Chapter 1 · Section 1.2

1.2 What Is a Solution?

In Section 1.1 we learned to recognise and classify differential equations. Now we ask the obvious next question: what does it mean to solve one? The answer is more subtle than "find a formula". A solution is a function, and a function always lives on some set of $x$ -values. We will see that the interval matters, that a solution may be given only implicitly, that some "solutions" are fakes, and that an ODE usually has not one solution but a whole family of them.

Explicit solutions

Throughout this section we write an ODE of order $n$ in its general form

F (x, y, y^{'}, \dots, y^{(n)}) = 0.

Roughly, a solution is a function that makes this equation true when we plug it in. The precise definition adds two details: where it must be true, and that all the derivatives must exist there.

Definition 1.2.1 (Explicit solution)

Let $I$ be an interval. A function $y = f (x)$ is an explicit solution of the ODE on $I$ when $f$ can be differentiated $n$ times at each point of $I$ , and the expression $F (x, f (x), f^{'} (x), \dots, f^{(n)} (x))$ makes sense and equals $0$ at each point of $I$ .

So plugging in is not enough at a few chosen points: the equation must hold at every $x$ of the interval. Checking this is a purely mechanical task.

Method in a nutshell: verifying a solution

Note where the candidate function and the equation are defined.
Compute every derivative the equation needs.
Substitute into the left-hand side and simplify.
Confirm that the result equals the right-hand side for every $x$ in the interval.

Example 1.2.1

Show that $y = 3 e^{- x} cos 4 x$ is an explicit solution of $y^{''} + 2 y^{'} + 17 y = 0$ on $(- \infty, \infty)$ .

Solution

Domain. The function is defined and infinitely differentiable for every real $x$ , and so is the equation. So we may take $I = (- \infty, \infty)$ .
First derivative. By the product rule, with $(e^{- x})^{'} = - e^{- x}$ and $(cos 4 x)^{'} = - 4 sin 4 x$ : $y^{'} = 3 [- e^{- x} cos 4 x + e^{- x} (- 4 sin 4 x)] = - 3 e^{- x} cos 4 x - 12 e^{- x} sin 4 x .$
Second derivative. Differentiate each term of $y^{'}$ separately: $(- 3 e^{- x} cos 4 x)^{'} (- 12 e^{- x} sin 4 x)^{'} = 3 e^{- x} cos 4 x + 12 e^{- x} sin 4 x, = 12 e^{- x} sin 4 x - 48 e^{- x} cos 4 x .$ Adding, $y^{''} = - 45 e^{- x} cos 4 x + 24 e^{- x} sin 4 x .$
Substitute. Collect the $e^{- x} cos 4 x$ terms and the $e^{- x} sin 4 x$ terms of $y^{''} + 2 y^{'} + 17 y$ : $cos terms: sin terms: - 45 + 2 (- 3) + 17 (3) = - 45 - 6 + 51 = 0, - 24 + 2 (- 12) + 17 (0) = 24 - 24 = 0.$ So $y^{''} + 2 y^{'} + 17 y = 0$ for every real $x$ . $✓$

The interval in the definition is not a formality. A formula can satisfy an equation everywhere it makes sense and still fail to be a solution on the whole real line.

Example 1.2.2 (The interval matters)

Show that $y = \frac{1}{9 - x ^{2}}$ satisfies $y^{'} = 2 x y^{2}$ . On which intervals is it an explicit solution?

Solution

Derivative. Write $y = (9 - x^{2})^{- 1}$ . By the chain rule, $y^{'} = - 1 \cdot (9 - x^{2})^{- 2} \cdot (- 2 x) = \frac{2 x}{( 9 - x ^{2} ) ^{2}} .$
Right-hand side. $2 x y^{2} = 2 x \cdot \frac{1}{( 9 - x ^{2} ) ^{2}}$ , which is the same expression. $✓$
Where? The formula is undefined at $x = 3$ and $x = - 3$ . A solution must be defined (and differentiable) on a whole interval, so the interval cannot contain $\pm 3$ . The possible intervals are $(- \infty, - 3), (- 3, 3), (3, \infty),$ or any subinterval of one of these. The formula is not a solution on $(- \infty, \infty)$ . In fact, the three pieces should be thought of as three different solutions that happen to share one formula.

Figure 1.2.1. The formula

y = 1/ (9 - x^{2})

has three separate pieces, one on each of the intervals

(- \infty, - 3)

(- 3, 3)

and

(3, \infty)

. Each piece is a solution of

y^{'} = 2 x y^{2}

on its own interval. No solution can cross the vertical asymptotes

x = \pm 3

Why insist on an interval?

An ODE describes how a quantity evolves continuously. If a "solution" blows up at $x = 3$ , we cannot follow it from $x = 2$ to $x = 4$ : the physical system it models would break down at $x = 3$ . Restricting to intervals keeps solutions in one connected piece.

Implicit solutions

Many methods in Chapter 2 do not produce $y$ as a formula in $x$ . Instead they produce a relation $G (x, y) = 0$ connecting $x$ and $y$ . Sometimes we can solve it for $y$ ; often we cannot. Such relations still count as solutions, provided they really do define a function.

Definition 1.2.2 (Implicit solution)

A relation $G (x, y) = 0$ is an implicit solution of an ODE on an interval $I$ if we can find a function $f$ , defined on $I$ , whose graph lies on the curve $G (x, y) = 0$ (that is, $G (x, f (x)) = 0$ for every $x$ in $I$ ) and which is an explicit solution of the ODE on $I$ .

In practice we seldom bother with the adjectives: both kinds are just solutions.

Example 1.2.3 (An ellipse)

Show that the relation $x^{2} + 4 y^{2} = 16$ is an implicit solution of $x + 4 y y^{'} = 0$ on the interval $(- 4, 4)$ .

Solution

Solve the relation for $y$ . $4 y^{2} y = 16 - x^{2} = \pm \frac{1}{2} 16 - x^{2} .$ So the relation defines two functions: $f_{1} (x) = \frac{1}{2} 16 - x^{2}$ (upper half of the ellipse) and $f_{2} (x) = - \frac{1}{2} 16 - x^{2}$ (lower half).
Differentiate $f_{1}$ . For $- 4 < x < 4$ , $f_{1}^{'} (x) = \frac{1}{2} \cdot \frac{- 2 x}{2 16 - x ^{2}} = \frac{- x}{2 16 - x ^{2}} .$
Substitute. $x + 4 f_{1} f_{1}^{'} = x + 4 \cdot \frac{1}{2} 16 - x^{2} \cdot \frac{- x}{2 16 - x ^{2}} = x - x = 0. ✓$ The same computation works for $f_{2}$ (both signs flip, and the product is unchanged).
Why the open interval? At $x = \pm 4$ the derivative $f_{1}^{'}$ has a zero denominator: the tangent to the ellipse is vertical there. So $f_{1}$ and $f_{2}$ are solutions on $(- 4, 4)$ but not on $[- 4, 4]$ .

Figure 1.2.2. The ellipse

x^{2} + 4 y^{2} = 16

is not the graph of one function, but it contains the graphs of two: the upper half

f_{1}

and the lower half

f_{2}

. Both solve

x + 4 y y^{'} = 0

(- 4, 4)

. At

x = \pm 4

the tangent is vertical, so the endpoints are excluded.

In Example 1.2.3 we could solve for $y$ . The next example shows a relation that we cannot conveniently solve, yet we can still prove it is a genuine implicit solution.

Example 1.2.4 (A relation we cannot easily solve)

Show that $y^{3} + 3 y - x^{2} = 5$ is an implicit solution of $3 (1 + y^{2}) y^{'} = 2 x$ on $(- \infty, \infty)$ .

Solution

Implicit differentiation. Treat $y$ as a function of $x$ and differentiate both sides with respect to $x$ : $3 y^{2} y^{'} + 3 y^{'} - 2 x 3 (y^{2} + 1) y^{'} = 0 = 2 x .$ This is exactly the ODE. So if the relation defines a differentiable function, that function is a solution.
Does it define a function? Write the relation as $h (y) = x^{2} + 5$ with $h (y) = y^{3} + 3 y$ . Since $h^{'} (y) = 3 y^{2} + 3 > 0$ , $h$ is strictly increasing. It also goes from $- \infty$ to $+ \infty$ . Therefore, for each real $x$ , there is exactly one real $y$ with $h (y) = x^{2} + 5$ . Call it $f (x)$ .
Is $f$ differentiable? The partial derivative of $G (x, y) = y^{3} + 3 y - x^{2} - 5$ with respect to $y$ is $3 y^{2} + 3$ , which is never zero. By the implicit function theorem from calculus, $f$ is differentiable everywhere, and its derivative is the one found in Step 1.
Conclusion. The relation defines a function $f$ on $(- \infty, \infty)$ that satisfies the ODE, so it is an implicit solution there. For instance, at $x = 3$ the relation reads $y^{3} + 3 y = 14$ , whose only real root is $y = 2$ ; so $f (3) = 2$ , even though we never wrote down a formula for $f$ .

A warning: formal solutions

Implicit differentiation is a quick test, but passing it is not enough. A relation might satisfy the ODE "on paper" and yet contain no real function at all.

Example 1.2.5 (A fake solution)

Consider the ODE $y^{'} + 2 x e^{- y} = 0$ . (a) Show that the relation $e^{y} + x^{2} = 0$ passes the implicit-differentiation test. (b) Is it an implicit solution? (c) Compare with the relation $e^{y} + x^{2} = 3$ .

Solution

(a) The test. Differentiate $e^{y} + x^{2} = 0$ with respect to $x$ : $e^{y} y^{'} + 2 x y^{'} = 0 = - 2 x e^{- y} .$ Substituting this $y^{'}$ into the ODE gives $- 2 x e^{- y} + 2 x e^{- y} = 0$ , an identity. So the relation formally satisfies the equation.
(b) Does it define a real function? We would need $e^{y} = - x^{2}$ . But $e^{y} > 0$ for every real $y$ , while $- x^{2} \leq 0$ . There is no real $y$ at all, for any $x$ . The relation defines no function, so it is not an implicit solution. A relation like this, which passes the differentiation test but describes no real curve, is called a formal solution.
(c) A genuine one. The same differentiation works for $e^{y} + x^{2} = 3$ . This time we can solve: $e^{y} = 3 - x^{2}$ , so $y = ln (3 - x^{2})$ , which is real for $- 3 < x < 3$ . Check directly: $y^{'} = \frac{- 2 x}{3 - x ^{2}}, 2 x e^{- y} = \frac{2 x}{3 - x ^{2}},$ and their sum is $0$ . $✓$ So $e^{y} + x^{2} = 3$ is an implicit solution on $(- 3, 3)$ .

Common mistake

Do not stop after implicit differentiation. Also make sure the relation actually defines a real, differentiable function on some interval. In routine exercises this step is usually skipped, because the relations our methods produce almost always pass it; when a result will be used for something important, do the check.

Families of solutions

Consider the very simple equation $y^{'} = 6 x$ . Integrating, $y = 3 x^{2} + c$ , and every real value of $c$ gives a solution. So this ODE has infinitely many solutions, indexed by one constant. This is typical.

Figure 1.2.3. Five members of the family

y = 3 x^{2} + c

of solutions of

y^{'} = 6 x

. Changing

c

shifts the curve up or down. Above any given

x

all members have the same slope

6 x

(short black segments at

x = 1

), and that is all the equation asks for.

Definition 1.2.3 (n-parameter family)

A formula $y = f (x, c_{1}, c_{2}, \dots, c_{n})$ , or a relation $G (x, y, c_{1}, \dots, c_{n}) = 0$ , containing $n$ arbitrary constants is an $n$ -parameter family of solutions of an ODE if it gives a solution for every admissible choice of the constants. The constants are called parameters. A solution obtained by giving the parameters specific values is a particular solution.

Section 1.1 suggested the rule of thumb: an ODE of order $n$ usually has an $n$ -parameter family of solutions. Each integration step brings in one new constant. For linear equations (Chapter 4) this rule becomes a theorem, and the family contains every solution; it is then called the general solution.

For nonlinear equations two surprises are possible.

Extra solutions. Take $y^{'} = 2 x y^{2}$ from Example 1.2.2. Repeating that computation with $9$ replaced by any constant $c$ shows that $y = \frac{1}{c - x ^{2}}$ is a one-parameter family of solutions. But $y \equiv 0$ is also a solution (both sides are zero), and no value of $c$ produces it, because $1/ (c - x^{2})$ is never zero. Solutions that are not members of a family like this are sometimes called singular solutions.
No solutions at all. The equation $(y^{'})^{2} + e^{y} = 0$ has no real solution: $(y^{'})^{2} \geq 0$ and $e^{y} > 0$ , so the left side is always positive.

Example 1.2.6 (A three-parameter family)

Show that $y = c_{1} + c_{2} x + c_{3} e^{2 x}$ is a three-parameter family of solutions of $y^{'''} - 2 y^{''} = 0$ on $(- \infty, \infty)$ .

Solution

Derivatives. $y^{'} y^{''} y^{'''} = c_{2} + 2 c_{3} e^{2 x}, = 4 c_{3} e^{2 x}, = 8 c_{3} e^{2 x} .$
Substitute. $y^{'''} - 2 y^{''} = 8 c_{3} e^{2 x} - 2 \cdot 4 c_{3} e^{2 x} = 0$ for every $x$ and every choice of $c_{1}, c_{2}, c_{3}$ . $✓$
Three genuine parameters. The three constants multiply three different functions $1$ , $x$ , $e^{2 x}$ , so none of them can be absorbed into the others. A third-order equation and three parameters: the rule of thumb holds.

Counting parameters honestly

The formula $y = c_{1} e^{x + c_{2}}$ seems to have two constants, but $c_{1} e^{x + c_{2}} = (c_{1} e^{c_{2}}) e^{x}$ , which is just $K e^{x}$ . It is really a one-parameter family. When we say " $n$ parameters", we mean $n$ constants that cannot be merged into fewer.

Example 1.2.7 (Guessing a solution)

Find all values of $m$ for which $y = x^{m}$ is a solution of $x^{2} y^{''} - 2 x y^{'} - 4 y = 0$ on $(0, \infty)$ .

Solution

Derivatives. $y^{'} = m x^{m - 1}$ and $y^{''} = m (m - 1) x^{m - 2}$ .
Substitute. Each term becomes a multiple of $x^{m}$ : $x^{2} \cdot m (m - 1) x^{m - 2} - 2 x \cdot m x^{m - 1} - 4 x^{m} = [m (m - 1) - 2 m - 4] x^{m} = (m^{2} - 3 m - 4) x^{m} .$
Set to zero. Since $x^{m} \neq = 0$ for $x > 0$ , we need $m^{2} - 3 m - 4 = 0$ , that is $(m - 4) (m + 1) = 0$ . So $m = 4$ or $m = - 1$ .
Answer. $y = x^{4}$ and $y = x^{- 1}$ are solutions on $(0, \infty)$ . Because the equation is linear and homogeneous, $y = c_{1} x^{4} + c_{2} x^{- 1}$ is then a two-parameter family of solutions on $(0, \infty)$ . (You can also check this family on $(- \infty, 0)$ ; only $x^{- 1}$ fails at $x = 0$ .)

Use the checker below to verify a candidate solution step by step. It starts with the two-parameter family that contains Example 1.2.1. Try your own: type the family from Example 1.2.6, or a deliberately wrong guess, and watch where the check fails.

The geometric picture: direction fields and integral curves

Think of a first-order equation $y^{'} = f (x, y)$ as a rule that hands out slopes: give it a point $(x, y)$ and it returns the number $f (x, y)$ . Any solution curve passing through that point must have exactly that slope there.

Draw a short line segment with slope $f (x, y)$ at many points of the plane. The picture is called a direction field (or slope field). The graphs of solutions are called integral curves. Each integral curve is tangent, at every one of its points, to the little segment drawn there. Without solving anything, the direction field shows how solutions behave.

Take $y^{'} = y - x$ as an example. We can check that

y = x + 1 + c e^{x}

is a one-parameter family of solutions: $y^{'} = 1 + c e^{x}$ and $y - x = 1 + c e^{x}$ , which agree. Now look at the field.

Along any line $y - x = k$ , every segment has the same slope $k$ . Such lines are called isoclines (equal-slope lines).
On the isocline $y = x + 1$ the slope is $k = 1$ , which equals the slope of the line itself. So this line is an integral curve: it is the member $c = 0$ of the family.
Curves above this line ( $c > 0$ ) bend upward and grow like $e^{x}$ . Curves below it ( $c < 0$ ) bend downward. As $x \to - \infty$ , all curves approach the line, because $c e^{x} \to 0$ .

Figure 1.2.4. Direction field of

y^{'} = y - x

with several integral curves

y = x + 1 + c e^{x}

. The straight line

y = x + 1

(

c = 0

) separates curves that bend upward (

c > 0

) from curves that bend downward (

c < 0

). To the left, all curves crowd onto this line. Along the dashed isocline

y = x

every segment is horizontal.

Click inside the field below to draw integral curves through points of your choice. Then change $f$ to explore other equations, for example $f = 2 x y^{2}$ from Example 1.2.2.

What does "solving" mean in practice?

Chapters 2 and 4 teach recipes that turn certain types of equations into formulas. Do not be misled by their success: those types were chosen precisely because formulas exist for them. Pick an ODE at random and, almost surely, its solutions cannot be written with the functions you know.

Even a very innocent equation shows this. The solutions of $y^{'} = e^{- x^{2}}$ are

y = c + \int_{0}^{x} e^{- t^{2}} d t,

by the fundamental theorem of calculus. This is a perfectly good solution, yet the integral cannot be written using polynomials, exponentials, logarithms and trigonometric functions. (It is so useful that it has its own name: $\int_{0}^{x} e^{- t^{2}} d t = \frac{π}{2} erf (x)$ .)

So what do we do then? We still learn a lot about the solutions. A direction field, like the one above, shows their shape at a glance. In Chapter 6 we build solutions as power series, term by term. In Chapter 8 we let a computer march along the solution in small steps and produce a table of accurate values; this is what engineering software does every day.

Before trusting any of these tools, it is wise to know that a solution exists and is unique. That is the subject of Section 1.3.

Exercises

Show that $y = 2 e^{5 x} - 7 e^{- 3 x}$ is a solution of $y^{''} - 2 y^{'} - 15 y = 0$ on $(- \infty, \infty)$ .
Answer
$y^{'} = 10 e^{5 x} + 21 e^{- 3 x}$ , $y^{''} = 50 e^{5 x} - 63 e^{- 3 x}$ ; then $y^{''} - 2 y^{'} - 15 y = (50 - 20 - 30) e^{5 x} + (- 63 - 42 + 105) e^{- 3 x} = 0$ .
Show that every function $y = x^{3} + \frac{c}{x}$ is a solution of $x y^{'} + y = 4 x^{3}$ on $(0, \infty)$ .
Answer
$x y^{'} = 3 x^{3} - c / x$ ; adding $y$ gives $4 x^{3}$ .
Show that $y = x ln x - x$ is a solution of $x y^{'} - y = x$ on $(0, \infty)$ . Why not on $(- \infty, \infty)$ ?
Answer
$y^{'} = ln x$ , so $x y^{'} - y = x ln x - x ln x + x = x$ . The function is undefined for $x \leq 0$ .
Show that $x^{2} + x y + y^{2} = 7$ is an implicit solution of $(x + 2 y) y^{'} + 2 x + y = 0$ . Find the explicit solution through the point $(1, 2)$ and its largest open interval.
Answer
$y = \frac{- x + 28 - 3 x ^{2}}{2}$ on $- 28/3 < x < 28/3$ .
Show that $y^{4} + x^{2} + 1 = 0$ formally satisfies $2 y^{3} y^{'} + x = 0$ . Is it an implicit solution?
Answer
Implicit differentiation gives $4 y^{3} y^{'} + 2 x = 0$ , i.e. the ODE. But $y^{4} = - (x^{2} + 1) < 0$ has no real solution, so it is only a formal solution.
Find all values of $m$ for which $y = e^{m x}$ is a solution of $y^{'''} - y^{''} - 6 y^{'} = 0$ .
Answer
$m = 0, 3, - 2$
Find all values of $m$ for which $y = x^{m}$ is a solution of $x^{2} y^{''} + 4 x y^{'} - 10 y = 0$ on $(0, \infty)$ .
Answer
$m = 2, - 5$
Show that $y = c_{1} cos 5 x + c_{2} sin 5 x + \frac{x}{25}$ is a two-parameter family of solutions of $y^{''} + 25 y = x$ .
Answer
The trigonometric terms give $0$ in $y^{''} + 25 y$ , and $x /25$ gives $0 + 25 \cdot x /25 = x$ .
Show that $y = tan 2 x$ satisfies $y^{'} = 2 (1 + y^{2})$ . What is the largest interval containing $x = 0$ on which it is a solution?
Answer
$y^{'} = 2 sec^{2} 2 x = 2 (1 + tan^{2} 2 x)$ ; interval $(- π /4, π /4)$ .
Show that $y = tanh (x + c)$ is a one-parameter family of solutions of $y^{'} = 1 - y^{2}$ . Then show that $y \equiv 1$ and $y \equiv - 1$ are also solutions, but are not members of the family.
Answer
$y^{'} = sech^{2} (x + c) = 1 - tanh^{2} (x + c)$ . Constants $\pm 1$ give $0 = 0$ ; but $∣ tanh ∣ < 1$ , so no $c$ produces them.
Explain why $(y^{'})^{2} + 3 = cos y$ has no real solution.
Answer
The left side is at least $3$ , the right side at most $1$ .
Show that $y = 2 + \int_{0}^{x} sin (t^{2}) d t$ is a solution of $y^{'} = sin (x^{2})$ on $(- \infty, \infty)$ , and find $y (0)$ .
Answer
By the fundamental theorem of calculus $y^{'} = sin (x^{2})$ ; $y (0) = 2$ .