Chapter 4 · Section 4.6

4.6 Theorems on Second-Order Homogeneous Linear Equations

Since Section 4.1 we have relied on a handful of facts. A second-order homogeneous linear equation has two independent solutions. The Wronskian of two solutions tells us whether they are independent. Every solution is a combination $c_{1} y_{1} + c_{2} y_{2}$ . We used these facts constantly in Sections 4.2–4.5, but we did not prove them.

This section supplies the proofs for second-order equations. They are short, and each step uses only the existence–uniqueness theorem, the product rule and a little algebra of $2 \times 2$ systems. Reading them carefully pays off. You will see why the hypotheses (continuity, a nonzero leading coefficient, an interval) matter, and you will be able to spot a wrong claim when you meet one. The same arguments carry over to order $n$ with $n \times n$ determinants.

Setting and standing assumptions

Throughout, we study

a_{0} (x) y^{''} + a_{1} (x) y^{'} + a_{2} (x) y = 0, (H)

where $a_{0}, a_{1}, a_{2}$ are continuous on an open interval $I$ and $a_{0} (x) \neq = 0$ for every $x \in I$ . Dividing by $a_{0}$ gives the normalized form

y^{''} + p (x) y^{'} + q (x) y = 0, p = \frac{a _{1}}{a _{0}}, q = \frac{a _{2}}{a _{0}} .

Because $a_{0}$ never vanishes on $I$ , the functions $p$ and $q$ are continuous on $I$ . Both forms have exactly the same solutions on $I$ , so we may use whichever is convenient. A solution on $I$ means a twice-differentiable function that satisfies the equation at every point of $I$ .

Definition 4.6.1 (Linear dependence and the Wronskian)

Two functions $f_{1}, f_{2}$ defined on $I$ are linearly dependent on $I$ if there are constants $c_{1}, c_{2}$ , not both zero, such that $c_{1} f_{1} (x) + c_{2} f_{2} (x) = 0$ for every $x \in I$ . Otherwise they are linearly independent on $I$ . For differentiable $f_{1}, f_{2}$ , the Wronskian is

W (f_{1}, f_{2}) (x) = f_{1} (x) f_{1}^{'} (x) f_{2} (x) f_{2}^{'} (x) = f_{1} (x) f_{2}^{'} (x) - f_{1}^{'} (x) f_{2} (x) .

Definition 4.6.2 (Fundamental set)

A fundamental set of solutions of (H) on $I$ is a pair $y_{1}, y_{2}$ of solutions of (H) on $I$ that are linearly independent on $I$ .

The two tools we need

The first tool is the existence–uniqueness theorem of Section 4.1, specialized to (H). We do not prove it here; one proof applies the successive approximations of Section 8.3 to the equivalent first-order system for $(y, y^{'})$ .

Theorem 4.6.1 (Existence and uniqueness for (H))

Assume the standing assumptions. Let $x_{0} \in I$ and let $K_{0}, K_{1}$ be any real numbers. Then:

There is exactly one solution $y$ of (H) with $y (x_{0}) = K_{0}$ and $y^{'} (x_{0}) = K_{1}$ , and it is defined on the whole interval $I$ .
In particular, if a solution satisfies $y (x_{0}) = 0$ and $y^{'} (x_{0}) = 0$ , then $y (x) = 0$ for all $x \in I$ .

Part 2 follows from Part 1: the zero function is a solution with these initial values, and there is only one such solution.

The second tool is a fact about two linear equations in two unknowns.

Theorem 4.6.2 (2×2 linear systems)

Consider the system

α_{11} c_{1} + α_{12} c_{2} α_{21} c_{1} + α_{22} c_{2} = β_{1}, = β_{2}, Δ = α_{11} α_{22} - α_{12} α_{21} .

If $Δ \neq = 0$ , the system has exactly one solution $(c_{1}, c_{2})$ for every right-hand side $(β_{1}, β_{2})$ .
The homogeneous system ( $β_{1} = β_{2} = 0$ ) has a solution with $c_{1}, c_{2}$ not both zero if and only if $Δ = 0$ .

Why Theorem 4.6.2 is true

Part 1 is Cramer's rule: $c_{1} = (β_{1} α_{22} - α_{12} β_{2}) /Δ$ and $c_{2} = (α_{11} β_{2} - β_{1} α_{21}) /Δ$ , and substitution shows this is the only solution. For Part 2: if $Δ \neq = 0$ , Part 1 says the only solution of the homogeneous system is $(0, 0)$ . If $Δ = 0$ , we can write down a nonzero solution. If the first row is not all zero, take $(c_{1}, c_{2}) = (α_{12}, - α_{11})$ . The first equation gives $α_{11} α_{12} - α_{12} α_{11} = 0$ , and the second gives $α_{21} α_{12} - α_{22} α_{11} = - Δ = 0$ . If the first row is all zero, take $(α_{22}, - α_{21})$ instead, or $(1, 0)$ if every coefficient is zero.

Superposition

Theorem 4.6.3 (Superposition)

If $y_{1}$ and $y_{2}$ are solutions of (H) on $I$ , then $c_{1} y_{1} + c_{2} y_{2}$ is a solution of (H) on $I$ for any constants $c_{1}, c_{2}$ .

Proof. Let $y = c_{1} y_{1} + c_{2} y_{2}$ . Differentiating term by term, $y^{'} = c_{1} y_{1}^{'} + c_{2} y_{2}^{'}$ and $y^{''} = c_{1} y_{1}^{''} + c_{2} y_{2}^{''}$ . Substitute into the left side of (H) and group by $c_{1}$ and $c_{2}$ :

a_{0} y^{''} + a_{1} y^{'} + a_{2} y = a_{0} (c_{1} y_{1}^{''} + c_{2} y_{2}^{''}) + a_{1} (c_{1} y_{1}^{'} + c_{2} y_{2}^{'}) + a_{2} (c_{1} y_{1} + c_{2} y_{2}) = c_{1} (a_{0} y_{1}^{''} + a_{1} y_{1}^{'} + a_{2} y_{1}) + c_{2} (a_{0} y_{2}^{''} + a_{1} y_{2}^{'} + a_{2} y_{2}) = c_{1} \cdot 0 + c_{2} \cdot 0 = 0.

This holds at every $x \in I$ , so $y$ is a solution. $■$

A fundamental set always exists

Theorem 4.6.4 (Existence of a fundamental set)

Under the standing assumptions, (H) has a fundamental set of solutions on $I$ .

Proof. We build two solutions explicitly and then show they are independent.

Construction. Fix a point $x_{0} \in I$ . By Theorem 4.6.1 there is a solution $y_{1}$ on $I$ with $y_{1} (x_{0}) = 1, y_{1}^{'} (x_{0}) = 0,$ and a solution $y_{2}$ on $I$ with $y_{2} (x_{0}) = 0, y_{2}^{'} (x_{0}) = 1.$
Suppose they were dependent. Then there would be constants $c_{1}, c_{2}$ , not both zero, with $c_{1} y_{1} (x) + c_{2} y_{2} (x) = 0$ for all $x \in I$ .
Differentiate the identity. Because it holds for all $x$ in an interval, we may differentiate it: $c_{1} y_{1}^{'} (x) + c_{2} y_{2}^{'} (x) = 0$ for all $x \in I$ .
Evaluate at $x_{0}$ . Using the initial values, $c_{1} \cdot 1 + c_{2} \cdot 0 = 0, c_{1} \cdot 0 + c_{2} \cdot 1 = 0,$ so $c_{1} = 0$ and $c_{2} = 0$ . This contradicts "not both zero".
Conclusion. $y_{1}, y_{2}$ are linearly independent on $I$ , so they form a fundamental set. $■$

The pair built in this proof is called the normalized fundamental set at $x_{0}$ . It is handy because the solution with $y (x_{0}) = K_{0}$ , $y^{'} (x_{0}) = K_{1}$ is simply $K_{0} y_{1} + K_{1} y_{2}$ (Example 4.6.4).

Abel's formula

The next result is the heart of the section. It says that the Wronskian of two solutions satisfies a simple first-order equation, so we can find it without solving (H).

Theorem 4.6.5 (Abel's formula)

Let $y_{1}, y_{2}$ be any two solutions of $y^{''} + p (x) y^{'} + q (x) y = 0$ on $I$ , where $p, q$ are continuous on $I$ , and let $W = W (y_{1}, y_{2})$ . Then for any fixed $x_{0} \in I$ ,

W (x) = W (x_{0}) exp (- \int_{x_{0}}^{x} p (s) d s), x \in I .

In terms of the original coefficients, $p = a_{1} / a_{0}$ .

Proof.

Differentiate $W$ . With $W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ and the product rule, $W^{'} = (y_{1}^{'} y_{2}^{'} + y_{1} y_{2}^{''}) - (y_{1}^{''} y_{2} + y_{1}^{'} y_{2}^{'}) = y_{1} y_{2}^{''} - y_{1}^{''} y_{2} .$ The two terms $y_{1}^{'} y_{2}^{'}$ cancel.
Use the equation. Each $y_{i}$ solves the equation, so $y_{i}^{''} = - p y_{i}^{'} - q y_{i}$ for $i = 1, 2$ . Substitute: $W^{'} = y_{1} (- p y_{2}^{'} - q y_{2}) - (- p y_{1}^{'} - q y_{1}) y_{2} = - p y_{1} y_{2}^{'} - q y_{1} y_{2} + p y_{1}^{'} y_{2} + q y_{1} y_{2} = - p (y_{1} y_{2}^{'} - y_{1}^{'} y_{2}) .$ The $q$ -terms cancel.
A first-order equation for $W$ . The last line says $W^{'} + p (x) W = 0.$ This is a first-order linear equation (Section 2.3).
Integrating factor. Let $μ (x) = exp (\int_{x_{0}}^{x} p (s) d s)$ . Then $μ^{'} = p μ$ , $μ (x_{0}) = 1$ , and $μ (x) > 0$ . By the product rule, $(μ W)^{'} = μ W^{'} + μ^{'} W = μ (W^{'} + p W) = 0.$
Conclude. A function with zero derivative on an interval is constant, so $μ (x) W (x) = μ (x_{0}) W (x_{0}) = W (x_{0})$ . Dividing by $μ (x)$ gives the formula. $■$

Notice that the proof never used independence of $y_{1}$ and $y_{2}$ . Abel's formula holds for any two solutions.

Theorem 4.6.6 (The Wronskian is identically zero or never zero)

If $y_{1}, y_{2}$ are solutions of (H) on $I$ , then either $W (y_{1}, y_{2}) (x) = 0$ for every $x \in I$ , or $W (y_{1}, y_{2}) (x) \neq = 0$ for every $x \in I$ .

Proof. In Abel's formula the exponential factor is never zero. So if $W (x_{0}) = 0$ at one point, then $W (x) = 0$ at every point. If $W (x_{0}) \neq = 0$ , then $W (x)$ is a nonzero number times a positive number, hence nonzero, at every point. $■$

The Wronskian test for independence

Theorem 4.6.7 (Independence of solutions)

Let $y_{1}, y_{2}$ be solutions of (H) on $I$ . The following are equivalent:

$y_{1}, y_{2}$ are linearly independent on $I$ ;
$W (y_{1}, y_{2}) (x_{0}) \neq = 0$ for some $x_{0} \in I$ ;
$W (y_{1}, y_{2}) (x) \neq = 0$ for every $x \in I$ .

Proof. Statements 2 and 3 are equivalent by Theorem 4.6.6. It remains to connect statement 1 with them. We prove two facts and then put them together.

Fact A: if $y_{1}, y_{2}$ are dependent, then $W \equiv 0$ on $I$ .

Dependence gives $c_{1}, c_{2}$ , not both zero, with $c_{1} y_{1} (x) + c_{2} y_{2} (x) = 0$ for all $x \in I$ .
Differentiating, $c_{1} y_{1}^{'} (x) + c_{2} y_{2}^{'} (x) = 0$ for all $x \in I$ .
Fix any $x \in I$ . The two lines say that the homogeneous system with coefficient matrix $(y_{1} (x) y_{1}^{'} (x) y_{2} (x) y_{2}^{'} (x))$ has the nonzero solution $(c_{1}, c_{2})$ .
By Theorem 4.6.2, Part 2, its determinant is zero. That determinant is $W (x)$ . Since $x$ was arbitrary, $W \equiv 0$ .

Fact B: if $W (x_{0}) = 0$ at some $x_{0} \in I$ , then $y_{1}, y_{2}$ are dependent.

The system $c_{1} y_{1} (x_{0}) + c_{2} y_{2} (x_{0}) c_{1} y_{1}^{'} (x_{0}) + c_{2} y_{2}^{'} (x_{0}) = 0, = 0$ has determinant $W (x_{0}) = 0$ . By Theorem 4.6.2, Part 2, it has a solution $(c_{1}, c_{2})$ with $c_{1}, c_{2}$ not both zero.
Define $y = c_{1} y_{1} + c_{2} y_{2}$ . By superposition (Theorem 4.6.3), $y$ solves (H) on $I$ .
By the choice of $c_{1}, c_{2}$ , we have $y (x_{0}) = 0$ and $y^{'} (x_{0}) = 0$ .
By uniqueness (Theorem 4.6.1, Part 2), $y (x) = 0$ for every $x \in I$ . That is, $c_{1} y_{1} + c_{2} y_{2} \equiv 0$ with $c_{1}, c_{2}$ not both zero: the solutions are dependent.

Putting it together. If $y_{1}, y_{2}$ are independent, Fact B (read backwards) says $W$ cannot vanish anywhere, which is statement 3. Conversely, if $W (x_{0}) \neq = 0$ at some point, then $W \neq \equiv 0$ , and Fact A (read backwards) says the solutions cannot be dependent, which is statement 1. $■$

Fact B needs the differential equation

Fact A holds for any two differentiable functions. Fact B does not: it used uniqueness for (H). Without an equation behind them, functions can have $W \equiv 0$ and still be independent. Take $f_{1} = x^{3}$ and $f_{2} = ∣ x ∣^{3}$ on $(- 1, 1)$ . For $x \geq 0$ they coincide, so $W = 0$ . For $x < 0$ , $f_{2} = - f_{1}$ , so again $W = 0$ . Yet they are independent: $c_{1} x^{3} + c_{2} ∣ x ∣^{3} = 0$ at $x = 1$ and $x = - 1$ gives $c_{1} + c_{2} = 0$ and $- c_{1} + c_{2} = 0$ , so $c_{1} = c_{2} = 0$ . By Theorem 4.6.7, these two functions cannot both solve one equation of type (H) on $(- 1, 1)$ .

Similarly, $x$ and $x^{3}$ have $W = 2 x^{3}$ , which vanishes at $x = 0$ only. By Theorem 4.6.6 they cannot both solve an equation of type (H) on an interval containing $0$ . They do solve $x^{2} y^{''} - 3 x y^{'} + 3 y = 0$ , but there the leading coefficient $x^{2}$ vanishes at $0$ , so the standing assumptions fail exactly at that point.

Every solution is a combination of a fundamental set

Theorem 4.6.8 (General solution of (H))

Let $y_{1}, y_{2}$ be a fundamental set of solutions of (H) on $I$ . Then every solution $φ$ of (H) on $I$ can be written as $φ = c_{1} y_{1} + c_{2} y_{2}$ for suitable constants $c_{1}, c_{2}$ (and these constants are unique).

Proof.

Match the data at one point. Fix $x_{0} \in I$ and consider the system for the unknowns $c_{1}, c_{2}$ : $c_{1} y_{1} (x_{0}) + c_{2} y_{2} (x_{0}) c_{1} y_{1}^{'} (x_{0}) + c_{2} y_{2}^{'} (x_{0}) = φ (x_{0}), = φ^{'} (x_{0}) .$
Solve it. Its determinant is $W (y_{1}, y_{2}) (x_{0})$ , which is nonzero by Theorem 4.6.7 since $y_{1}, y_{2}$ are independent. By Theorem 4.6.2, Part 1, there is exactly one solution $(c_{1}, c_{2})$ .
Compare two solutions. Let $ψ = c_{1} y_{1} + c_{2} y_{2}$ . By superposition, $ψ$ solves (H) on $I$ . By step 2, $ψ (x_{0}) = φ (x_{0})$ and $ψ^{'} (x_{0}) = φ^{'} (x_{0})$ .
Uniqueness. Both $φ$ and $ψ$ solve the same initial-value problem at $x_{0}$ . By Theorem 4.6.1 they are equal on all of $I$ : $φ = c_{1} y_{1} + c_{2} y_{2}$ . Uniqueness of the constants follows because they were determined uniquely in step 2. $■$

Theorem 4.6.8 is what justifies the phrase "general solution". The formula $c_{1} y_{1} + c_{2} y_{2}$ contains every solution, not just many of them. In the language of linear algebra, the solutions of (H) form a vector space of dimension $2$ , and any fundamental set is a basis.

One more consequence connects these results with Sections 4.3–4.5. Suppose the normalized right-hand side $f$ is continuous on $I$ , and $y_{p}$ is any particular solution of $a_{0} y^{''} + a_{1} y^{'} + a_{2} y = g$ . If $y$ is another solution, then $y - y_{p}$ solves (H), by the same grouping argument as in the superposition proof. By Theorem 4.6.8, $y - y_{p} = c_{1} y_{1} + c_{2} y_{2}$ .

Theorem 4.6.9 (General solution of the nonhomogeneous equation)

Let $a_{0}, a_{1}, a_{2}, g$ be continuous on $I$ with $a_{0} \neq = 0$ there. Let $y_{p}$ be one solution of $a_{0} y^{''} + a_{1} y^{'} + a_{2} y = g$ on $I$ , and let $y_{1}, y_{2}$ be a fundamental set for (H). Then every solution on $I$ has the form $y = c_{1} y_{1} + c_{2} y_{2} + y_{p}$ .

Worked examples

Example 4.6.1 (Abel's formula without solving)

Without solving it, find the Wronskian of any two solutions of $x y^{''} + 2 y^{'} - x y = 0$ on $x > 0$ . Then check your answer with the solutions $y_{1} = e^{x} / x$ and $y_{2} = e^{- x} / x$ .

Solution

Normalize. Divide by $x$ (nonzero on $x > 0$ ): $y^{''} + \frac{2}{x} y^{'} - y = 0$ , so $p (x) = 2/ x$ .
Abel's formula. $- \int p d x = - 2 ln x$ , so $W (x) = C e^{- 2 l n x} = \frac{C}{x ^{2}}$ for some constant $C$ .
Check. $y_{1}^{'} = e^{x} (\frac{1}{x} - \frac{1}{x ^{2}})$ , $y_{2}^{'} = e^{- x} (- \frac{1}{x} - \frac{1}{x ^{2}})$ . Then $W = \frac{e ^{x}}{x} e^{- x} (- \frac{1}{x} - \frac{1}{x ^{2}}) - e^{x} (\frac{1}{x} - \frac{1}{x ^{2}}) \frac{e ^{- x}}{x} = - \frac{1}{x ^{2}} - \frac{1}{x ^{3}} - \frac{1}{x ^{2}} + \frac{1}{x ^{3}} = - \frac{2}{x ^{2}} .$ This is $C / x^{2}$ with $C = - 2$ . Since $W \neq = 0$ , the two functions form a fundamental set on $x > 0$ .

The same idea gives a famous result. For Bessel's equation $x^{2} y^{''} + x y^{'} + (x^{2} - ν^{2}) y = 0$ (Chapter 6), $p = 1/ x$ , so every Wronskian of two solutions is $C / x$ . For the standard Bessel functions one finds $W (J_{ν}, Y_{ν}) = 2/ (π x)$ .

Example 4.6.2 (a second solution from Abel's formula)

One solution of $x y^{''} - (x + 1) y^{'} + y = 0$ on $x > 0$ is $y_{1} = e^{x}$ . Use Abel's formula to find a second, independent solution.

Solution

Normalize. $p (x) = - \frac{x + 1}{x} = - 1 - \frac{1}{x}$ .
Abel's formula. $- \int p d x = x + ln x$ , so $W (y_{1}, y_{2}) = C e^{x + l n x} = C x e^{x}$ . We are free to choose $C = 1$ .
A first-order equation for $y_{2}$ . The quotient rule gives $(\frac{y _{2}}{y _{1}})^{'} = \frac{y _{1} y _{2}^{'} - y _{1}^{'} y _{2}}{y _{1}^{2}} = \frac{W}{y _{1}^{2}} .$ Hence $(\frac{y _{2}}{e ^{x}})^{'} = \frac{x e ^{x}}{e ^{2 x}} = x e^{- x}$ .
Integrate. By parts, $\int x e^{- x} d x = - (x + 1) e^{- x}$ (constant omitted). So $y_{2} = e^{x} \cdot (- (x + 1) e^{- x}) = - (x + 1)$ .
Answer. Dropping the sign, $y_{2} = x + 1$ . Direct substitution confirms it: $0 - (x + 1) \cdot 1 + (x + 1) = 0$ . This is the pair used in Example 4.4.4.

This is reduction of order (Section 4.1) in a compact form: Abel's formula hands us $W$ , and $(y_{2} / y_{1})^{'} = W / y_{1}^{2}$ finishes the job.

Example 4.6.3 (ruling out a candidate pair)

Can $y_{1} = x$ and $y_{2} = sin x$ both be solutions of an equation $y^{''} + p (x) y^{'} + q (x) y = 0$ with $p, q$ continuous on an interval containing $0$ ?

Solution

Wronskian. $W = x cos x - 1 \cdot sin x = x cos x - sin x$ .
Values. $W (0) = 0$ , but $W (π) = - π \neq = 0$ . So $W$ is zero at some points and not at others.
Apply Theorem 4.6.6. For two solutions of such an equation, $W$ is either identically zero or never zero on the interval. Neither happens on an interval containing both $0$ and points where $W \neq = 0$ . And any interval containing $0$ contains such points: $x cos x - sin x = - x^{3} /3 + \dots$ is nonzero for small $x \neq = 0$ . So the answer is no.
What equation do they satisfy? Setting the $3 \times 3$ determinant with rows $(y, x, sin x)$ , $(y^{'}, 1, cos x)$ , $(y^{''}, 0, - sin x)$ equal to zero gives $(x cos x - sin x) y^{''} + x sin x y^{'} - sin x y = 0.$ Its leading coefficient is exactly $W$ , which vanishes at $x = 0$ . So the standing assumptions fail there, in agreement with the theorem.

Figure 4.6.1. Left: the Wronskian of two solutions of

x y^{''} + 2 y^{'} - x y = 0

(Example 4.6.1) is

- 2/ x^{2}

, never zero on

x > 0

, as Theorem 4.6.6 requires. Right:

W (x, sin x) = x cos x - sin x

vanishes at

x = 0

but not at nearby points, so

x

and

sin x

cannot both solve an equation (H) on an interval containing

0

(Example 4.6.3).

Example 4.6.4 (normalized fundamental set)

For $x y^{''} - (x + 1) y^{'} + y = 0$ on $x > 0$ , find the normalized fundamental set at $x_{0} = 1$ (as in the proof of Theorem 4.6.4). Use it to write down the solution with $y (1) = 4$ , $y^{'} (1) = - 3$ .

Solution

General solution. From Example 4.6.2 and Theorem 4.6.8, every solution is $y = a (x + 1) + b e^{x}$ , with $y^{'} = a + b e^{x}$ .
First member $Y_{1}$ : $Y_{1} (1) = 1$ , $Y_{1}^{'} (1) = 0$ . We need $2 a + b e = 1$ and $a + b e = 0$ . Subtracting, $a = 1$ , then $b = - 1/ e$ . So $Y_{1} = x + 1 - e^{x - 1}$ .
Second member $Y_{2}$ : $Y_{2} (1) = 0$ , $Y_{2}^{'} (1) = 1$ . We need $2 a + b e = 0$ and $a + b e = 1$ . Subtracting, $a = - 1$ , then $b = 2/ e$ . So $Y_{2} = - (x + 1) + 2 e^{x - 1}$ .
Check independence. $W (Y_{1}, Y_{2}) (1) = Y_{1} (1) Y_{2}^{'} (1) - Y_{1}^{'} (1) Y_{2} (1) = 1 \cdot 1 - 0 \cdot 0 = 1 \neq = 0$ .
The requested solution. It is simply $4 Y_{1} - 3 Y_{2}$ : $y = 4 (x + 1 - e^{x - 1}) - 3 (- (x + 1) + 2 e^{x - 1}) = 7 (x + 1) - 10 e^{x - 1} .$ Check: $y (1) = 14 - 10 = 4$ and $y^{'} (1) = 7 - 10 = - 3$ .

Figure 4.6.2. The normalized fundamental set at

x_{0} = 1

of Example 4.6.4.

Y_{1}

starts with value

1

and slope

0

Y_{2}

with value

0

and slope

1

(short segments). The solution with

y (1) = 4

y^{'} (1) = - 3

is simply

4 Y_{1} - 3 Y_{2}

Exercises

Use Abel's formula to find, up to a constant factor, the Wronskian of two solutions of $(1 - x^{2}) y^{''} - x y^{'} + 4 y = 0$ on $- 1 < x < 1$ .
Answer
$W = \frac{C}{1 - x ^{2}}$ (for example, $W (2 x^{2} - 1, x 1 - x^{2}) = - 1/ 1 - x^{2}$ )
Find, up to a constant factor, the Wronskian of two solutions of $y^{''} - 2 x y^{'} + 6 y = 0$ .
Answer
$W = C e^{x^{2}}$
Two solutions of $x^{2} y^{''} + 3 x y^{'} + 5 y = 0$ on $x > 0$ have Wronskian $3$ at $x = 1$ . Find their Wronskian at $x = 2$ without solving the equation.
Answer
$W (x) = 3 x^{- 3}$ , so $W (2) = \frac{3}{8}$
(a) Verify that $y_{1} = x$ and $y_{2} = e^{3 x}$ solve $(3 x - 1) y^{''} - 9 x y^{'} + 9 y = 0$ . (b) Compute $W (y_{1}, y_{2})$ and compare with Abel's formula. (c) $W$ vanishes at $x = \frac{1}{3}$ . Why does this not contradict Theorem 4.6.6?
Answer
(b) $W = (3 x - 1) e^{3 x}$ , in agreement with Abel's formula for $p = - 9 x / (3 x - 1)$ . (c) The leading coefficient $3 x - 1$ is zero at $x = \frac{1}{3}$ , so the theorem applies only on $x < \frac{1}{3}$ or on $x > \frac{1}{3}$ , and on each of these $W \neq = 0$ .
Prove: if two solutions of (H) on $I$ vanish at the same point $x_{0} \in I$ , they are linearly dependent on $I$ .
Answer
$W (x_{0}) = 0 \cdot y_{2}^{'} (x_{0}) - y_{1}^{'} (x_{0}) \cdot 0 = 0$ , so they are dependent by Theorem 4.6.7.
Prove: if two solutions of (H) on $I$ both have a horizontal tangent at the same point $x_{0} \in I$ , they are linearly dependent on $I$ .
Answer
$y_{1}^{'} (x_{0}) = y_{2}^{'} (x_{0}) = 0$ gives $W (x_{0}) = 0$ ; apply Theorem 4.6.7.
Show that $y = x^{2} sin x$ cannot be a solution of any equation $y^{''} + p (x) y^{'} + q (x) y = 0$ with $p, q$ continuous on an interval containing $0$ .
Answer
$y (0) = 0$ and $y^{'} (0) = 0$ , so by Theorem 4.6.1 it would be identically zero on the interval. But $y \neq = 0$ at small $x \neq = 0$ .
Let $y$ be a solution of (H) on $I$ that is not identically zero. Prove that at every zero $x_{0}$ of $y$ we have $y^{'} (x_{0}) \neq = 0$ . (So the graph crosses the axis at each zero rather than touching it.)
Answer
If $y (x_{0}) = y^{'} (x_{0}) = 0$ , uniqueness forces $y \equiv 0$ , contrary to the assumption.
Find the normalized fundamental set at $x_{0} = 0$ for $y^{''} - 4 y^{'} - 5 y = 0$ , and use it to write the solution with $y (0) = 3$ , $y^{'} (0) = - 2$ .
Answer
$Y_{1} = \frac{1}{6} (e^{5 x} + 5 e^{- x})$ , $Y_{2} = \frac{1}{6} (e^{5 x} - e^{- x})$ ; $y = 3 Y_{1} - 2 Y_{2} = \frac{1}{6} (e^{5 x} + 17 e^{- x})$
Show that $f_{1} = x ∣ x ∣$ and $f_{2} = x^{2}$ have $W (f_{1}, f_{2}) \equiv 0$ on $(- 1, 1)$ but are linearly independent there. What does Theorem 4.6.7 tell you about them?
Answer
On $x \geq 0$ , $f_{1} = f_{2}$ ; on $x < 0$ , $f_{1} = - f_{2}$ ; in both cases $W = 0$ (also at $0$ , where both derivatives vanish). Testing $c_{1} f_{1} + c_{2} f_{2} = 0$ at $x = \pm 1$ gives $c_{1} = c_{2} = 0$ . They cannot both solve one equation of type (H) on $(- 1, 1)$ .
For $y^{'''} + p_{1} (x) y^{''} + p_{2} (x) y^{'} + p_{3} (x) y = 0$ with continuous coefficients, show that the Wronskian of three solutions satisfies $W^{'} = - p_{1} W$ , and hence $W (x) = W (x_{0}) exp (- \int_{x_{0}}^{x} p_{1} (s) d s)$ . (Hint: differentiate the $3 \times 3$ determinant one row at a time.)
Answer
Differentiating rows 1 and 2 produces determinants with two equal rows, which vanish. Differentiating row 3 replaces $y_{i}^{''}$ by $y_{i}^{'''} = - p_{1} y_{i}^{''} - p_{2} y_{i}^{'} - p_{3} y_{i}$ ; the $p_{2}$ and $p_{3}$ parts are multiples of rows 2 and 1 and drop out, leaving $- p_{1} W$ .
Prove that if $y_{1}, y_{2}$ is a fundamental set of (H) on $I$ , then so is $y_{1} + y_{2}, y_{1} - y_{2}$ .
Answer
Both are solutions by superposition, and $W (y_{1} + y_{2}, y_{1} - y_{2}) = - 2 W (y_{1}, y_{2}) \neq = 0$ .