P3: Differential Equations

The Secret Language of Change 🤫

Introduction

1. Introduction

Welcome to one of the coolest topics in A-Level Maths! Ever wondered how we can mathematically describe things that are constantly changing, like a cooling cup of coffee or a growing population? That's where differential equations come in. They're basically equations that involve rates of change (hello, derivatives!). In this chapter, we're going to learn three key skills. First, we'll learn how to form these special equations by translating real-world scenarios into the language of calculus. Then, we'll master a powerful technique called separation of variables to solve them. Finally, we'll see how to find particular solutions - the one specific answer that fits a given situation. It might sound complex, but you'll soon see it's a logical and super useful process. Let's get started!

\frac{dy}{dt}

Worked example

Your Pizza Getting Cold, But Make It Math 🍕

\theta

\frac{dy}{dx}

+C

Worked example

Let's Solve This Thing 🚀

\frac{dy}{dx} = \frac{3x^2}{e^y}

Alright, so you've mastered separating variables and integrating to find a general solution to a differential equation. That solution always has a cheeky little '+ C' tagging along at the end, right? Think of that general solution like a generic playlist on Spotify called 'Chill Vibes'. It has the right structure, but there are infinite variations. The constant 'C' is like a slider that changes the whole playlist - shifting it up or down, making it totally different each time. This means a single differential equation actually represents an entire family of solution curves.

(x, y)

Worked example

Let's Pinpoint This Solution! 📍

\frac{dy}{dx} = 4x\sqrt{x^2 + 3}

1
$y$
$\int dy = \int 4x(x^2 + 3)^{\frac{1}{2}} dx$
2
$u = x^2 + 3$
$y = \int 2(x^2 + 3)^{\frac{1}{2}} (2x dx) = \int 2u^{\frac{1}{2}} du = 2 \times \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{4}{3}u^{\frac{3}{2}} + C$
3
$u = x^2 + 3$
$y = \frac{4}{3}(x^2 + 3)^{\frac{3}{2}} + C$
4
$(1, 8)$
$8 = \frac{4}{3}((1)^2 + 3)^{\frac{3}{2}} + C$
5
$Let's simplify and solve for C. This is the key to unlocking our unique answer.$
$8 = \frac{4}{3}(4)^{\frac{3}{2}} + C \implies 8 = \frac{4}{3}(8) + C \implies 8 = \frac{32}{3} + C \implies C = 8 - \frac{32}{3} = \frac{24-32}{3} = -\frac{8}{3}$
6
$C = -\frac{8}{3}$
$y = \frac{4}{3}(x^2 + 3)^{\frac{3}{2}} - \frac{8}{3}$

Answer

y = \frac{4}{3}(x^2 + 3)^{\frac{3}{2}} - \frac{8}{3}

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