IB Analysis and Approaches HL · Integral calculus · Section 5 of 13

Separable differential equations

3 worked examples · By Álvaro López

Key idea

A differential equation relates a function to its derivative, and its solution is a function. They appear when you know how a quantity changes (growth, cooling, decay) and want the quantity itself. Those with separable variables are solved by integrating, and they are among the types most often asked in IB exams.

What a differential equation is

A first-order differential equation is an equality involving $x$ , a function $y (x)$ and its first derivative $\frac{d y}{d x}$ . You can almost always isolate the derivative and write $\frac{d y}{d x} = F (x, y) .$ For example, $x \frac{d y}{d x} = 2 y$ or $\frac{d y}{d x} = - \frac{x}{y}$ .

Seeing the equation: the slope field

The equation $\frac{d y}{d x} = F (x, y)$ tells you the gradient of any solution at each point $(x, y)$ . If you draw a short tangent segment at many points, you obtain the slope field, and each solution is a curve that follows those segments. For $\frac{d y}{d x} = - \frac{x}{y}$ the solutions are circles centred at the origin; the one through $(0, 3)$ is the upper semicircle of radius $3$ .

Slope field of dy/dx = -x/y: circles centred at the origin; the solution through (0, 3) is the upper semicircle of radius 3

Separable variables

The idea is to put everything involving $y$ with $d y$ on one side, and everything involving $x$ with $d x$ on the other, and then integrate each side with respect to its own variable.

Dividing by $h (y)$ loses the constant solutions $y = k$ with $h (k) = 0$ . In IB problems the initial condition usually rules them out, but it is worth keeping them in mind.

Context problems

Many situations are modelled by an equation of this type. The most common is "the rate of change of $y$ is proportional to $y$ ", $\frac{d y}{d t} = k y$ , with solution $y = y_{0} e^{k t}$ . In Newton's law of cooling, the temperature $T$ of an object in surroundings at temperature $T_{a}$ satisfies $\frac{d T}{d t} = - k (T - T_{a})$ , with $k > 0$ : the object cools faster the greater the difference from its surroundings.

Worked examples

Example 1 Basic

Proportional growth

Solve $\frac{d y}{d x} = 2 x y$ , with $y > 0$ , given that $y = 3$ when $x = 0$ . Give the solution in the form $y = f (x)$ .

Solution

Separate the variables: $\frac{d y}{y} = 2 x d x$ . Integrate: $ln y = x^{2} + C .$ With $y = 3$ at $x = 0$ : $ln 3 = C$ . Then $ln y = x^{2} + ln 3$ , and taking exponentials, $y = 3 e^{x^{2}} .$

Example 2 Medium

With a tangent in the result

Solve $\frac{d y}{d x} = x (1 + y^{2})$ given that $y = 1$ when $x = 0$ . Give the solution as $y = f (x)$ , for values of $x$ close to $0$ .

Solution

Separate the variables: $\frac{d y}{1 + y ^{2}} = x d x$ . Integrate: $arctan y = \frac{x ^{2}}{2} + C .$ With $y = 1$ at $x = 0$ : $arctan 1 = \frac{π}{4} = C$ . Then $arctan y = \frac{x ^{2}}{2} + \frac{π}{4}$ and $y = tan (\frac{x ^{2}}{2} + \frac{π}{4}) .$

Example 3 Hard

Cooling a coffee

The temperature $T$ (in degrees Celsius) of a coffee $t$ minutes after it is poured satisfies the equation $\frac{d T}{d t} = - k (T - 20), k > 0,$ where $20$ is the room temperature. At the initial instant the coffee is at $80$ degrees and after $5$ minutes it is at $50$ degrees.

a) Find the exact value of $k$ .

b) Find the temperature after $10$ minutes.

c) Find the instant at which the coffee reaches $25$ degrees.

Solution

Separate the variables, with $T > 20$ : $\frac{d T}{T - 20} = - k d t$ . Integrate: $ln (T - 20) = - k t + C \Rightarrow T - 20 = A e^{- k t}, A = e^{C} .$ With $T (0) = 80$ : $A = 60$ . So $T = 20 + 60 e^{- k t}$ .

a) $T (5) = 50 \Rightarrow 30 = 60 e^{- 5 k} \Rightarrow e^{- 5 k} = \frac{1}{2} \Rightarrow k = \frac{ln 2}{5}$ .

b) $e^{- 10 k} = (e^{- 5 k})^{2} = \frac{1}{4}$ , so $T (10) = 20 + 60 \cdot \frac{1}{4} = 35$ degrees.

c) $T = 25 \Rightarrow 5 = 60 e^{- k t} \Rightarrow e^{- k t} = \frac{1}{12} \Rightarrow t = \frac{ln 12}{k} = \frac{5 ln 12}{ln 2}$ , approximately $17.9$ minutes.

In the exam: each step is marked separately. After integrating, use the initial condition before solving for anything else: if you lose the constant $A$ , you cannot continue. Keep $k$ exact until the end and round only the numerical answer.

Common mistakes

Summary

A first-order differential equation relates $x$ , $y$ and $\frac{d y}{d x}$ ; its solution is a function. The general solution includes $C$ ; the particular solution comes from an initial condition.
Separable variables: $\frac{d y}{d x} = g (x) h (y)$ , then $\int \frac{d y}{h ( y )} = \int g (x) d x$ .
A single constant $C$ ; find it from the initial condition and solve for $y$ at the end.
Models: $\frac{d y}{d t} = k y$ gives $y_{0} e^{k t}$ ; Newton's law of cooling gives $T = T_{a} + (T_{0} - T_{a}) e^{- k t}$ .
The slope field shows the direction of the solutions at each point.

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