8.3.5 Solving & Interpreting Differential Equations

Solving and interpreting differential equations is a crucial aspect of A Level Maths, particularly in modelling real-world scenarios. Let's break down the process into clear steps, covering both the mathematical techniques for solving differential equations and how to interpret the solutions in context.

1. Solving Differential Equations

There are several types of differential equations, and each type has specific methods for finding the solution. Below are the common types you might encounter:

a) First-Order Differential Equations

These involve the first derivative of the unknown function.

i. Separable Differential Equations:

When a differential equation can be written in the form:

$\frac{dy}{dx} = g(x)h(y)$

Steps to Solve:

Separate the Variables: Move all terms involving $y$ to one side and all terms involving $x$ to the other.
Integrate Both Sides: Integrate with respect to $x$ and $y$ .
Solve for $y$ : If possible, solve explicitly for $y$ .

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Example: Solve the differential equation:

$\frac{dy}{dx} = 3y$

Solution:

Separate variables:

$\frac{1}{y} \, dy = 3 \, dx$

Integrate both sides:

$\ln |y| = 3x + C$

Solve for $y$ :

$y = C_1 e^{3x} \quad \text{(where $ C_1 = e^C $)}$

This is the general solution.

ii. Linear First-Order Differential Equations:

These are of the form:

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

Steps to Solve:

Find the Integrating Factor (IF):

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

Multiply the whole equation by the IF to make the left side an exact derivative.
Integrate the resulting equation.
Solve for $y$ .

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Example: Solve the differential equation:

$\frac{dy}{dx} + 2y = e^{-x}$

Solution:

Find the integrating factor:

$IF = e^{\int 2 \, dx} = e^{2x}$

Multiply the entire equation by $e^{2x}$ :

$e^{2x} \frac{dy}{dx} + 2e^{2x}y = e^{x}$

Notice that the left side is now the derivative of $y \cdot e^{2x}$ :

$\frac{d}{dx}(y \cdot e^{2x}) = e^{x}$

Integrate both sides:

$y \cdot e^{2x} = \int e^{x} \, dx = e^{x} + C$

Solve for $y$ :

$y = e^{-2x}(e^x + C) = e^{-x} + Ce^{-2x}$

This is the general solution.

b) Second-Order Differential Equations

These involve the second derivative of the unknown function and often appear in problems related to motion and oscillations.

i. Homogeneous Equations:

These are of the form:

$\frac{d^2y}{dx^2} + p\frac{dy}{dx} + qy = 0$

Steps to Solve:

Find the characteristic equation: This is a quadratic equation obtained by substituting $y = e^{mx}$ .
Solve the characteristic equation: The nature of the roots (real and distinct, real and repeated, or complex) determines the form of the solution.
Form the general solution using the roots of the characteristic equation.

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Example: Solve the differential equation:

$\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = 0$

Solution:

Characteristic equation:

$m^2 - 3m + 2 = 0$

Factor the quadratic:

$(m - 1)(m - 2) = 0$

Roots: $m = 1$ and $m = 2$ .
General solution:

$y(x) = C_1e^{x} + C_2e^{2x}$

2. Interpreting Solutions

After solving a differential equation, it's important to interpret the solution in the context of the problem. Here's how you might approach it:

a) Initial Conditions

Often, the problem will provide initial conditions, such as $y(0) = y_0$ , to find a particular solution.
Substitute the initial conditions into the general solution to find the specific values of the constants $C_1$ and $C_2$ .

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Example: Given the initial condition $y(0) = 5$ for the previous example $y(x) = C_1e^{x} + C_2e^{2x}$ , determine the particular solution.

Substitute $x = 0$ and $y(0) = 5$ :

$5 = C_1e^{0} + C_2e^{0} = C_1 + C_2$

Hence, $C_1 + C_2 = 5$ . You'd need another condition to determine both $C_1$ and $C_2$ .

b) Physical Interpretation

The solution should be related back to the physical scenario described by the problem.
For example, in a population growth model where $y(t) = P_0e^{kt}$ , $y(t)$ might represent the population at time $t$ , with $P_0$ being the initial population and $k$ the growth rate.

c) Long-Term Behaviour

Consider what happens to $y(x)$ as $x$ becomes very large or very small.
This can give insight into the stability of the system or the long-term predictions of the model.

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Example: For the solution $y(t) = e^{-kt}(A \cos(\omega t) + B \sin(\omega t))$ , as $t$ increases:

If $k > 0$ , $e^{-kt}$ tends to zero, so $y(t)$ tends towards zero, indicating that the oscillations dampen over time.

Summary

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Solving differential equations involves applying specific methods depending on the type of equation (separable, linear, second-order, etc.).
Interpreting solutions requires understanding the context of the problem, applying initial conditions to find particular solutions, and analysing the behaviour of the solution over time or other variables.
The final solution should always be connected back to the real-world scenario to ensure it makes sense within that context.

Solving & Interpreting Differential Equations Simplified Revision Notes for A-Level OCR Maths Pure

8.3.5 Solving & Interpreting Differential Equations

1. Solving Differential Equations

a) First-Order Differential Equations

b) Second-Order Differential Equations

2. Interpreting Solutions

a) Initial Conditions

b) Physical Interpretation

c) Long-Term Behaviour

Summary

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