Differential Equations · Lesson 3 of 6

Separation of Variables and Growth Models

“Separate variable-dependent factors, integrate with care, and build curve and continuous-growth models.”

Learning Objectives

• Recognise and solve a first-order equation with separable variables. • Explain the integration step and interpret the arbitrary constant. • Use a point or an initial value to select a particular solution. • Translate tangent-slope and rate statements into differential equations. • Model continuous interest, population growth, and constant volume increase. • Check excluded factors, constant solutions, domains, and inverse-function branches.

Turn a rate rule into an integral

We now move from checking proposed solutions to finding them. The simplest useful method works when the dependence on the two variables can be separated into factors. Once that separation is made, each side can be integrated using one variable only.

For a first-order equation $\frac{d y}{d x} = F (x, y)$ , suppose $F (x, y) = g (x) h (y)$ . If $h (y) \neq = 0$ , division by $h (y)$ puts every factor involving $y$ beside its differential and every factor involving $x$ on the other side. The aim is complete separation, not merely moving some symbols across the equals sign.

Definition

Separable equation

A first-order equation is separable when it can be written as $\frac{d y}{d x} = g (x) h (y)$ and therefore, where the division is valid, as $\frac{1}{h ( y )} d y = g (x) d x$ .

The reason this method works is the chain rule. If $H^{'} (y) = 1/ h (y)$ , then $\frac{d}{d x} H (y (x)) = \frac{1}{h ( y )} \frac{d y}{d x} = g (x)$ . Therefore $H (y)$ differs from an antiderivative of $g (x)$ by a constant. Treating the differential notation as a separation is a convenient way to express this calculation.

Separated integralLaTeX

\int \frac{1}{h ( y )} d y = \int g (x) d x + C

Here

g

depends only on

x

h

only on

y

, and

C

is one arbitrary constant. A constant from each antiderivative can be combined into this single constant.

The final result may be implicit. An equation such as $H (y) = G (x) + C$ can already describe a solution family; you do not have to force it into an explicit formula if doing so obscures the result. After integrating, check whether any division excluded solutions and whether the proposed interval avoids undefined expressions.

Example — Separate polynomial expressions

Problem
Find the general solution of $\frac{d y}{d x} = \frac{x + 1}{2 - y}$ , where $y \neq = 2$ .

1.Multiply by $(2 - y) d x$ to obtain $(2 - y) d y = (x + 1) d x$ . Each side now involves only its own variable.
2.Integrate term by term: $2 y - \frac{y ^{2}}{2} = \frac{x ^{2}}{2} + x + C_{1}$ .
3.Multiply by $2$ and rearrange: $x^{2} + y^{2} + 2 x - 4 y + 2 C_{1} = 0$ .
4.Rename $2 C_{1}$ as $C$ . The implicit family is $x^{2} + y^{2} + 2 x - 4 y + C = 0$ , understood on solution portions where $y \neq = 2$ .
5.Differentiating this implicit family gives $(2 y - 4) y^{'} = - 2 x - 2$ , which recovers the original quotient when $y \neq = 2$ .

Example — Use inverse-tangent antiderivatives

Problem
Solve $\frac{d y}{d x} = \frac{1 + y ^{2}}{1 + x ^{2}}$ .

1.Since $1 + y^{2} > 0$ for real $y$ , divide safely to obtain $\frac{d y}{1 + y ^{2}} = \frac{d x}{1 + x ^{2}}$ .
2.Integrating gives $tan^{- 1} y = tan^{- 1} x + C$ .
3.This implicit relation is sufficient. Where an explicit branch is valid, it gives $y = tan (tan^{- 1} x + C)$ . Intervals containing a tangent pole must be excluded.

Apply a value and retain excluded solutions

A condition such as $y (0) = 1$ is applied after integration to find the constant. However, the algebra used to separate variables can exclude entire solutions. Check those excluded values in the original equation before deciding that the family is complete.

Example — Choose a particular rational solution

Problem
Solve $y^{'} = - 4 x y^{2}$ with $y (0) = 1$ .

1.On a nonzero solution, separate to obtain $y^{- 2} d y = - 4 x d x$ .
2.Integration gives $- \frac{1}{y} = - 2 x^{2} + C$ , or $y = \frac{1}{2 x ^{2} - C}$ .
3.At $x = 0$ , $y = 1$ , so $1 = - 1/ C$ and $C = - 1$ . Thus $y = \frac{1}{2 x ^{2} + 1}$ .
4.The denominator is positive for all real $x$ . Differentiation gives $y^{'} = - 4 x / (2 x^{2} + 1)^{2} = - 4 x y^{2}$ .
5.The constant solution $y = 0$ also satisfies the original equation, but it does not satisfy the condition $y (0) = 1$ .

Division can lose a constant solution

For $y^{'} = g (x) h (y)$ , every real zero $y = a$ of $h (y)$ should be checked as a possible constant solution. Dividing by $h (y)$ excludes those values. Also, $\int d y / y = ln ∣ y ∣$ , not simply $ln y$ unless $y > 0$ is known.

Example — A constant solution and a family

Problem
Find the solutions of $y^{'} = 1 - y$ .

1.First notice that $y = 1$ gives derivative zero and right-hand side zero, so it is a constant solution.
2.For $y \neq = 1$ , separate: $\frac{d y}{1 - y} = d x$ . Integrate: $- ln ∣1 - y ∣ = x + C_{1}$ .
3.Exponentiating and absorbing sign and scale into a real constant gives $1 - y = C e^{- x}$ . Hence $y = 1 - C e^{- x}$ .
4.Allowing $C = 0$ in this final family restores the constant solution $y = 1$ . Never discard it simply because an intermediate logarithm excluded it.

Find a curve from its tangent slope

At a point $(x, y)$ , the derivative $d y / d x$ is the slope of the tangent when the curve is written as $y (x)$ . A verbal rule about this slope therefore supplies a differential equation. A point on the curve then determines which member of the solution family is wanted.

Example — Integrate a slope depending only on position

Problem
Find the curve through $(1, 1)$ satisfying $x d y = (2 x^{2} + 1) d x$ , with $x \neq = 0$ .

1.Divide by $x$ to write $d y = (2 x + 1/ x) d x$ .
2.Integrate: $y = x^{2} + ln ∣ x ∣ + C$ .
3.At $(1, 1)$ , $1 = 1 + 0 + C$ , so $C = 0$ . The required curve is $y = x^{2} + ln ∣ x ∣$ .
4.The solution interval containing the specified point lies in $x > 0$ ; the equation cannot be continued across $x = 0$ in this form.

Example — Translate the slope statement first

Problem
The slope at $(x, y)$ is $2 x / y^{2}$ . Find the curve through $(- 2, 3)$ .

1.The slope statement gives $y^{'} = 2 x / y^{2}$ , with $y \neq = 0$ . Separate: $y^{2} d y = 2 x d x$ .
2.Integrate: $y^{3} /3 = x^{2} + C$ .
3.Substitute the point: $27/3 = 4 + C$ , so $C = 5$ . Thus $y^{3} = 3 x^{2} + 15$ , or $y = (3 x^{2} + 15)^{1/3}$ .
4.The expression inside the cube root is positive. Differentiating the implicit result gives $3 y^{2} y^{'} = 6 x$ , the required slope rule.

A slope can also be described relative to a line segment. The slope of the segment from $(- 4, - 3)$ to $(x, y)$ is $(y + 3) / (x + 4)$ when $x \neq = - 4$ . “Twice this slope” means $y^{'} = 2 (y + 3) / (x + 4)$ , a separable equation. Translating the geometry correctly is often the most important step.

Continuous percentage growth

When the rate of increase is proportional to the amount already present, equal time intervals produce equal multiplying factors. This is a continuous model: the rate law acts at every instant. Time units and percentage units must be converted before integration.

Let $P (t)$ be an amount, and let $k$ be the proportionality constant per unit time. The statement “rate is proportional to the amount” becomes $d P / d t = k P$ . For a positive amount, separation gives $d P / P = k d t$ , so $ln P = k t + C_{1}$ . Exponentiating and using $P (0) = P_{0}$ fixes the positive multiplier.

Continuous proportional growthLaTeX

P (t) = P_{0} e^{k t}

P_{0}

is the initial amount and

k

is the rate constant. A continuously compounded rate of

r

percent per year means

k = r /100

when

t

is measured in years.

Equal time gives equal growth factors— The curve grows faster in absolute amount as the principal increases, although its proportional rate stays fixed.

Example — Find a doubling time

Problem
A principal of ₹1000 increases continuously at 5 percent per year. When will it double?

1.The rate law is $d P / d t = 0.05 P = P /20$ , with $P (0) = 1000$ .
2.The model is $P = 1000 e^{t /20}$ . For doubling, set $2000 = 1000 e^{t /20}$ .
3.Divide by $1000$ and take the natural logarithm: $ln 2 = t /20$ . Thus $t = 20 ln 2 \approx 13.86$ years.
4.The initial principal cancels. Under this same proportional rate, every positive initial principal has the same doubling time.

Example — Infer a rate from two measurements

Problem
A bacterial culture initially contains 100000 bacteria and grows by 10 percent in two hours. Assuming rate proportional to count, when will it reach 200000?

1.Write $N (t) = 100000 e^{k t}$ , with $t$ in hours.
2.After two hours, $110000 = 100000 e^{2 k}$ , so $k = ln (1.1) /2$ .
3.For doubling, $2 = e^{k t}$ , so $t = ln 2/ k = \frac{2 l n 2}{l n 1.1} \approx 14.55$ hours from the initial measurement.
4.A 10 percent increase in two hours is a multiplying factor of $1.1$ ; it does not mean that the continuous rate constant is exactly $0.05$ per hour.

Identify which quantity changes at a constant rate

A constant rate for one quantity does not imply a constant rate for every related quantity. If volume increases uniformly, radius does not: a larger spherical balloon needs more added volume for the same increase in radius. Write the rate law for the quantity named in the problem before changing variables.

Example — An inflating spherical balloon

Problem
A spherical balloon has constant volume-increase rate. Its radius is 3 units initially and 6 units after 3 seconds. Find its radius after t seconds.

1.Let $V$ be volume. A constant rate gives $d V / d t = K$ , so $V = K t + V_{0}$ .
2.Since $V = (4 π /3) r^{3}$ , the cube of the radius is linear in time: $r^{3} = A t + 27$ .
3.At $t = 3$ , $r = 6$ , so $216 = 3 A + 27$ . Thus $A = 63$ .
4.The model is $r (t) = (27 + 63 t)^{1/3}$ for $t \geq 0$ in this inflation scenario.
5.Differentiating $r^{3} = 27 + 63 t$ gives $3 r^{2} d r / d t = 63$ . Therefore $d r / d t = 21/ r^{2}$ , which decreases as the balloon grows.

Volume and radius behave differently— The straight graph is for the cube of the radius. The radius itself follows a curved graph.

Rearrange before deciding the method

An equation may be separable without looking like a product at first. Exponentials, trigonometric identities, and inverse functions can reveal the required form. Those transformations must preserve the real domain and all relevant branches.

Example — A logarithm of the derivative

Problem
Solve $ln (y^{'}) = 3 x + 4 y$ with $y (0) = 0$ .

1.The logarithm requires $y^{'} > 0$ . Exponentiate to obtain $y^{'} = e^{3 x} e^{4 y}$ .
2.Separate: $e^{- 4 y} d y = e^{3 x} d x$ . Integrate to get $- \frac{1}{4} e^{- 4 y} = \frac{1}{3} e^{3 x} + C$ .
3.At $(0, 0)$ , $- 1/4 = 1/3 + C$ , giving $C = - 7/12$ . Rearrangement gives $4 e^{3 x} + 3 e^{- 4 y} = 7$ .
4.An explicit version is $y = - \frac{1}{4} ln (\frac{7 - 4 e ^{3 x}}{3})$ . For the interval containing zero, require $x < \frac{1}{3} ln (7/4)$ .
5.On this interval, differentiating gives $y^{'} = e^{3 x + 4 y} > 0$ , consistent with the original logarithm.

Example — An inverse-cosine equation has branches

Problem
Find real differentiable solutions of $cos (y^{'}) = a$ passing through $(0, 1)$ .

1.Real solutions require $- 1 \leq a \leq 1$ . A cosine equation does not uniquely determine its angle.
2.The possible constant slopes are $m = 2 nπ \pm cos^{- 1} a$ , where $n$ is an integer. On a connected differentiable solution interval, the derivative stays on one such branch.
3.Integrate $y^{'} = m$ to obtain $y = m x + C$ . The point $(0, 1)$ gives $C = 1$ .
4.Thus the real solution lines are $y = 1 + m x$ with slopes from that set. If $∣ a ∣ > 1$ , there is no real solution.
5.The equation has order one but undefined degree as originally written. Solving it after a trigonometric rearrangement does not change that original classification.

One integration constant is enough

If separation gives $H (y) + C_{1} = G (x) + C_{2}$ , rewrite it as $H (y) = G (x) + C$ with $C = C_{2} - C_{1}$ . Do not retain two independent arbitrary constants for this first-order integration.

Check your understanding

Choose a valid separation before doing any integration. For applications, first identify the changing quantity and the time unit, then use the supplied information to determine constants.

Quick check

Which is a correct separation of $y^{'} = x y^{2}$ for $y \neq = 0$ ?

Quick check

Which solution may be lost when dividing $y^{'} = y (1 - y)$ by $y (1 - y)$ ?

Quick check

If $P^{'} = 0.05 P$ , with time in years, what is the doubling time?

Quick check

For $y = x^{2} + ln ∣ x ∣ + C$ passing through $(1, 1)$ , what is $C$ ?

Quick check

Which quantity is linear in time when a spherical balloon has constant volume-increase rate?

Quick check

Which pair gives a general solution of $y^{'} = \frac{1 + y ^{2}}{1 + x ^{2}}$ ?

Quick check

What does $ln (y^{'}) = 3 x + 4 y$ imply?

Quick check

If a positive quantity grows by 10 percent in two hours, which relation determines k in $N = N_{0} e^{k t}$ ?

Practice Problems

Solve $y^{'} = (1 - cos x) / (1 + cos x)$ on an interval where the denominator is nonzero. Use a half-angle identity.
Solve $y^{'} = 4 - y^{2}$ for $- 2 < y < 2$ . Also identify constant solutions of the unrestricted equation.
Solve $sec^{2} x tan y d x + sec^{2} y tan x d y = 0$ where the trigonometric functions are defined.
Solve $(e^{x} + e^{- x}) d y = (e^{x} - e^{- x}) d x$ .
Solve $y^{'} = (1 + x^{2}) (1 + y^{2})$ and state an implicit answer.
Solve $y ln y d x - x d y = 0$ for $y > 0$ on an interval where $x \neq = 0$ .
Solve $x^{5} y^{'} = - y^{5}$ for nonzero x and y, then check whether y = 0 is also a solution.
Find the curve through $(0, 0)$ with $y^{'} = e^{x} sin x$ .
Find the curve through $(0, - 2)$ for which the tangent slope multiplied by the y-coordinate equals the x-coordinate.
Find the curve through $(- 2, 1)$ whose tangent slope is twice the slope of the segment joining $(x, y)$ to $(- 4, - 3)$ .
A continuously increasing principal doubles in ten years. Find the percentage rate per year, using $ln 2 = 0.6931$ .
Find the value after ten years of ₹1000 at a continuous rate of 5 percent per year, using $e^{0.5} = 1.648$ .
Solve $e^{x} tan y d x + (1 - e^{x}) sec^{2} y d y = 0$ where the expressions are defined.
Solve $x (x^{2} - 1) y^{'} = 1$ with $y (2) = 0$ . State an interval containing the given point.
Solve $y^{'} = sin^{- 1} x$ on $- 1 < x < 1$ . Use integration by parts.
Solve $(x^{3} + x^{2} + x + 1) y^{'} = 2 x^{2} + x$ with $y (0) = 1$ . Factor the cubic before using partial fractions.
Solve $y^{'} = y tan x$ with $y (0) = 1$ .
Find the solution curve of $x y y^{'} = (x + 2) (y + 2)$ through $(1, - 1)$ .
Solve $y^{'} = e^{x + y}$ and check the signs after integration.

The quotient is $tan^{2} (x /2) = sec^{2} (x /2) - 1$ . Integrate to get $y = 2 tan (x /2) - x + C$ .

Key Takeaways

• Separation turns a product rate rule into two one-variable integrals. • Keep one arbitrary constant, then use a given value or point to determine it. • Check constant solutions before dividing by a variable-dependent factor. • Tangent slope is $d y / d x$ ; a continuous proportional growth rate gives an exponential model. • Constant volume increase makes the cube of a sphere’s radius linear in time. • Rearrangements, logarithms, and inverse functions require valid domains and branches.

Previous · Lesson 2

Solutions, Families, and Initial Conditions

Next · Lesson 4

Homogeneous Equations and Ratio Substitutions

Turn a rate rule into an integral

Apply a value and retain excluded solutions

Find a curve from its tangent slope

Continuous percentage growth

Identify which quantity changes at a constant rate

Rearrange before deciding the method

Check your understanding

Quiz

Practice Problems

Key Takeaways