Lesson 8.1: First-order Separable Differential Equations

Introduction

Welcome to Lesson 8.1 of Foundation Further Mathematics! Today, we'll dive into the world of differential equations, specifically focusing on first-order separable differential equations. 🌍✨

Learning Objectives:

By the end of this lesson, students should be able to:

• Recognise and solve separable equations.
• Apply initial/boundary conditions to find particular solutions.
• Model growth, decay, and cooling using differential equations.
• Solve a separable first-order differential equation.
• Use conditions to determine the constant of integration.

Why is this Important?

Differential equations are crucial for understanding how things change over time. Whether it's predicting population growth, modeling the cooling of an object, or even describing how products decay in economics, mastering this topic opens up countless applications in real life! 💡

What is a Separable Differential Equation?

A first-order separable differential equation is an equation that can be expressed in the form:

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

This means you can isolate the variables $y$ and $x$ on different sides of the equation. Let's break this down step by step.

Step 1: Rearranging the Equation

To solve a separable differential equation, the first step is to move all the $y$ terms to one side and $x$ terms to the other. For example, consider the equation:

$$\frac{dy}{dx} = y \cdot \sin(x)$$

We can rewrite this as:

$$\frac{1}{y} dy = \sin(x) dx$$

Step 2: Integrating Both Sides

Next, we integrate both sides:

$$\int \frac{1}{y} dy = \int \sin(x) dx$$

Doing these integrals gives us:

$$\ln|y| = -\cos(x) + C$$

where $C$ is the constant of integration.

Step 3: Solving for y

To isolate $y$, we exponentiate both sides:

$$y = e^{-\cos(x) + C} = e^C e^{-\cos(x)}$$

We can rewrite $e^C$ as a new constant $C_1$:

$$y = C_1 e^{-\cos(x)}$$

This is the general solution to our differential equation!

Example: Population Growth Model

A common example of a separable differential equation in the real world is the population growth model, described by:

$$\frac{dP}{dt} = kP$$

where $P$ is the population at time $t$ and $k$ is a constant representing the growth rate.

1. Rearranging:

$$\frac{1}{P} dP = k dt$$

1. Integrating:

$$\int \frac{1}{P} dP = \int k dt$$

This leads to:

$$\ln|P| = kt + C$$

1. Exponentiating:

$$P = e^{kt + C} = C_2 e^{kt}$$

This equation models exponential growth of a population over time! 🦠

Applying Initial Conditions

To find a particular solution, we often use initial conditions. For instance, if at $t=0$ the population $P = P_0$, we can substitute to find $C_2$:

$$P_0 = C_2 e^{k \cdot 0} \Rightarrow C_2 = P_0$$

Thus, the particular solution becomes:

$$P(t) = P_0 e^{kt}$$

Example: Cooling Object

Consider Newton's Law of Cooling, which can be modeled by:

$$\frac{dT}{dt} = -k(T - T_a)$$

where $T$ is the temperature of the object, $T_a$ is the ambient temperature, and $k$ is a positive constant.

1. Rearranging gives:

$$\frac{1}{T - T_a} dT = -k dt$$

1. Integrating:

$$\int \frac{1}{T - T_a} dT = -k \int dt$$

Thus, we get:

$$\ln|T - T_a| = -kt + C$$

1. Exponentiating:

$$T - T_a = C_3 e^{-kt}$$

To find $C_3$, we use an initial condition, say at $ t = 0$, the initial temperature $T_0 = T(0)$.

Hence:

$$T_0 - T_a = C_3$$

The particular solution would then be:

$$T(t) = T_a + (T_0 - T_a) e^{-kt}$$

This equation describes how the temperature of an object approaches the ambient temperature over time. 🕒

Conclusion

In this lesson, we've learned how to recognise, rearrange, integrate, and apply initial conditions to first-order separable differential equations. These equations provide incredible insights into real-world phenomena, such as population growth and cooling processes. Remember, mastering these steps will be key as we progress further into the topic of differential equations!

Study Notes

• A first-order separable differential equation can be written as $\frac{dy}{dx} = g(y)h(x)$.
• The three key steps to solving separable equations: rearranging, integrating both sides, and solving for $y$.
• Initial conditions allow us to find particular solutions.
• Real-world applications include population growth, cooling, and decay modeling.
• Key formulae:
• Population growth: $P(t) = P_0 e^{kt}$.
• Newton's Law of Cooling: $T(t) = T_a + (T_0 - T_a) e^{-kt}$.