0$ for at least one point $x_{0} \in I$ , then $f_{1}, \dots, f_{n}$ are linearly independent on $I$ .

The converse is false for arbitrary functions

If the Wronskian of some arbitrary functions is identically zero, you may not conclude that they are dependent. Example 4.1.4 below shows two independent functions whose Wronskian is zero everywhere. The converse does hold, however, when the functions are solutions of the same homogeneous linear equation; that is Theorem 4.1.5.

Example 4.1.3

Compute $W (e^{- x}, e^{2 x}, e^{5 x})$ and decide whether the three functions are independent on $(- \infty, \infty)$ .

Solution

Write the determinant. Each row is the derivative of the row above: $W = e^{- x} - e^{- x} e^{- x} e^{2 x} 2 e^{2 x} 4 e^{2 x} e^{5 x} 5 e^{5 x} 25 e^{5 x} .$
Factor the exponentials out of the columns. Column 1 has a common factor $e^{- x}$ , column 2 has $e^{2 x}$ , column 3 has $e^{5 x}$ . Their product is $e^{- x + 2 x + 5 x} = e^{6 x}$ : $W = e^{6 x} 1 - 1 1 1241525 .$
Evaluate the numerical determinant by expanding along the first row: $1 \cdot (2 \cdot 25 - 5 \cdot 4) - 1 \cdot ((- 1) \cdot 25 - 5 \cdot 1) + 1 \cdot ((- 1) \cdot 4 - 2 \cdot 1) = 30 + 30 - 6 = 54.$
Conclusion. $W = 54 e^{6 x}$ , which is never zero. By Theorem 4.1.4 the functions are linearly independent on $(- \infty, \infty)$ .

Use the calculator below to compute Wronskians of your own sets of two or three functions. Try, for example, adding a function that is a combination of the others and watch the Wronskian collapse to zero.

Example 4.1.4 (A zero Wronskian does not imply dependence)

Let $f (x) = x^{4}$ and $g (x) = x^{3} ∣ x ∣$ on $I = (- 1, 1)$ . Show that $W (f, g) = 0$ for every $x \in I$ , yet $f$ and $g$ are linearly independent on $I$ .

Solution

Understand $g$ . For $x \geq 0$ , $g (x) = x^{4}$ ; for $x < 0$ , $g (x) = - x^{4}$ .
Derivative of $g$ . For $x > 0$ , $g^{'} (x) = 4 x^{3}$ ; for $x < 0$ , $g^{'} (x) = - 4 x^{3}$ . Both can be written as $g^{'} (x) = 4 x^{2} ∣ x ∣$ . At $x = 0$ the difference quotient is $g (h) / h = h^{2} ∣ h ∣ \to 0$ , so $g^{'} (0) = 0 = 4 \cdot 0^{2} \cdot ∣0∣$ as well. Thus $g^{'} (x) = 4 x^{2} ∣ x ∣$ on all of $I$ .
Wronskian. $W (f, g) = f g^{'} - g f^{'} = x^{4} \cdot 4 x^{2} ∣ x ∣ - x^{3} ∣ x ∣ \cdot 4 x^{3} = 4 x^{6} ∣ x ∣ - 4 x^{6} ∣ x ∣ = 0 for every x \in I .$
Independence. Suppose $c_{1} x^{4} + c_{2} x^{3} ∣ x ∣ = 0$ for all $x \in I$ . At $x = \frac{1}{2}$ : $\frac{1}{16} (c_{1} + c_{2}) = 0$ . At $x = - \frac{1}{2}$ : $\frac{1}{16} (c_{1} - c_{2}) = 0$ . Adding and subtracting gives $c_{1} = c_{2} = 0$ . So $f$ and $g$ are independent.
No contradiction. On $(0, 1)$ alone the two functions are equal, and on $(- 1, 0)$ alone one is the negative of the other. Over the whole interval, however, no single constant ratio works. Theorem 4.1.5 below cannot be applied because $f$ and $g$ are not solutions of a common equation $y^{''} + p (x) y^{'} + q (x) y = 0$ with $p, q$ continuous on $I$ . (Indeed, both satisfy $y (0) = y^{'} (0) = 0$ , so by Theorem 4.1.2 such solutions would have to be identically zero.)

Figure 4.1.2. The functions of Example 4.1.4. On

(0, 1)

the two graphs coincide; on

(- 1, 0)

one is the mirror image of the other in the

x

-axis. On each half the ratio is constant, which is why the Wronskian vanishes everywhere, but no single constant works on the whole interval

(- 1, 1)

For solutions of a linear homogeneous equation the situation is much cleaner.

Theorem 4.1.5 (Wronskian of solutions)

Let $y_{1}, \dots, y_{n}$ be solutions of the homogeneous equation $L [y] = 0$ of order $n$ on an interval $I$ , where $a_{0}, \dots, a_{n}$ are continuous and $a_{0} (x) \neq = 0$ on $I$ . Then:

Either $W (y_{1}, \dots, y_{n}) (x) = 0$ for every $x \in I$ , or $W (y_{1}, \dots, y_{n}) (x) \neq = 0$ for every $x \in I$ . There is no middle ground.
$y_{1}, \dots, y_{n}$ are linearly independent on $I$ if and only if $W (y_{1}, \dots, y_{n}) (x_{0}) \neq = 0$ at some (and hence every) point $x_{0} \in I$ .

Why the converse now holds (sketch for n = 2)

Suppose $W (y_{1}, y_{2}) (x_{0}) = 0$ . Then the system $c_{1} y_{1} (x_{0}) + c_{2} y_{2} (x_{0}) = 0$ , $c_{1} y_{1}^{'} (x_{0}) + c_{2} y_{2}^{'} (x_{0}) = 0$ has a solution with $c_{1}, c_{2}$ not both zero. The function $y = c_{1} y_{1} + c_{2} y_{2}$ solves the homogeneous equation (superposition) and has $y (x_{0}) = y^{'} (x_{0}) = 0$ . By Theorem 4.1.2, $y \equiv 0$ on $I$ . So $y_{1}, y_{2}$ are dependent. The uniqueness theorem is what makes the converse true, and it is exactly what is missing in Example 4.1.4.

In practice this means: to check that $n$ solutions are independent, compute their Wronskian at one convenient point (often $x_{0} = 0$ or $x_{0} = 1$ ).

Fundamental sets and the general solution

Definition 4.1.4 (Fundamental set)

A set of $n$ linearly independent solutions $y_{1}, \dots, y_{n}$ of an $n$ th-order homogeneous linear equation on $I$ is called a fundamental set of solutions on $I$ .

Theorem 4.1.6 (General solution of the homogeneous equation)

Under the hypotheses of Theorem 4.1.1 (with $F \equiv 0$ ):

the homogeneous equation $L [y] = 0$ of order $n$ always has a fundamental set of $n$ solutions on $I$ ;
if $y_{1}, \dots, y_{n}$ is any fundamental set, then every solution on $I$ can be written as $y = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n}$ for suitable constants $c_{1}, \dots, c_{n}$ . This expression is called the general solution.

Why every solution is covered. Let $φ$ be any solution and pick $x_{0} \in I$ . We want constants with $c_{1} y_{1} + \dots + c_{n} y_{n}$ having the same initial data as $φ$ at $x_{0}$ . Writing this out gives $n$ linear equations:

c_{1} y_{1} (x_{0}) + \dots + c_{n} y_{n} (x_{0}) c_{1} y_{1}^{'} (x_{0}) + \dots + c_{n} y_{n}^{'} (x_{0}) c_{1} y_{1}^{(n - 1)} (x_{0}) + \dots + c_{n} y_{n}^{(n - 1)} (x_{0}) = φ (x_{0}), = φ^{'} (x_{0}), ⋮ = φ^{(n - 1)} (x_{0}) .

The determinant of this system is precisely $W (y_{1}, \dots, y_{n}) (x_{0})$ , which is nonzero because the $y_{k}$ are independent. So the system has a solution $c_{1}, \dots, c_{n}$ . The combination and $φ$ then solve the same initial-value problem, so by uniqueness they are the same function. This argument is also exactly how we solve initial-value problems in practice: find the general solution, then solve a linear system for the constants.

The message is important enough to repeat: an $n$ th-order linear homogeneous equation has exactly $n$ "building blocks". Find $n$ independent solutions and you have them all.

Example 4.1.5

Consider $x^{2} y^{''} - 3 x y^{'} - 5 y = 0$ on $I = (0, \infty)$ .

Verify that $y_{1} = x^{5}$ and $y_{2} = x^{- 1}$ are solutions.
Show that they form a fundamental set on $I$ and write the general solution.
Solve the initial-value problem with $y (1) = 2$ , $y^{'} (1) = 4$ .

Solution

Hypotheses. In normalized form the equation is $y^{''} - \frac{3}{x} y^{'} - \frac{5}{x ^{2}} y = 0$ , whose coefficients are continuous on $(0, \infty)$ . So Theorems 4.1.1–4.1.6 apply on $I$ .
Check $y_{1} = x^{5}$ . $y_{1}^{'} = 5 x^{4}$ , $y_{1}^{''} = 20 x^{3}$ . Then $x^{2} (20 x^{3}) - 3 x (5 x^{4}) - 5 x^{5} = 20 x^{5} - 15 x^{5} - 5 x^{5} = 0.$
Check $y_{2} = x^{- 1}$ . $y_{2}^{'} = - x^{- 2}$ , $y_{2}^{''} = 2 x^{- 3}$ . Then $x^{2} (2 x^{- 3}) - 3 x (- x^{- 2}) - 5 x^{- 1} = 2 x^{- 1} + 3 x^{- 1} - 5 x^{- 1} = 0.$
Wronskian. $W (x^{5}, x^{- 1}) = x^{5} \cdot (- x^{- 2}) - x^{- 1} \cdot 5 x^{4} = - x^{3} - 5 x^{3} = - 6 x^{3} .$ This is nonzero for every $x > 0$ , so ${x^{5}, x^{- 1}}$ is a fundamental set on $I$ and the general solution is $y = c_{1} x^{5} + c_{2} x^{- 1}$ .
Initial conditions. $y^{'} = 5 c_{1} x^{4} - c_{2} x^{- 2}$ . At $x = 1$ : $c_{1} + c_{2} 5 c_{1} - c_{2} = 2, = 4.$ Adding the equations gives $6 c_{1} = 6$ , so $c_{1} = 1$ and then $c_{2} = 1$ .
Answer. $y = x^{5} + x^{- 1}$ on $(0, \infty)$ .

Figure 4.1.3. The fundamental set

{x^{5}, x^{- 1}}

of Example 4.1.5 and the solution of the initial-value problem,

y = x^{5} + x^{- 1}

. It is the only solution that passes through

(1, 2)

with slope

4

(short segment).

Reduction of order

Suppose we have managed to find one nonzero solution $y_{1}$ of a second-order homogeneous equation, perhaps by a lucky guess. Theorem 4.1.6 says we need a second, independent one. There is a neat trick that produces it: look for the second solution in the form $y = v (x) y_{1} (x)$ , where $v$ is an unknown function. (If $v$ were a constant we would only get multiples of $y_{1}$ , so $v$ must genuinely vary.) The method is called reduction of order because it turns a second-order equation into a first-order one.

Start from the normalized equation

y^{''} + P (x) y^{'} + Q (x) y = 0,

and assume $y_{1}$ is a known solution with $y_{1} (x) \neq = 0$ on $I$ .

Substitute $y = v y_{1}$ . By the product rule, $y y^{'} y^{''} = v y_{1}, = v^{'} y_{1} + v y_{1}^{'}, = v^{''} y_{1} + 2 v^{'} y_{1}^{'} + v y_{1}^{''} .$
Plug into the equation and group the terms by $v^{''}$ , $v^{'}$ and $v$ : $v^{''} y_{1} + v^{'} (2 y_{1}^{'} + P y_{1}) + v (y_{1}^{''} + P y_{1}^{'} + Q y_{1}) = 0.$
Use that $y_{1}$ is a solution. The bracket multiplying $v$ is exactly $L [y_{1}] = 0$ . The $v$ term disappears: $y_{1} v^{''} + (2 y_{1}^{'} + P y_{1}) v^{'} = 0.$
Reduce the order. Put $w = v^{'}$ . Then $y_{1} w^{'} + (2 y_{1}^{'} + P y_{1}) w = 0$ , a first-order equation for $w$ .
Separate variables. Divide by $y_{1} w$ : $\frac{w ^{'}}{w} = - 2 \frac{y _{1}^{'}}{y _{1}} - P .$
Integrate. $ln ∣ w ∣ = - 2 ln ∣ y_{1} ∣ - \int P d x + const .$ We need only one $w$ , so we drop the constant (take it zero) and exponentiate: $w = \frac{e ^{- \int P d x}}{y _{1}^{2}} .$
Integrate once more to get $v = \int w d x$ , and multiply by $y_{1}$ .

Theorem 4.1.7 (Reduction of order formula)

If $y_{1}$ is a solution of $y^{''} + P (x) y^{'} + Q (x) y = 0$ that does not vanish on $I$ (and $P$ , $Q$ are continuous on $I$ ), then

y_{2} (x) = y_{1} (x) \int \frac{e ^{- \int P (x) d x}}{y _{1} ( x ) ^{2}} d x

is a second solution, and ${y_{1}, y_{2}}$ is a fundamental set on $I$ .

Why independent? With $y_{2} = v y_{1}$ we get $W (y_{1}, y_{2}) = y_{1} (v^{'} y_{1} + v y_{1}^{'}) - v y_{1} y_{1}^{'} = y_{1}^{2} v^{'} = e^{- \int P d x}$ , an exponential, which is never zero.

Method in a nutshell: reduction of order

Write the equation in normalized form $y^{''} + P y^{'} + Q y = 0$ and read off $P (x)$ .
Compute $\int P d x$ (no constant needed) and then $e^{- \int P d x}$ .
Divide by $y_{1}^{2}$ and integrate (again no constant needed): this is $v$ .
Set $y_{2} = v y_{1}$ and drop any constant factor. The general solution is $y = c_{1} y_{1} + c_{2} y_{2}$ .

Common mistake

Forgetting to normalize. In $x y^{''} + \dots$ the coefficient $P$ is the coefficient of $y^{'}$ after dividing by $x$ . Using the coefficient from the original equation gives a wrong $y_{2}$ . You can always catch this error by substituting your $y_{2}$ back into the equation.

Example 4.1.6

Verify that $y_{1} = e^{x}$ solves $x y^{''} - (2 x + 1) y^{'} + (x + 1) y = 0$ for $x > 0$ , and find the general solution.

Solution

Check $y_{1}$ . All derivatives of $e^{x}$ equal $e^{x}$ , so the left side is $e^{x} [x - (2 x + 1) + (x + 1)] = e^{x} \cdot 0 = 0$ .
Normalize. Dividing by $x$ : $y^{''} - \frac{2 x + 1}{x} y^{'} + \frac{x + 1}{x} y = 0, P (x) = - \frac{2 x + 1}{x} = - 2 - \frac{1}{x} .$
Integrate $P$ . $\int P d x = - 2 x - ln x$ (we are on $x > 0$ ).
Exponentiate. $e^{- \int P d x} = e^{2 x + l n x} = e^{2 x} \cdot e^{l n x} = x e^{2 x}$ .
Divide by $y_{1}^{2} = e^{2 x}$ . $v^{'} = \frac{x e ^{2 x}}{e ^{2 x}} = x$ .
Integrate. $v = \frac{x ^{2}}{2}$ .
Second solution. $y_{2} = v y_{1} = \frac{1}{2} x^{2} e^{x}$ . Dropping the factor $\frac{1}{2}$ , we take $y_{2} = x^{2} e^{x}$ .
General solution. $y = c_{1} e^{x} + c_{2} x^{2} e^{x}$ for $x > 0$ .

Check. $y_{2}^{'} = (x^{2} + 2 x) e^{x}$ and $y_{2}^{''} = (x^{2} + 4 x + 2) e^{x}$ . Then $x (x^{2} + 4 x + 2) - (2 x + 1) (x^{2} + 2 x) + (x + 1) x^{2} = (x^{3} + 4 x^{2} + 2 x) - (2 x^{3} + 5 x^{2} + 2 x) + (x^{3} + x^{2}) = 0$ . ✓

Nonhomogeneous equations: y = y_c + y_p

Now let $F (x) \neq = 0$ . The homogeneous equation $L [y] = 0$ obtained by replacing $F$ with zero is called the associated homogeneous equation (or complementary equation). Its general solution is called the complementary function and denoted $y_{c}$ . Any single solution of the full equation, with no arbitrary constants, is called a particular solution $y_{p}$ .

Theorem 4.1.8 (Structure of the general solution)

Assume the hypotheses of Theorem 4.1.1 on $I$ . Let $y_{p}$ be any particular solution of $L [y] = F$ on $I$ , and let $y_{c} = c_{1} y_{1} + \dots + c_{n} y_{n}$ be the general solution of $L [y] = 0$ on $I$ . Then every solution of $L [y] = F$ on $I$ has the form

y = y_{c} + y_{p} = c_{1} y_{1} + \dots + c_{n} y_{n} + y_{p} .

Why. If $y$ is any solution, then $L [y - y_{p}] = L [y] - L [y_{p}] = F - F = 0$ . So $y - y_{p}$ solves the homogeneous equation, and by Theorem 4.1.6 it equals some $c_{1} y_{1} + \dots + c_{n} y_{n}$ .

Theorem 4.1.8 splits every nonhomogeneous problem into two independent jobs: (1) solve the homogeneous equation to get $y_{c}$ ; (2) find one particular solution $y_{p}$ , by any method at all, even guessing. The next one is a very handy tool for job (2).

Theorem 4.1.9 (Superposition for the right-hand side)

If $y_{p 1}$ solves $L [y] = F_{1}$ and $y_{p 2}$ solves $L [y] = F_{2}$ , then for any constants $k_{1}, k_{2}$ the function $k_{1} y_{p 1} + k_{2} y_{p 2}$ solves

L [y] = k_{1} F_{1} + k_{2} F_{2} .

The proof is again one line of linearity: $L [k_{1} y_{p 1} + k_{2} y_{p 2}] = k_{1} F_{1} + k_{2} F_{2}$ . In engineering language: the response to a sum of inputs is the sum of the responses. This is why we can treat each term of a complicated forcing function separately.

Common mistake

When solving an initial-value problem for a nonhomogeneous equation, apply the initial conditions to the complete solution $y = y_{c} + y_{p}$ , never to $y_{c}$ alone. The particular solution contributes to $y (x_{0})$ and $y^{'} (x_{0})$ too.

Example 4.1.7

Let $L [y] = x^{2} y^{''} - 3 x y^{'} - 5 y$ on $(0, \infty)$ . Compute $L [x^{2}]$ and $L [x]$ , and use them to find the general solution of

x^{2} y^{''} - 3 x y^{'} - 5 y = 9 x^{2} + 24 x .

Solution

Compute $L [x^{2}]$ . With $y = x^{2}$ : $y^{'} = 2 x$ , $y^{''} = 2$ , so $L [x^{2}] = 2 x^{2} - 6 x^{2} - 5 x^{2} = - 9 x^{2}$ .
Compute $L [x]$ . With $y = x$ : $y^{'} = 1$ , $y^{''} = 0$ , so $L [x] = 0 - 3 x - 5 x = - 8 x$ .
Match each piece of the right-hand side. Since $L [x^{2}] = - 9 x^{2}$ , we have $L [- x^{2}] = 9 x^{2}$ . Since $L [x] = - 8 x$ , we have $L [- 3 x] = 24 x$ .
Superpose. By Theorem 4.1.9, $y_{p} = - x^{2} - 3 x$ satisfies $L [y_{p}] = 9 x^{2} + 24 x$ .
Complementary function. From Example 4.1.5, $y_{c} = c_{1} x^{5} + c_{2} x^{- 1}$ .
General solution. By Theorem 4.1.8, $y = c_{1} x^{5} + c_{2} x^{- 1} - x^{2} - 3 x, x > 0.$

Summary

For a linear equation with continuous coefficients and $a_{0} \neq = 0$ on $I$ , every initial-value problem has exactly one solution, defined on all of $I$ .
Linear combinations of solutions of a homogeneous linear equation are solutions.
An $n$ th-order homogeneous linear equation has $n$ independent solutions, and their combinations give every solution.
For solutions, independence is checked by evaluating the Wronskian at one point. For arbitrary functions, a nonzero Wronskian proves independence, but a zero Wronskian proves nothing.
One known solution of a second-order equation gives a second one by reduction of order.
Nonhomogeneous: $y = y_{c} + y_{p}$ , and right-hand sides can be split into pieces.

Exercises

Find the largest interval on which Theorem 4.1.1 guarantees a unique solution of $(x - 5) y^{''} + \frac{1}{x + 1} y^{'} + y = ln (x + 3)$ , $y (0) = 1$ , $y^{'} (0) = 2$ .
Answer
$(- 1, 5)$
Consider $(x^{2} - 7 x + 10) y^{''} + e^{x} y^{'} + y = 0$ with $y (3) = 0$ , $y^{'} (3) = 0$ . On which interval does Theorem 4.1.1 apply, and what is the solution there?
Answer
The interval is $(2, 5)$ ; by Theorem 4.1.2 the solution is $y \equiv 0$ on $(2, 5)$ .
Show that $x^{2} + 2 x$ , $3 x - 1$ and $2 x^{2} + x + 1$ are linearly dependent on $(- \infty, \infty)$ by finding a nontrivial relation.
Answer
$2 (x^{2} + 2 x) - (3 x - 1) - (2 x^{2} + x + 1) = 0$ .
Compute $W (1, e^{- x}, e^{4 x})$ . Are the functions independent?
Answer
$W = - 20 e^{3 x} \neq = 0$ ; independent on $(- \infty, \infty)$ .
Compute $W (e^{2 x} cos 3 x, e^{2 x} sin 3 x)$ .
Answer
$W = 3 e^{4 x}$
Compute $W (x, x^{2}, x^{3})$ and explain why $x, x^{2}, x^{3}$ are independent on every interval, even one containing $0$ .
Answer
$W = 2 x^{3}$ . Every interval contains a point $x_{0} \neq = 0$ where $W (x_{0}) \neq = 0$ , so Theorem 4.1.4 gives independence.
Show that $y_{1} = x^{2}$ and $y_{2} = x^{- 4}$ form a fundamental set of $x^{2} y^{''} + 3 x y^{'} - 8 y = 0$ on $(0, \infty)$ , and solve the IVP $y (1) = 3$ , $y^{'} (1) = 0$ .
Answer
$W = - 6 x^{- 3} \neq = 0$ on $(0, \infty)$ ; $y = 2 x^{2} + x^{- 4}$ .
Given that $y_{1} = x^{- 3}$ solves $x^{2} y^{''} + 7 x y^{'} + 9 y = 0$ for $x > 0$ , use reduction of order to find the general solution.
Answer
$y = c_{1} x^{- 3} + c_{2} x^{- 3} ln x$
Given that $y_{1} = e^{3 x}$ solves $(3 x + 5) y^{''} - 9 (x + 2) y^{'} + 9 y = 0$ for $x > - \frac{5}{3}$ , find a second, independent solution. (Hint: $\frac{x + 2}{3 x + 5} = \frac{1}{3} + \frac{1}{3 ( 3 x + 5 )}$ .)
Answer
$y_{2} = x + 2$ (reduction of order gives $v = - (x + 2) e^{- 3 x}$ , so $v y_{1} = - (x + 2)$ ); general solution $y = c_{1} e^{3 x} + c_{2} (x + 2)$ .
Let $L [y] = x^{2} y^{''} + 3 x y^{'} - 8 y$ on $(0, \infty)$ . Compute $L [x]$ and $L [1]$ , and then find the general solution of $L [y] = 10 x - 4$ . (Use Exercise 7.)
Answer
$L [x] = - 5 x$ , $L [1] = - 8$ ; $y = c_{1} x^{2} + c_{2} x^{- 4} - 2 x + \frac{1}{2}$ .
Explain why $y = x^{3}$ cannot be a solution of any equation $y^{''} + p (x) y^{'} + q (x) y = 0$ with $p$ and $q$ continuous on an interval containing $0$ .
Answer
$x^{3}$ has $y (0) = 0$ and $y^{'} (0) = 0$ . By Theorem 4.1.2 such a solution must be identically zero on the interval, but $x^{3}$ is not.
True or false: "If $W (f, g) (x) = 0$ for every $x$ in $I$ , then $f$ and $g$ are linearly dependent on $I$ ." Explain, and state an extra hypothesis that makes the statement true.
Answer
False in general (Example 4.1.4: $x^{4}$ and $x^{3} ∣ x ∣$ on $(- 1, 1)$ ). It becomes true if $f$ and $g$ are solutions of the same equation $y^{''} + p y^{'} + q y = 0$ with $p, q$ continuous on $I$ (Theorem 4.1.5).

The nth-order linear equation

Normalized form

Operator notation

Existence and uniqueness

The superposition principle

Linear dependence and independence

The Wronskian

Fundamental sets and the general solution

Reduction of order

Nonhomogeneous equations: y = yc + yp

Summary

Exercises

Nonhomogeneous equations: y = y_c + y_p