Chapter 2 · Section 2.2

2.2 Separable Equations and Equations Reducible to Separable Form

The simplest first-order equations are those where the right-hand side splits into a piece that depends only on $x$ and a piece that depends only on $y$ . Then we can put every $y$ on one side, every $x$ on the other, and integrate each side on its own. This idea is called separation of variables. It is the workhorse behind growth and decay laws, cooling, draining tanks, and simple chemical kinetics.

In the second half of the section we meet homogeneous equations. They are not separable as given, but one clever substitution makes them separable.

Separable equations

Definition 2.2.1 (Separable equation)

A first-order equation is separable if it can be written as

y^{'} = f (x) g (y),

or, in differential form, as $p_{1} (x) q_{1} (y) d x + p_{2} (x) q_{2} (y) d y = 0$ , where each coefficient is a product of a function of $x$ and a function of $y$ .

For example, $y^{'} = x^{2} y^{3}$ and $y^{'} = e^{x - y} = e^{x} e^{- y}$ are separable. The equation $y^{'} = x + y$ is not: no algebra splits $x + y$ into a product of a function of $x$ and a function of $y$ .

Why separation works

Students often learn to "multiply by $d x$ and divide by $g (y)$ ". That is a handy shorthand, but it helps to see what is really going on.

Let $y = φ (x)$ be a solution on an interval where $g (φ (x)) \neq = 0$ . Then $\frac{φ ^{'} ( x )}{g ( φ ( x ))} = f (x) .$
Let $H (y)$ be any antiderivative of $1/ g (y)$ , so $H^{'} (y) = 1/ g (y)$ .
By the chain rule, the left side is a derivative: $\frac{d}{d x} H (φ (x)) = H^{'} (φ (x)) φ^{'} (x) = \frac{φ ^{'} ( x )}{g ( φ ( x ))} .$
So $\frac{d}{d x} H (φ (x)) = f (x)$ . Integrate both sides with respect to $x$ : $H (φ (x)) = \int f (x) d x + c .$

This is exactly what the shorthand produces:

\frac{d y}{g ( y )} = f (x) d x ⟹ \int \frac{d y}{g ( y )} = \int f (x) d x + c .

Connection with Section 2.1

After dividing by $g (y)$ , the equation reads $f (x) d x - \frac{1}{g ( y )} d y = 0$ . Here $M = f (x)$ depends only on $x$ and $N = - 1/ g (y)$ only on $y$ , so $M_{y} = 0 = N_{x}$ . In other words, $1/ g (y)$ is an integrating factor, and the separated equation is exact.

The danger: lost solutions

Step 1 above needed $g (y) \neq = 0$ . What happens where $g$ vanishes? Suppose $g (y_{0}) = 0$ for some number $y_{0}$ . Try the constant function $y \equiv y_{0}$ :

y^{'} = 0 and f (x) g (y_{0}) = f (x) \cdot 0 = 0.

Both sides agree. So every root of $g$ gives a constant solution $y \equiv y_{0}$ (an equilibrium). The separation step divided by zero there and simply did not see these solutions. Sometimes they reappear in the general formula for a special value of the constant; sometimes they do not. Either way, you must look for them separately.

In the differential form $p_{1} (x) q_{1} (y) d x + p_{2} (x) q_{2} (y) d y = 0$ , we separate by dividing by $q_{1} (y) p_{2} (x)$ . Roots of $q_{1}$ give lost horizontal lines $y = y_{0}$ , and roots of $p_{2}$ give lost vertical lines $x = x_{0}$ (solutions if we let $x$ be a function of $y$ ).

A word about constants

When you integrate both sides you get two constants, one on each side. Their difference is again an arbitrary constant, so a single $c$ on one side is enough.

Logarithms appear often, and the constant then changes shape. Suppose you reach $ln ∣ y ∣ = 3 x + c$ . Exponentiating gives $∣ y ∣ = e^{c} e^{3 x}$ , so $y = \pm e^{c} e^{3 x}$ . The number $\pm e^{c}$ can be any nonzero constant; call it $K$ . Thus $y = K e^{3 x}$ with $K \neq = 0$ . Very often $K = 0$ also gives a solution (here $y \equiv 0$ , a root of $g$ ), and then we simply allow every real $K$ . Renaming constants like this is routine. Just keep track of which values are allowed.

Finally, the answer may stay implicit, such as $y^{3} + y = x^{2} + c$ . That is acceptable. Solve for $y$ when you can do so cleanly, and always when an initial condition asks for a specific function.

Method in a nutshell

Write the equation as $y^{'} = f (x) g (y)$ .
Find all roots $y_{0}$ of $g (y) = 0$ . Each gives a constant solution $y \equiv y_{0}$ . Write them down now.
For $g (y) \neq = 0$ , separate: $\frac{d y}{g ( y )} = f (x) d x$ .
Integrate both sides; one constant $c$ is enough.
Solve for $y$ if you can. Check whether the constant solutions from step 2 are included for some value of $c$ ; list those that are not.
For an initial condition, determine $c$ and state an interval on which the solution exists.

The solver below handles $y^{'} = f (x) g (y)$ where $g$ is a polynomial of degree at most two in $y$ , or a power $y^{n}$ . It lists the constant solutions first and then separates. Enter your own $f$ and $g$ and compare with the examples.

Worked examples: separable equations

Example 2.2.1

Solve $y^{'} = 2 x (y^{2} + 9)$ .

Solution

Constant solutions. $g (y) = y^{2} + 9 > 0$ never vanishes, so there are none, and nothing can be lost.
Separate. $\frac{d y}{y ^{2} + 9} = 2 x d x$ .
Integrate. Using $\int \frac{d y}{y ^{2} + a ^{2}} = \frac{1}{a} arctan \frac{y}{a}$ with $a = 3$ : $\frac{1}{3} arctan \frac{y}{3} = x^{2} + c_{1} .$
Solve for $y$ . Multiply by 3 and rename $3 c_{1} = c$ : $arctan \frac{y}{3} = 3 x^{2} + c ⟹ y = 3 tan (3 x^{2} + c) .$

Example 2.2.2 (a lost solution)

Solve $y^{'} = (x + 1) (y^{2} - 4)$ .

Solution

Constant solutions. $y^{2} - 4 = 0$ gives $y \equiv 2$ and $y \equiv - 2$ . Both satisfy the equation.
Separate (for $y \neq = \pm 2$ ): $\frac{d y}{y ^{2} - 4} = (x + 1) d x$ .
Partial fractions. $\frac{1}{y ^{2} - 4} = \frac{1}{4} (\frac{1}{y - 2} - \frac{1}{y + 2})$ .
Integrate. $\frac{1}{4} ln \frac{y - 2}{y + 2} = \frac{x ^{2}}{2} + x + c_{1} .$
Remove the logarithm. Multiply by 4 and exponentiate: $\frac{y - 2}{y + 2} = e^{4 c_{1}} e^{2 x^{2} + 4 x} ⟹ \frac{y - 2}{y + 2} = K e^{2 x^{2} + 4 x},$ where $K = \pm e^{4 c_{1}}$ is any nonzero constant.
Solve for $y$ . Write $E = e^{2 x^{2} + 4 x}$ . Then $y - 2 y (1 - K E) y = K E (y + 2) = 2 + 2 K E = \frac{2 ( 1 + K e ^{2 x^{2} + 4 x} )}{1 - K e ^{2 x^{2} + 4 x}} .$
Account for the constant solutions. $K = 0$ gives $y \equiv 2$ , so we may allow $K = 0$ and that solution is recovered. But $y \equiv - 2$ corresponds to no finite $K$ . It is a lost solution. The complete answer is the family above together with $y \equiv - 2$ .

Figure 2.2.1. Direction field and solution curves of

y^{'} = (x + 1) (y^{2} - 4)

. The two constant solutions appear as horizontal lines. Curves between them, or below

y = - 2

, approach the lost solution

y \equiv - 2

x

moves away from

- 1

in either direction. Curves above

y = 2

move away from it and blow up.

Example 2.2.3 (a draining tank)

Water leaves a tank through a hole in the bottom. By Torricelli's law the outflow speed is proportional to $h$ , where $h$ is the water depth. For a particular tank, with $h$ in decimetres and $t$ in minutes, $\frac{d h}{d t} = - \frac{1}{2} h, h (0) = 16.$ When is the tank empty?

Solution

Constant solutions. $h = 0$ gives $h \equiv 0$ : an empty tank stays empty.
Separate (for $h > 0$ ): $h^{- 1/2} d h = - \frac{1}{2} d t$ .
Integrate. $2 h = - \frac{1}{2} t + c$ .
Initial condition. At $t = 0$ : $216 = 8 = c$ . So $h = 4 - \frac{t}{4}$ .
Solve. Since $h \geq 0$ , this formula is valid only while $4 - t /4 \geq 0$ , that is, for $0 \leq t \leq 16$ : $h (t) = (4 - \frac{t}{4})^{2}, 0 \leq t \leq 16.$
After the tank empties. At $t = 16$ we have $h = 0$ and $h^{'} (16) = 0$ . From then on the physical solution is $h \equiv 0$ . The two pieces join smoothly, and the tank is empty after 16 minutes.
A remark on uniqueness. The function $h$ has an infinite derivative at $h = 0$ , so the uniqueness theorem of Chapter 1 does not apply there. Indeed, knowing only that the tank is empty at $t = 20$ does not tell you when it became empty. Physically this makes sense.

Figure 2.2.2. The draining tank of Example 2.2.3. The depth

h (t) = (4 - t /4)^{2}

falls more and more slowly and reaches zero at

t = 16

min, where it joins the solution

h \equiv 0

smoothly. The dashed curves are tanks that empty at other times. They also satisfy

h (20) = 0

, so that condition alone cannot tell them apart.

Example 2.2.4 (differential form)

Solve $2 x y^{2} d x - (x^{2} + 5) d y = 0$ with $y (2) = 1$ .

Solution

Separate. Divide by $y^{2} (x^{2} + 5)$ , which requires $y \neq = 0$ (note that $x^{2} + 5 > 0$ always): $\frac{2 x}{x ^{2} + 5} d x - \frac{d y}{y ^{2}} = 0.$
Lost solution check. $y \equiv 0$ makes both terms zero, so it is a solution. It does not satisfy $y (2) = 1$ , so it plays no role in this problem.
Integrate. $\int \frac{2 x}{x ^{2} + 5} d x = ln (x^{2} + 5)$ and $\int - \frac{d y}{y ^{2}} = \frac{1}{y}$ . Hence $ln (x^{2} + 5) + \frac{1}{y} = c .$
Initial condition. $c = ln 9 + 1$ .
Solve. $\frac{1}{y} = 1 + ln 9 - ln (x^{2} + 5) ⟹ y = \frac{1}{1 + ln \frac{9}{x ^{2} + 5}} .$
Interval. The denominator vanishes when $x^{2} + 5 = 9 e$ , i.e. $∣ x ∣ = 9 e - 5 \approx 4.41$ . The solution exists on $- 4.41 < x < 4.41$ and blows up at the ends.

Homogeneous equations

Look at $y^{'} = \frac{x + 3 y}{3 x + y}$ . It is not separable. But divide the top and bottom by $x$ :

y^{'} = \frac{1 + 3 ( y / x )}{3 + ( y / x )} .

The right side depends only on the ratio $y / x$ . Equations with this property can be cracked.

Definition 2.2.2 (Homogeneous function and equation)

A function $F (x, y)$ is homogeneous of degree $n$ if $F (t x, t y) = t^{n} F (x, y)$ for all $t \neq = 0$ (and all $(x, y)$ where both sides are defined).

The equation $y^{'} = f (x, y)$ is called homogeneous if $f$ is homogeneous of degree zero:

f (t x, t y) = f (x, y) .

The equation $M d x + N d y = 0$ is homogeneous if $M$ and $N$ are homogeneous of the same degree.

The two definitions agree. If $M$ and $N$ both have degree $n$ , then $f = - M / N$ satisfies

f (t x, t y) = - \frac{t ^{n} M ( x , y )}{t ^{n} N ( x , y )} = f (x, y),

so $f$ has degree zero. A quick way to spot homogeneity in polynomials: every term of $M$ and of $N$ has the same total degree. In $(x^{2} + 3 x y) d x + (y^{2} - 5 x^{2}) d y$ every term has degree 2, so it is homogeneous. In $(x^{2} + y) d x + x d y$ the terms have degrees 2, 1, 1, so it is not.

Two meanings of "homogeneous"

This use of the word has nothing to do with a "homogeneous linear equation" (one with zero right-hand side), which you will meet in Chapter 4. Same word, different idea.

Why the ratio $y / x$ appears

Take $f$ homogeneous of degree zero and $x \neq = 0$ . Choose $t = 1/ x$ in the definition:

f (x, y) = f (\frac{x}{x}, \frac{y}{x}) = f (1, \frac{y}{x}) .

So $f$ really is a function of $y / x$ alone. This suggests making $v = y / x$ the new unknown.

Theorem 2.2.1 (The substitution

y = vx

)

If $y^{'} = f (x, y)$ is homogeneous, the substitution $y = vx$ , where $v$ is a new unknown function of $x$ , transforms it into the separable equation

x \frac{d v}{d x} = f (1, v) - v .

The derivation takes three short steps.

Differentiate the substitution. $y = vx$ is a product, so by the product rule $y^{'} = v + x v^{'} .$
Rewrite the right side. Since $f (x, y) = f (1, y / x) = f (1, v)$ , the equation becomes $v + x v^{'} = f (1, v) .$
Isolate $v^{'}$ . $x v^{'} = f (1, v) - v$ . The right side depends only on $v$ , and the factor $x$ only on $x$ : this is separable. Separating (where $f (1, v) - v \neq = 0$ and $x \neq = 0$ ): $\frac{d v}{f ( 1 , v ) - v} = \frac{d x}{x} .$

Lost solutions appear again. Every root $v_{0}$ of $f (1, v) - v = 0$ gives a constant $v \equiv v_{0}$ , that is, a straight line $y = v_{0} x$ through the origin that solves the equation.

Method in a nutshell

Check that $f (t x, t y) = f (x, y)$ (or that $M$ and $N$ have the same degree).
Substitute $y = vx$ and $y^{'} = v + x v^{'}$ . Simplify the right side to a function of $v$ .
Separate: $\frac{d v}{f ( 1 , v ) - v} = \frac{d x}{x}$ , and integrate.
Replace $v$ by $y / x$ .
Find the roots $v_{0}$ of $f (1, v) - v = 0$ ; the lines $y = v_{0} x$ are solutions. Note which are not in your family.

Two practical tips. First, simplify $f (1, v) - v$ completely before separating; the resulting fraction is usually ready for partial fractions. Second, if $N$ is much simpler than $M$ , the substitution $x = u y$ (with $u$ a function of $y$ ) works in exactly the same way and sometimes leads to easier integrals.

The solver below treats $y^{'} = \frac{a x + b y}{c x + d y}$ with the substitution $y = vx$ . Change the four coefficients and watch how the separated integral and the line solutions change.

Worked examples: homogeneous equations

Example 2.2.5

Solve $y^{'} = \frac{x + 3 y}{3 x + y}$ .

Solution

Homogeneous? Numerator and denominator are both of degree 1. Yes.
Substitute $y = vx$ , $y^{'} = v + x v^{'}$ : $v + x v^{'} = \frac{x + 3 vx}{3 x + vx} = \frac{1 + 3 v}{3 + v} .$
Isolate $x v^{'}$ . $x v^{'} = \frac{1 + 3 v}{3 + v} - v = \frac{1 + 3 v - 3 v - v ^{2}}{3 + v} = \frac{1 - v ^{2}}{3 + v} .$
Separate (for $v \neq = \pm 1$ ): $\frac{3 + v}{1 - v ^{2}} d v = \frac{d x}{x}$ .
Partial fractions. Write $\frac{3 + v}{( 1 - v ) ( 1 + v )} = \frac{A}{1 - v} + \frac{B}{1 + v}$ . Then $3 + v = A (1 + v) + B (1 - v)$ . Putting $v = 1$ gives $A = 2$ ; putting $v = - 1$ gives $B = 1$ .
Integrate. $- 2 ln ∣1 - v ∣ + ln ∣1 + v ∣ = ln ∣ x ∣ + c_{1} ⟹ \frac{1 + v}{( 1 - v ) ^{2}} = K x .$
Back to $y$ . With $v = y / x$ : $\frac{1 + v}{( 1 - v ) ^{2}} = \frac{( x + y ) / x}{( x - y ) ^{2} / x ^{2}} = \frac{x ( x + y )}{( x - y ) ^{2}}$ . Setting this equal to $K x$ and cancelling $x$ : $x + y = K (x - y)^{2} .$
Line solutions. $1 - v^{2} = 0$ gives $v = \pm 1$ , i.e. $y = x$ and $y = - x$ . The line $y = - x$ is the case $K = 0$ . The line $y = x$ (check: $y^{'} = 1$ and $\frac{4 x}{4 x} = 1$ ) is not in the family; it is a lost solution.

Figure 2.2.3. Solution curves

x + y = K (x - y)^{2}

of Example 2.2.5. They are parabolas through the origin whose common axis is the lost solution

y = x

; the line

y = - x

is the case

K = 0

. Because the equation is homogeneous, enlarging a solution curve by any factor about the origin gives another solution curve.

Example 2.2.6

Solve $(y^{2} + 3 x y) d x - x^{2} d y = 0$ with $y (1) = 1$ .

Solution

Homogeneous? $M = y^{2} + 3 x y$ and $N = - x^{2}$ are both of degree 2. Yes. In derivative form, $y^{'} = \frac{y ^{2} + 3 x y}{x ^{2}} = (\frac{y}{x})^{2} + 3 \frac{y}{x} .$
Substitute. $v + x v^{'} = v^{2} + 3 v$ , so $x v^{'} = v^{2} + 2 v = v (v + 2)$ .
Separate (for $v \neq = 0, - 2$ ): $\frac{d v}{v ( v + 2 )} = \frac{d x}{x}$ .
Integrate. Since $\frac{1}{v ( v + 2 )} = \frac{1}{2} (\frac{1}{v} - \frac{1}{v + 2})$ , $\frac{1}{2} ln \frac{v}{v + 2} = ln ∣ x ∣ + c_{1} ⟹ \frac{v}{v + 2} = K x^{2} .$
Back to $y$ . $\frac{v}{v + 2} = \frac{y}{y + 2 x}$ . So $y = K x^{2} (y + 2 x)$ , and solving for $y$ : $y (1 - K x^{2}) = 2 K x^{3} ⟹ y = \frac{2 K x ^{3}}{1 - K x ^{2}} .$
Line solutions. $v = 0$ gives $y \equiv 0$ (the case $K = 0$ ). $v = - 2$ gives $y = - 2 x$ , which is not in the family.
Initial condition. $1 = \frac{2 K}{1 - K}$ gives $K = \frac{1}{3}$ , so $y = \frac{2 x ^{3}}{3 - x ^{2}} .$ The derivative form needs $x \neq = 0$ and the formula needs $x \neq = \pm 3$ ; the interval containing $x = 1$ is $0 < x < 3$ .

Check your answer

For Example 2.2.6: $y = \frac{2 x ^{3}}{3 - x ^{2}}$ gives $y^{'} = \frac{6 x ^{2} ( 3 - x ^{2} ) + 4 x ^{4}}{( 3 - x ^{2} ) ^{2}} = \frac{18 x ^{2} - 2 x ^{4}}{( 3 - x ^{2} ) ^{2}}$ , while $\frac{y ^{2} + 3 x y}{x ^{2}} = \frac{y ( y + 3 x )}{x ^{2}} = \frac{2 x ^{3} ( 9 x - x ^{3} )}{x ^{2} ( 3 - x ^{2} ) ^{2}} = \frac{18 x ^{2} - 2 x ^{4}}{( 3 - x ^{2} ) ^{2}}$ . They match. A thirty-second check like this catches most algebra slips.

Exercises

In Exercises 1–7, solve the separable equation. Mention any constant solutions not contained in your formula.

$y^{'} = 4 x^{3} y$
Answer
$y = c e^{x^{4}}$ (the solution $y \equiv 0$ is $c = 0$ ).
$y^{2} y^{'} = x^{2} + 1$
Answer
$y^{3} = x^{3} + 3 x + c$ .
$y^{'} = 3 e^{x + y}$
Answer
$y = - ln (c - 3 e^{x})$ , where $c - 3 e^{x} > 0$ .
$y^{'} = (y - 1) (y - 5)$
Answer
$y = \frac{5 - K e ^{4 x}}{1 - K e ^{4 x}}$ ( $K = 0$ gives $y \equiv 5$ ); also the lost solution $y \equiv 1$ .
$y^{'} = (1 + y^{2}) cos x$ , $y (0) = 1$
Answer
$y = tan (sin x + \frac{π}{4})$ , valid while $sin x + \frac{π}{4} < \frac{π}{2}$ .
$y^{'} = (1 + 2 x) y$ , $y (0) = 5$
Answer
$y = 5 e^{x + x^{2}}$ .
$e^{x} (y^{2} + 4) d x - 2 (e^{x} + 1) d y = 0$
Answer
$ln (1 + e^{x}) - arctan \frac{y}{2} = c$ , i.e. $y = 2 tan (ln (1 + e^{x}) - c)$ .
Solve the homogeneous equation $y^{'} = \frac{x - y}{x + 3 y}$ .
Answer
$3 y^{2} + 2 x y - x^{2} = c$ . (The equation is also exact; check that both methods agree.)
Solve $y^{'} = \frac{2 x ^{2} + y ^{2}}{x y}$ .
Answer
$y^{2} = 4 x^{2} ln ∣ x ∣ + c x^{2}$ .
Solve $x y^{'} = y + 2 x e^{- y / x}$ , $y (1) = 0$ .
Answer
$y = x ln (1 + 2 ln x)$ , for $x > e^{- 1/2}$ .
Which of these functions are homogeneous of degree zero? (a) $\frac{x ^{3} + y ^{3}}{x y ^{2}}$ (b) $\frac{x ^{2} + y}{x ^{2}}$
Answer
(a) yes; (b) no, since $\frac{t ^{2} x ^{2} + t y}{t ^{2} x ^{2}} \neq = \frac{x ^{2} + y}{x ^{2}}$ in general.
Consider $y^{'} = 4 y^{3/4}$ , $y (0) = 0$ . Show that both $y \equiv 0$ and $y = x^{4}$ (for $x \geq 0$ ) are solutions. Which one does separation of variables find, and why is the other one missed?
Answer
Separation gives $y^{1/4} = x + c$ , so $y = (x + c)^{4}$ with $x + c \geq 0$ ; $c = 0$ gives $y = x^{4}$ . The solution $y \equiv 0$ is lost because we divided by $y^{3/4}$ . Uniqueness fails because $\partial f / \partial y = 3 y^{- 1/4}$ is unbounded near $y = 0$ .

Separable equations

Why separation works

The danger: lost solutions

A word about constants

Worked examples: separable equations

Homogeneous equations

Why the ratio y/x appears

Worked examples: homogeneous equations

Exercises

Why the ratio $y / x$ appears