IB Mathematics AA AHL Differential equations Study Notes - New Syllabus
IB Mathematics AA AHL Differential equations Study Notes
LEARNING OBJECTIVE
- Differential equations
Key Concepts:
- Differential equations
- Variables separable
- Homogeneous differential equation
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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First Order Differential Equations
First Order Differential Equations
First order differential equations are equations involving the first derivative of a function, typically written as:
$\boxed{ \frac{dy}{dx} = f(x, y) }$
These equations describe how the dependent variable \( y \) changes with respect to the independent variable \( x \), and are foundational in modeling dynamic systems in physics, biology, economics, and more.
Types of First Order Differential Equations
| Type | General Form | Solution Method |
|---|---|---|
| Separable | \(\frac{dy}{dx} = g(x)h(y)\) | Separate variables and integrate both sides |
| Linear | \(\frac{dy}{dx} + P(x)y = Q(x)\) | Use integrating factor \( \mu(x) = e^{\int P(x) dx} \) |
| Homogeneous | \(\frac{dy}{dx} = F\left(\frac{y}{x}\right)\) | Use substitution \( v = \frac{y}{x} \) |
| Exact | \( M(x, y)dx + N(x, y)dy = 0 \) | Check if exact; solve using potential function |
| Numerical (Euler’s Method) | Applicable to any \( \frac{dy}{dx} = f(x, y) \) | Use iterative approach for approximation |
General Solution Process
- Identify the type of the first order equation: separable, linear, etc.
- Rearrange the equation as needed to match the standard form.
- Solve using integration or the relevant method.
- Apply initial conditions if given, to find the particular solution.
Numerical Solution of \(\frac{dy}{dx} = f(x, y)\) using Euler’s Method
Euler’s Method
Euler’s Method is a numerical technique used to approximate the solution of first-order differential equations of the form:
$\boxed{ \frac{dy}{dx} = f(x, y) }$
It provides an estimated value of \( y \) at discrete points using an initial value and a fixed step size \( h \).
Euler’s Update Formula
Given: \( y(x_0) = y_0 \) and step size \( h \)
Then:
$ \boxed{x_{n+1} = x_n + h \quad \text{and} \quad y_{n+1} = y_n + h \cdot f(x_n, y_n)} $
Step-by-Step Algorithm
- Start with initial values: \( x_0 \), \( y_0 \), and step size \( h \).
- Compute \( y_1 = y_0 + h \cdot f(x_0, y_0) \)
- Compute \( x_1 = x_0 + h \)
- Repeat the process for the desired number of steps or until a specific \( x \)-value is reached.
Example:
Approximate the solution to
$ \frac{dy}{dx} = x + y, \quad y(0) = 1 $
using Euler’s method with step size \( h = 0.1 \), for 3 steps.
▶️ Answer/Explanation
| Step | \( x_n \) | \( y_n \) | \( f(x_n, y_n) = x_n + y_n \) | \( y_{n+1} = y_n + h f(x_n, y_n) \) |
|---|---|---|---|---|
| 0 | 0.0 | 1.000 | 1.000 | 1.000 + 0.1(1.000) = 1.100 |
| 1 | 0.1 | 1.100 | 1.200 | 1.100 + 0.1(1.200) = 1.220 |
| 2 | 0.2 | 1.220 | 1.420 | 1.220 + 0.1(1.420) = 1.362 |
Answer: The approximate value of \( y \) after 3 steps (at \( x = 0.3 \)) is:
$ \rm{y(0.3) \approx 1.362}$
Notes
- The smaller the step size \( h \), the more accurate the approximation (but more computation).
- Euler’s method is simple but can accumulate error quickly for large intervals or stiff equations.
Example: (GDC)
Use Euler’s method to approximate the solution to
$ \frac{dy}{dx} = x – y, \quad y(0) = 1 $
Use a step size of \( h = 0.2 \) to estimate \( y(0.6) \).
▶️ Answer/Explanation
Step 1: Open Spreadsheet Mode
- Go to the Lists & Spreadsheet app.
Step 2: Enter Headings
- Column A: `x` values
- Column B: `y` values
- Column C: \( f(x, y) = x – y \)
Step 3: Fill Initial Values
- A1: `0` (initial x)
- B1: `1` (initial y)
- C1: `=A1 – B1` → evaluates \( f(x, y) = x – y \)
Step 4: Fill Recursion Formulas
From row 2 downward:
- A2: `=A1 + 0.2`
- B2: `=B1 + 0.2 * C1`
- C2: `=A2 – B2`
Then drag these cells downward to compute successive steps.
Step 5: Read Result
- At \( x = 0.6 \) (Row 4), the y-value is approximately:
$ \rm{y(0.6) \approx 0.704} $
Variables Separable
Variables Separable
A first-order differential equation is separable if it can be written as:
$\boxed{ \frac{dy}{dx} = f(x) \cdot g(y)} $
This allows us to rearrange terms so that all \( y \)-dependent terms are on one side and all \( x \)-dependent terms are on the other:
$\boxed{ \frac{1}{g(y)} \, dy = f(x) \, dx }$
Then we integrate both sides to find the implicit or explicit solution.
Step-by-Step Procedure
- Rewrite the differential equation in the form \( \frac{dy}{dx} = f(x) g(y) \).
- Separate variables: move all terms with \( y \) to one side and \( x \) to the other, typically as \( \frac{1}{g(y)} dy = f(x) dx \).
- Integrate both sides: \( \int \frac{1}{g(y)} dy = \int f(x) dx + C \).
- Solve the resulting equation for \( y \) if possible.
Example
Solve the logistic differential equation \( \frac{dn}{dt} = 2n(5-n) \) with initial condition \( n(0) = 1 \).
▶️ Answer/Explanation
Separate variables and write:
\( \frac{1}{n(5-n)} dn = 2 \, dt \)
Partial fractions:
\( \frac{1}{n(5-n)} = \frac{A}{n} + \frac{B}{5-n} \)
- Set \( n=0 \): \( 1 = 5A \implies A = \frac{1}{5} \)
- Set \( n=5 \): \( 1 = 5B \implies B = \frac{1}{5} \)
So:
\( \frac{1}{n(5-n)} = \frac{1}{5} \left( \frac{1}{n} + \frac{1}{5-n} \right) \)
Integrate:
\( \int \frac{1}{n(5-n)} dn = \frac{1}{5} (\ln|n| – \ln|5-n|) + C \)
Equate and integrate right side:
\( \frac{1}{5} \ln \left| \frac{n}{5-n} \right| = 2t + C_1 \)
Exponentiate:
\( \left| \frac{n}{5-n} \right| = Ce^{10t} \), where \( C = e^{5C_1} \)
Express \( n \):
\( n = \frac{5Ce^{10t}}{1 + Ce^{10t}} \)
Use initial condition \( n(0) = 1 \) to find \( C \):
\( 1 = \frac{5C}{1 + C} \Rightarrow 1 + C = 5C \Rightarrow 1 = 4C \Rightarrow C = \frac{1}{4} \)
$ \rm{ n(t) = \frac{5 \cdot \frac{1}{4} e^{10t}}{1 + \frac{1}{4} e^{10t}} = \frac{ \frac{5}{4} e^{10t}}{1 + \frac{1}{4} e^{10t}} = \frac{5 e^{10t}}{4 + e^{10t}} } $
Homogeneous Equations
Homogeneous Equations
A first-order differential equation is called homogeneous if it can be written in the form:
$\boxed{\frac{dy}{dx} = f\left(\frac{y}{x}\right) }$
where the right-hand side is a function of the ratio \( \frac{y}{x} \) only.
Method: Substitution \( y = vx \)
To solve such equations, use the substitution:
$\boxed{ y = v x \quad \Rightarrow \quad \frac{dy}{dx} = v + x \frac{dv}{dx} }$
Substitute into the original differential equation to obtain an equation involving \( v \) and \( x \):
$\boxed{ v + x \frac{dv}{dx} = f(v) }$
This can be rearranged into a separable equation for \( v \) and \( x \):
$\boxed{ x \frac{dv}{dx} = f(v) – v \quad \Rightarrow \quad \frac{dv}{f(v) – v} = \frac{dx}{x} }$
Integrate both sides to find \( v \) in terms of \( x \), then substitute back \( y = vx \) to get the solution.
Example
Solve the differential equation:
\( \displaystyle \frac{dy}{dx} = \frac{y + x}{x} \)
▶️ Answer/Explanation
Solution:
Rewrite the right side in terms of \( \frac{y}{x} \):
\( \displaystyle \frac{y + x}{x} = \frac{y}{x} + 1 = v + 1 \)
Use substitution \( y = v x \) so \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\( \displaystyle v + x \frac{dv}{dx} = v + 1 \)
\( \displaystyle x \frac{dv}{dx} = 1 \quad \Rightarrow \quad \frac{dv}{dx} = \frac{1}{x} \)
Separate variables and integrate:
\( \displaystyle dv = \frac{1}{x} dx \quad \Rightarrow \quad \int dv = \int \frac{1}{x} dx \)
So:
\( \displaystyle v = \ln|x| + C \)
\( v = \frac{y}{x} \), so:
\( \displaystyle \frac{y}{x} = \ln|x| + C \quad \Rightarrow \quad y = x (\ln|x| + C) \)
Solution Using Integrating Factor
Solution Using Integrating Factor
For a first order linear differential equation of the form:
\(\boxed{ \displaystyle \frac{dy}{dx} + P(x) y = Q(x)} \),
the method of integrating factor (IF) is applied as follows:
Calculate the integrating factor:
\(\boxed{ \displaystyle \mu(x) = e^{\int P(x) \, dx}} \)
Multiply the entire differential equation by \( \mu(x) \), so that the left side becomes the derivative of \( \mu(x) y \):
\(\boxed{ \displaystyle \mu(x) \frac{dy}{dx} + \mu(x) P(x) y = \frac{d}{dx} \left( \mu(x) y \right)} \)
Rewrite the equation as:
\(\boxed{ \displaystyle \frac{d}{dx} \left( \mu(x) y \right) = \mu(x) Q(x)} \)
Integrate both sides with respect to \( x \):
\(\boxed{ \displaystyle \mu(x) y = \int \mu(x) Q(x) \, dx + C } \)
Solve for \( y \):
\(\boxed{ \displaystyle y = \frac{1}{\mu(x)} \left( \int \mu(x) Q(x) \, dx + C \right) } \)
This method converts the original differential equation into an exact derivative form, making it straightforward to integrate and find the solution.
Example
Solve the differential equation:
\( \displaystyle y’ + 2y = e^{-x} \)
▶️ Answer/Explanation
Solution:
Identify \( P(x) = 2 \) and \( Q(x) = e^{-x} \).
integrating factor (IF):
\( \displaystyle \mu(x) = e^{\int 2 dx} = e^{2x} \)
Multiply both sides of the equation by \( \mu(x) \):
\( e^{2x} y’ + 2 e^{2x} y = e^{x} \)
left side as the derivative of \( e^{2x} y \):
\( \displaystyle \frac{d}{dx} \left( e^{2x} y \right) = e^{x} \)
Integrate both sides:
\( \displaystyle e^{2x} y = \int e^{x} dx + C = e^{x} + C \)
Solve \( y \):
\( \displaystyle y = e^{-2x} (e^{x} + C) = e^{-x} + C e^{-2x} \)
