First-Order Differential Equations & Euler’s Method

Figure 5.4: Euler’s Numerical Step Method vs. Exact Solution Curve

1. Foundational Anatomy of Differential Equations

A first-order ordinary differential equation (ODE) expresses a mathematical relationship between an independent variable $x$, an unknown function $y(x)$, and its first derivative $\frac{dy}{dx}$:

$$\frac{dy}{dx} = f(x, y)$$

General Solution: A family of functions $y = g(x, C)$ containing an arbitrary constant of integration $C$, representing an infinite family of solution curves.
Particular Solution: A unique curve selected by applying an Initial Value Condition $(x_0, y_0)$ to solve for the specific numerical value of $C$.

2. Analytical Method 1: Separation of Variables (Topic 5.16)

An ODE is separable if the derivative factors into a product of a purely $x$-dependent function and a purely $y$-dependent function:

$$\frac{dy}{dx} = g(x) \, h(y) \quad \iff \quad \frac{1}{h(y)} \, dy = g(x) \, dx$$

Integrating both sides with respect to their respective variables yields:

$$\int \frac{1}{h(y)} \, dy = \int g(x) \, dx + C$$

Immediate Constant Placement Rule

Always add the constant of integration $+ C$ immediately upon integration. If the left side produces $\ln|y|$, exponentiate the entire sum:

$$\ln|y| = G(x) + C \implies |y| = e^{G(x) + C} = e^C \cdot e^{G(x)} \implies y = A e^{G(x)} \quad (A = \pm e^C)$$

Never write $y = e^{G(x)} + C$, which is mathematically invalid!

3. Analytical Method 2: Homogeneous Equations via Substitution $y = vx$ (Topic 5.16 AHL)

A differential equation is homogeneous of degree zero if $\frac{dy}{dx}$ can be written strictly as a function of the ratio $\frac{y}{x}$:

$$\frac{dy}{dx} = F\left(\frac{y}{x}\right)$$

The Homogeneous Solution Protocol:

Substitute: Let $y = v \cdot x$, where $v = v(x)$ is a new dependent variable.
Differentiate via Product Rule: $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Equate and Separate: $$v + x\frac{dv}{dx} = F(v) \implies x\frac{dv}{dx} = F(v) - v \implies \int \frac{1}{F(v) - v}\,dv = \int \frac{1}{x}\,dx$$
Integrate and Back-Substitute: Evaluate the integrals, then substitute $v = \frac{y}{x}$ to obtain the solution in terms of $x$ and $y$.

4. Analytical Method 3: First-Order Linear ODEs & Integrating Factors (Topic 5.16)

A linear first-order differential equation has the canonical standard form:

$$\frac{dy}{dx} + P(x)\,y = Q(x)$$

Critical Requirement: The leading coefficient of $\frac{dy}{dx}$ MUST be equal to $1$. If given $A(x)\frac{dy}{dx} + B(x)y = C(x)$, you must divide the entire equation by $A(x)$ before identifying $P(x)$ and $Q(x)$.

To solve, multiply through by the Integrating Factor $I(x)$ defined as:

$$I(x) = e^{\int P(x) \, dx}$$

By the product rule, the left-hand side condenses into the exact derivative of the product $y \cdot I(x)$:

$$\frac{d}{dx}\left[ y \cdot I(x) \right] = Q(x) \cdot I(x)$$

Integrating both sides with respect to $x$ yields the complete general solution:

$$y \cdot I(x) = \int Q(x) \, I(x) \, dx + C \implies y = \frac{1}{I(x)}\left( \int Q(x) \, I(x) \, dx + C \right)$$

5. Numerical Method: Euler’s Step Method & Error Analysis (Topic 5.17)

When an ODE $\frac{dy}{dx} = f(x, y)$ cannot be solved analytically, Euler’s Method constructs a piecewise-linear approximation over equal increments of step size $h$:

$$\begin{cases} x_{n+1} = x_n + h \\ y_{n+1} = y_n + h \cdot f(x_n, y_n) \end{cases}$$

Mathematical Concavity and Under/Over-Estimation

In IB DP AA HL examinations, you are frequently asked whether an Euler approximation represents an overestimate or underestimate. This is determined solely by the second derivative $y''$:

Second Derivative	Curve Geometry	Tangent Line Placement	Euler’s Estimation Result
$y''(x) > 0$	Concave Up ($\cup$)	Tangents lie below the true curve	Underestimate
$y''(x) < 0$	Concave Down ($\cap$)	Tangents lie above the true curve	Overestimate

To evaluate $y''$, differentiate the differential equation implicitly: $\frac{d^2y}{dx^2} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$.

Worked Examples

Worked Example 1: Homogeneous ODE via Substitution $y = vx$ (Paper 1 Standard)

Solve the differential equation $x\frac{dy}{dx} = y + \sqrt{x^2 + y^2}$ for $x > 0$, given that $y(1) = 0$.

Solution:

1. Divide through by $x$ to verify homogeneous form of degree zero: [M1]

$$\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2}$$

2. Substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$: [M1]

$$v + x\frac{dv}{dx} = v + \sqrt{1 + v^2}$$

3. Subtract $v$ from both sides and separate variables: [A1]

$$x\frac{dv}{dx} = \sqrt{1 + v^2} \implies \frac{1}{\sqrt{1 + v^2}}\,dv = \frac{1}{x}\,dx \quad \text{[M1]}$$

4. Integrate both sides (using standard booklet inverse hyperbolic / log antiderivative): [A1]

$$\ln\left( v + \sqrt{1 + v^2} \right) = \ln(x) + C$$

5. Apply the initial condition $y(1) = 0 \implies v(1) = \frac{0}{1} = 0$: [M1]

$$\ln\left( 0 + \sqrt{1 + 0} \right) = \ln(1) + C \implies \ln(1) = 0 + C \implies C = 0 \quad \text{[A1]}$$

6. Exponentiate and back-substitute $v = \frac{y}{x}$: [M1]

$$v + \sqrt{1 + v^2} = x \implies \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = x$$ $$y + \sqrt{x^2 + y^2} = x^2 \implies \sqrt{x^2 + y^2} = x^2 - y$$

Square both sides: $x^2 + y^2 = x^4 - 2x^2 y + y^2 \implies x^2 = x^4 - 2x^2 y$.
Divide by $x^2$ ($x > 0$): $1 = x^2 - 2y \implies 2y = x^2 - 1 \implies y = \frac{x^2 - 1}{2}$. [A1]

Worked Example 2: Non-Normalized Linear ODE with Integrating Factor (Paper 1 Standard)

Find the particular solution to the differential equation $\cos(x)\frac{dy}{dx} + y\sin(x) = \cos^3(x)$ satisfying $y(0) = 3$, for $-\frac{\pi}{2} < x < \frac{\pi}{2}$.

Solution:

1. Divide by $\cos(x)$ to establish the standard canonical form $\frac{dy}{dx} + P(x)y = Q(x)$: [M1]

$$\frac{dy}{dx} + \tan(x)y = \cos^2(x) \quad \text{[A1]}$$

Here $P(x) = \tan(x)$ and $Q(x) = \cos^2(x)$.

2. Calculate the Integrating Factor $I(x)$: [M1]

$$I(x) = e^{\int \tan(x)\,dx} = e^{\ln|\sec(x)|} = \sec(x) \quad \text{[A1]}$$

3. Multiply the differential equation by $I(x) = \sec(x)$: [M1]

$$\frac{d}{dx}\left[ y \cdot \sec(x) \right] = \cos^2(x) \cdot \sec(x) = \cos^2(x) \cdot \frac{1}{\cos(x)} = \cos(x)$$

4. Integrate with respect to $x$: [A1]

$$y \sec(x) = \int \cos(x)\,dx = \sin(x) + C$$

5. Apply the initial condition $x = 0, y = 3$: [M1]

$$3 \sec(0) = \sin(0) + C \implies 3(1) = 0 + C \implies C = 3 \quad \text{[A1]}$$

6. Solve explicitly for $y$: [A1]

$$y = \frac{\sin(x) + 3}{\sec(x)} = \sin(x)\cos(x) + 3\cos(x) = \frac{1}{2}\sin(2x) + 3\cos(x)$$

Worked Example 3: Euler’s Method Multistep with Concavity Proof (Paper 2 Standard)

Consider the initial value problem $\frac{dy}{dx} = 2x - y$ with $y(0) = 1$.
(a) Use Euler’s method with step size $h = 0.2$ to approximate $y(0.4)$. [4 marks]
(b) By finding an expression for $\frac{d^2y}{dx^2}$, determine whether your approximation is an overestimate or an underestimate. [3 marks]

Solution:

(a) Euler Iterations:

Step 1 ($n = 0$): $x_0 = 0, y_0 = 1$.

$$f(x_0, y_0) = 2(0) - 1 = -1 \quad \text{[M1]}$$ $$y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.2(-1) = 0.8 \quad (x_1 = 0.2) \quad \text{[A1]}$$

Step 2 ($n = 1$): $x_1 = 0.2, y_1 = 0.8$.

$$f(x_1, y_1) = 2(0.2) - 0.8 = 0.4 - 0.8 = -0.4 \quad \text{[M1]}$$ $$y_2 = y_1 + h \cdot f(x_1, y_1) = 0.8 + 0.2(-0.4) = 0.8 - 0.08 = 0.72 \quad (x_2 = 0.4) \quad \text{[A1]}$$ $$\implies y(0.4) \approx 0.72$$

(b) Concavity Analysis:

Differentiate $\frac{dy}{dx} = 2x - y$ implicitly with respect to $x$: [M1]

$$\frac{d^2y}{dx^2} = \frac{d}{dx}[2x - y] = 2 - \frac{dy}{dx} = 2 - (2x - y) = 2 - 2x + y \quad \text{[A1]}$$

On the interval $x \in [0, 0.4]$, $2 - 2x \ge 2 - 2(0.4) = 1.2 > 0$ and $y > 0$.
Therefore, $\frac{d^2y}{dx^2} > 0$ throughout the interval $[0, 0.4]$.
Since the solution curve is concave up, tangent lines lie strictly below the true curve, so Euler’s method yields an underestimate. [R1]

Worked Example 4: Applied Fluid Mixing Tank (Paper 2 Section B Standard)

A tank initially contains $200\text{ litres}$ of water in which $10\text{ kg}$ of salt is dissolved. Brine containing $0.2\text{ kg/L}$ of salt flows into the tank at a rate of $5\text{ L/min}$, and the thoroughly stirred mixture drains out at the same rate.
(a) Set up a differential equation for the mass of salt $S(t)$ in kg at time $t$ minutes. [2 marks]
(b) Solve the differential equation to find $S(t)$. [4 marks]
(c) Find the mass of salt in the tank after $40$ minutes, and state the limiting mass as $t \to \infty$. [3 marks]

Solution:

(a) Rate Equation:

$$\frac{dS}{dt} = \text{Rate In} - \text{Rate Out} \quad \text{[M1]}$$ $$\text{Rate In} = 5\text{ L/min} \times 0.2\text{ kg/L} = 1\text{ kg/min}$$ $$\text{Rate Out} = 5\text{ L/min} \times \frac{S(t)}{200}\text{ kg/L} = \frac{S}{40}\text{ kg/min}$$ $$\frac{dS}{dt} = 1 - \frac{S}{40} = \frac{40 - S}{40} \quad \text{[A1]}$$

(b) Solving the ODE:

Separate variables: [M1]

$$\frac{1}{40 - S}\,dS = \frac{1}{40}\,dt \implies -\ln|40 - S| = \frac{t}{40} + C \quad \text{[A1]}$$ $$40 - S = A e^{-t/40} \quad (A = \pm e^{-C})$$

Initial condition $S(0) = 10$: $40 - 10 = A e^0 \implies A = 30$. [M1]

$$S(t) = 40 - 30e^{-t/40} \quad \text{[A1]}$$

(c) Evaluation & Asymptotic Limit:

At $t = 40$: [M1]

$$S(40) = 40 - 30e^{-40/40} = 40 - 30e^{-1} \approx 40 - 11.036 = 28.96\text{ kg} \approx 29.0\text{ kg} \quad \text{[A1]}$$

As $t \to \infty$, $e^{-t/40} \to 0$:

$$\lim_{t \to \infty} S(t) = 40 - 0 = 40\text{ kg} \quad \text{[A1]}$$

(Check: Volume $200\text{ L} \times \text{inflow concentration } 0.2\text{ kg/L} = 40\text{ kg}$.)

Worked Example 5: Integrating Factor with Initial Value Problem (Paper 2 Standard)

Solve the initial value problem $x \frac{dy}{dx} + 2y = \cos x$ for $x > 0$, given that $y(\pi) = 0$. [7 marks]

Solution:

Step 1: Standard Linear Form

Divide by $x$ to achieve standard form $\frac{dy}{dx} + P(x)y = Q(x)$:

$$\frac{dy}{dx} + \frac{2}{x}y = \frac{\cos x}{x} \quad [M1]$$

Here $P(x) = \frac{2}{x}$ and $Q(x) = \frac{\cos x}{x}$.

Step 2: Integrating Factor $I(x)$

$$I(x) = e^{\int \frac{2}{x}\,dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2 \quad [A1]$$

Step 3: Multiply and Integrate Both Sides

$$\frac{d}{dx}\left( x^2 y \right) = x^2 \cdot \frac{\cos x}{x} = x\cos x \quad [M1]$$ $$x^2 y = \int x\cos x\,dx \quad [M1]$$

Apply integration by parts with $u = x, dv = \cos x\,dx \implies du = dx, v = \sin x$:

$$\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C \quad [A1]$$ $$x^2 y = x\sin x + \cos x + C \implies y = \frac{x\sin x + \cos x + C}{x^2}$$

Step 4: Apply Initial Condition $y(\pi) = 0$

$$0 = \frac{\pi\sin\pi + \cos\pi + C}{\pi^2} = \frac{0 - 1 + C}{\pi^2} \implies C - 1 = 0 \implies C = 1 \quad [M1]$$

Particular solution:

$$y = \frac{x\sin x + \cos x + 1}{x^2} \quad [A1]$$

Examiner Commentary: Candidates often forget to divide by $x$ initially, attempting an integrating factor directly on $x \frac{dy}{dx}$. The standard form requires leading coefficient 1 for $\frac{dy}{dx}$.

Practice Exercises (10 Tiered Questions)

Complete these 10 exam-style questions spanning core techniques to advanced HL challenge problems. Open each question to inspect the full step-by-step IB markscheme.

Exercise 1 · Separation of Variables with Arctangent

Find the general solution of the differential equation $\frac{dy}{dx} = 2x(y^2 + 1)$.

Show Full Solution

Separate variables: $\frac{1}{y^2 + 1}\,dy = 2x\,dx$. [M1]

$$\int \frac{1}{y^2 + 1}\,dy = \int 2x\,dx \implies \arctan(y) = x^2 + C \quad \text{[A1]}$$

Solve explicitly for $y$: [A1]

$$y = \tan(x^2 + C)$$

Exercise 2 · Initial Value Problem with Sign Branch

Solve the initial value problem $\frac{dy}{dx} = \frac{e^{2x}}{y}$, with initial condition $y(0) = -3$.

Show Full Solution

Separate variables: $y\,dy = e^{2x}\,dx$. [M1]

$$\int y\,dy = \int e^{2x}\,dx \implies \frac{1}{2}y^2 = \frac{1}{2}e^{2x} + C \implies y^2 = e^{2x} + K \quad \text{[A1]}$$

Apply initial condition $y(0) = -3$: $(-3)^2 = e^0 + K \implies 9 = 1 + K \implies K = 8$. [M1]

Because $y(0) = -3 < 0$, choose the negative root branch: [R1]

$$y = -\sqrt{e^{2x} + 8} \quad \text{[A1]}$$

Exercise 3 · Linear ODE with Power Integrating Factor

Find the general solution of the linear differential equation $\frac{dy}{dx} + \frac{2}{x}y = 5x^2$, for $x > 0$.

Show Full Solution

Standard form: $P(x) = \frac{2}{x}$, $Q(x) = 5x^2$.

Integrating Factor: $I(x) = e^{\int \frac{2}{x}dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2$. [M1][A1]

Multiply ODE by $x^2$: $\frac{d}{dx}[x^2 y] = 5x^4$. [M1]

$$x^2 y = \int 5x^4\,dx = x^5 + C \quad \text{[A1]}$$ $$y = x^3 + \frac{C}{x^2} \quad \text{[A1]}$$

Exercise 4 · Trigonometric Linear Initial Value Problem

Solve the initial value problem $\frac{dy}{dx} + y\tan x = \cos^2 x$, given that $y(0) = 2$, for $-\frac{\pi}{2} < x < \frac{\pi}{2}$.

Show Full Solution

Integrating factor: $I(x) = e^{\int \tan x\,dx} = e^{\ln|\sec x|} = \sec x$. [M1]

$$\frac{d}{dx}[y\sec x] = \cos^2 x \cdot \sec x = \cos x \quad \text{[M1]}$$ $$y\sec x = \int \cos x\,dx = \sin x + C \quad \text{[A1]}$$

Substitute $x=0, y=2$: $2\sec(0) = \sin(0) + C \implies 2(1) = 0 + C \implies C = 2$. [M1]

$$y = \frac{\sin x + 2}{\sec x} = \sin x \cos x + 2\cos x = \frac{1}{2}\sin 2x + 2\cos x \quad \text{[A1]}$$

Exercise 5 · Euler's Method Numerical Approximation

Given the differential equation $\frac{dy}{dx} = x + y^2$ with $y(0) = 1$, use Euler’s method with step size $h = 0.1$ to approximate $y(0.2)$ to 4 decimal places.

Show Full Solution

Step 1 ($x_0 = 0, y_0 = 1$): [M1]

$$f(0, 1) = 0 + 1^2 = 1 \implies y_1 = 1 + 0.1(1) = 1.1 \quad (x_1 = 0.1) \quad \text{[A1]}$$

Step 2 ($x_1 = 0.1, y_1 = 1.1$): [M1]

$$f(0.1, 1.1) = 0.1 + (1.1)^2 = 0.1 + 1.21 = 1.31$$ $$y_2 = 1.1 + 0.1(1.31) = 1.1 + 0.131 = 1.2310 \quad (x_2 = 0.2) \quad \text{[A1]}$$

$\implies y(0.2) \approx 1.2310$.

Exercise 6 · Newton's Law of Cooling

A hot cup of coffee with initial temperature $90^\circ\text{C}$ is placed in a room at constant temperature $20^\circ\text{C}$. Newton’s law of cooling gives $\frac{dT}{dt} = -k(T - 20)$. After 5 minutes, $T = 60^\circ\text{C}$. Find the temperature after 15 minutes.

Show Full Solution

Separate variables: $\frac{1}{T - 20}\,dT = -k\,dt \implies T - 20 = A e^{-kt}$. [M1]

Initial condition $T(0) = 90 \implies 70 = A e^0 \implies A = 70$. [A1]

At $t = 5$: $60 = 20 + 70e^{-5k} \implies e^{-5k} = \frac{40}{70} = \frac{4}{7}$. [M1]

At $t = 15$: $e^{-15k} = (e^{-5k})^3 = \left(\frac{4}{7}\right)^3 = \frac{64}{343}$. [M1]

$$T(15) = 20 + 70\left(\frac{64}{343}\right) = 20 + \frac{640}{49} \approx 33.1^\circ\text{C} \quad \text{[A1]}$$

Exercise 7 · Non-Normalized Linear ODE with Parts Integration

Find the particular solution to $x\frac{dy}{dx} - y = x^3 e^x$ satisfying $y(1) = 0$, for $x > 0$.

Show Full Solution

Divide by $x$: $\frac{dy}{dx} - \frac{1}{x}y = x^2 e^x$. [M1]

Integrating factor: $I(x) = e^{\int -1/x dx} = e^{-\ln x} = \frac{1}{x}$. [A1]

$$\frac{d}{dx}\left[ \frac{y}{x} \right] = \frac{1}{x}(x^2 e^x) = x e^x \implies \frac{y}{x} = \int x e^x\,dx \quad \text{[M1]}$$

By parts: $\int x e^x dx = (x - 1)e^x + C$. [A1]

Condition $y(1) = 0$: $0 = (1 - 1)e^1 + C \implies C = 0$. [M1]

$$y = x(x - 1)e^x \quad \text{[A1]}$$

Exercise 8 · Euler's Method Step & Concavity Error Analysis

Consider the differential equation $\frac{dy}{dx} = 2x - y$ with $y(0) = 1$. Use Euler’s method with step $h = 0.5$ to estimate $y(1)$, and prove whether it is an overestimate or underestimate using the second derivative.

Show Full Solution

Step 1 ($x_0 = 0, y_0 = 1$): $y_1 = 1 + 0.5(2(0) - 1) = 0.5$ ($x_1 = 0.5$). [M1]

Step 2 ($x_1 = 0.5, y_1 = 0.5$): $y_2 = 0.5 + 0.5(2(0.5) - 0.5) = 0.5 + 0.5(0.5) = 0.75$ ($x_2 = 1.0$). [A1]

Second derivative: $y'' = \frac{d}{dx}[2x - y] = 2 - y' = 2 - (2x - y) = 2 - 2x + y$. [M1]

On $[0, 1]$, $2 - 2x \ge 0$ and $y > 0$, hence $y''(x) > 0$ (curve is concave up). [R1]

Because the tangent lines lie below the curve, Euler's method gives an underestimate. [A1]

Exercise 9 · Mixing Tank Differential Equation

A tank initially contains $100\text{ L}$ of pure water. Brine containing $0.5\text{ kg/L}$ of salt flows in at $4\text{ L/min}$, and the thoroughly mixed solution drains out at $4\text{ L/min}$. Set up and solve the ODE for salt mass $S(t)$, and find $\lim_{t\to\infty} S(t)$.

Show Full Solution

Rate In = $4 \times 0.5 = 2\text{ kg/min}$; Rate Out = $4 \times \frac{S}{100} = \frac{S}{25}\text{ kg/min}$. [M1]

$$\frac{dS}{dt} = 2 - \frac{S}{25} = \frac{50 - S}{25} \implies \frac{1}{50 - S}\,dS = \frac{1}{25}\,dt \quad \text{[A1]}$$ $$-\ln|50 - S| = \frac{t}{25} + C \implies 50 - S = A e^{-t/25} \quad \text{[M1]}$$

Pure water at $t = 0 \implies S(0) = 0 \implies 50 = A$. [A1]

$$S(t) = 50(1 - e^{-t/25})\text{ kg} \quad \text{[A1]}$$ $$\lim_{t \to \infty} S(t) = 50(1 - 0) = 50\text{ kg} \quad \text{[A1]}$$

Exercise 10 · Terminal Velocity Linear Model

A falling object of mass $m$ experiences gravity $mg$ and air drag proportional to velocity $kv$. Starting from rest ($v(0) = 0$), solve $m\frac{dv}{dt} = mg - kv$. Show that the terminal velocity is $v_T = \frac{mg}{k}$.

Show Full Solution

Standard linear ODE form: $\frac{dv}{dt} + \frac{k}{m}v = g$. [M1]

Integrating factor: $I(t) = e^{\int (k/m)dt} = e^{kt/m}$. [A1]

$$\frac{d}{dt}\left[ v e^{kt/m} \right] = g e^{kt/m} \implies v e^{kt/m} = \frac{mg}{k}e^{kt/m} + C \quad \text{[M1]}$$ $$v(t) = \frac{mg}{k} + C e^{-kt/m}$$

Condition $v(0) = 0 \implies 0 = \frac{mg}{k} + C \implies C = -\frac{mg}{k}$. [M1]

$$v(t) = \frac{mg}{k}\left(1 - e^{-kt/m}\right) \quad \text{[A1]}$$

Since $k, m > 0$, $\lim_{t\to\infty} e^{-kt/m} = 0$, so $v_T = \frac{mg}{k}$. [AG]

Formula Booklet References (Topics 5.16 & 5.17)

Syllabus 5.16: Integrating Factor for Linear First-Order ODEs

$$\frac{dy}{dx} + P(x)y = Q(x) \implies I(x) = e^{\int P(x)\,dx}$$ $$y \cdot I(x) = \int Q(x) \cdot I(x)\,dx + C$$

Syllabus 5.17: Euler’s Numerical Step Method

$$x_{n+1} = x_n + h$$ $$y_{n+1} = y_n + h \cdot f(x_n, y_n)$$

Key Formulas & Identities

1. Homogeneous Differential Equations ($y = vx$ Substitution)

• Identify: $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$

• Substitute: $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$

• Separate: $x\frac{dv}{dx} = F(v) - v \implies \int \frac{1}{F(v) - v}\,dv = \int \frac{1}{x}\,dx$

2. Concavity & Euler Error Determination

• If $\frac{d^2y}{dx^2} > 0$ (Concave up $\cup$): Tangents lie below curve $\implies$ Underestimate

• If $\frac{d^2y}{dx^2} < 0$ (Concave down $\cap$): Tangents lie above curve $\implies$ Overestimate

Common Exam Pitfalls

Premature Exponentiation & Missing $+ C$

When solving $\int \frac{1}{y}dy = \int g(x)dx$, you must write $\ln|y| = G(x) + C$ before exponentiating: $y = A e^{G(x)}$ where $A = \pm e^C$. Writing $y = e^{G(x)} + C$ is an instant method failure.

Forgetting to Normalize Linear ODEs

If given $x\frac{dy}{dx} + 3y = x^2$, students often take $P(x) = 3$. You MUST divide by $x$ first to get $\frac{dy}{dx} + \frac{3}{x}y = x$, so $P(x) = \frac{3}{x}$ and $I(x) = x^3$.

Conflating First and Second Derivatives in Euler Error Analysis

The direction of error (over- or under-estimate) depends solely on the sign of the second derivative $y''$ (concavity), NOT the sign of $y'$ (increasing/decreasing).

Figure 5.4: Euler’s Numerical Step Method vs. Exact Solution Curve

1. Foundational Anatomy of Differential Equations

2. Analytical Method 1: Separation of Variables (Topic 5.16)

Immediate Constant Placement Rule

3. Analytical Method 2: Homogeneous Equations via Substitution $y = vx$ (Topic 5.16 AHL)

4. Analytical Method 3: First-Order Linear ODEs & Integrating Factors (Topic 5.16)

5. Numerical Method: Euler’s Step Method & Error Analysis (Topic 5.17)

Mathematical Concavity and Under/Over-Estimation

Worked Examples

Worked Example 1: Homogeneous ODE via Substitution $y = vx$ (Paper 1 Standard)

Worked Example 2: Non-Normalized Linear ODE with Integrating Factor (Paper 1 Standard)

Worked Example 3: Euler’s Method Multistep with Concavity Proof (Paper 2 Standard)

Worked Example 4: Applied Fluid Mixing Tank (Paper 2 Section B Standard)

Worked Example 5: Integrating Factor with Initial Value Problem (Paper 2 Standard)

Practice Exercises (10 Tiered Questions)

Formula Booklet References (Topics 5.16 & 5.17)

Syllabus 5.16: Integrating Factor for Linear First-Order ODEs

Syllabus 5.17: Euler’s Numerical Step Method

Key Formulas & Identities

1. Homogeneous Differential Equations ($y = vx$ Substitution)

2. Concavity & Euler Error Determination

Common Exam Pitfalls

Premature Exponentiation & Missing $+ C$

Forgetting to Normalize Linear ODEs

Conflating First and Second Derivatives in Euler Error Analysis

Topic Diagnostic Test

Section A: Diagnostic Multiple Choice Questions

Section B: Examination-Style Free Response Questions