Solving Differential Equations Revision Notes for Leaving Cert Applied Maths

Solving Differential Equations

What are differential equations?

Differential equations are mathematical equations that contain derivatives (such as $\frac{dy}{dx}$ ) within them. These equations describe relationships between a function and its rate of change, making them incredibly useful for modelling real-world situations where quantities change over time.

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For example: $4x \frac{dy}{dx} = 5y$ is a differential equation because it contains the derivative $\frac{dy}{dx}$ .

Key terminology:

A 1st order differential equation has a first derivative ( $\frac{dy}{dx}$ ) as its highest derivative
A 2nd order differential equation has a second derivative ( $\frac{d²y}{dx²}$ ) as its highest derivative

Types of differential equations

The differential equations in the Leaving Certificate Applied Maths course fall into three main categories:

Type 1: 1st order differential equations with general solutions
Type 2: 1st order differential equations with definite values (initial conditions)
Type 3: 2nd order separable differential equations

Let's examine each type with worked examples to understand the solution methods.

Type 1: First order differential equations with general solutions

This type focuses on finding the general solution to a differential equation, which includes an arbitrary constant of integration.

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Method:

Separate the variables (get all y terms on one side, all x terms on the other)
Integrate both sides
Rearrange to make y the subject

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Worked Example: Finding the General Solution

Find the general solution to $\frac{dy}{dx} = 5x²y$

Solution process:

First, multiply both sides by dx to eliminate fractions: $dy = 5x²y , dx$
Separate variables by dividing both sides by y: $\frac{1}{y} dy = 5x² dx$
Integrate both sides: $\int\frac{1}{y} dy = \int 5x² dx$
This gives: $\log\_e y = \frac{5x³}{3} + c$
Rearrange to get y on its own: $y = e^{(\frac{5x³}{3} + c)}$

The final answer is the general solution because it contains the arbitrary constant c.

Type 2: First order differential equations with definite values

This type involves finding a particular solution by using given initial conditions to determine the value of the constant of integration.

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Method:

Start as in Type 1 - separate variables and integrate
Use the given initial condition to find the value of c
Substitute this value back into the solution
Rearrange to get y as the subject

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Worked Example: Finding a Particular Solution

Find a function $y = f(x)$ such that $\frac{dy}{dx} = 3y²$ and $y = 2$ when $x = 0$

Solution process:

Separate variables: $\frac{dy}{dx} = 3y²$ becomes $\frac{1}{y²} dy = 3 dx$
Integrate both sides: $\int y^{-2} dy = \int 3 dx$
This gives: $\frac{y^{-1}}{-1} = 3x + c$ , so $-\frac{1}{y} = 3x + c$
Apply the initial condition ( $y = 2$ when $x = 0$ ): $-\frac{1}{2} = 3(0) + c$ , so $c = -\frac{1}{2}$
Substitute back: $-\frac{1}{y} = 3x - \frac{1}{2}$
Rearrange to solve for y: $y = \frac{2}{1-6x}$

This gives us a particular solution because we've used the initial conditions to find the specific value of c.

Type 3: Second order separable differential equations

These equations involve second derivatives ( $\frac{d²y}{dx²}$ ) and require a substitution technique to solve.

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Method:

Let $v = \frac{dy}{dx}$ , so $\frac{dv}{dx} = \frac{d²y}{dx²}$
Rewrite the equation in terms of v
Solve this first-order equation to find v
Use $\frac{dy}{dx} = v$ to find y through integration
Apply initial conditions to find constants

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Worked Example: Second Order Differential Equation

Solve $\frac{d²y}{dx²} = 6\frac{dy}{dx}$ given that $y = 1$ when $\frac{dy}{dx} = 1$ and $x = 0$

Solution process:

Let $v = \frac{dy}{dx}$ , so the equation becomes $\frac{dv}{dx} = 6v$
Solve this first-order equation: $\frac{1}{v} dv = 6 dx$
Integrate: $\log\_e v = 6x + c$
Apply initial condition: when $x = 0$ , $v = 1$ , so $c = 0$
Therefore: $v = e^{6x}$ , which means $\frac{dy}{dx} = e^{6x}$
Integrate to find y: $y = \frac{1}{6}e^{6x} + c$
Apply second initial condition: when $x = 0$ , $y = 1$ , so $c = \frac{5}{6}$
Final solution: $y = \frac{1}{6}e^{6x} + \frac{5}{6}$

Key techniques and exam tips

Variable separation: This is the fundamental technique for most differential equations. Always aim to get all terms involving y (including dy) on one side and all terms involving x (including dx) on the other side.

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Integration: Remember your integration rules, especially:

$\int\frac{1}{x} dx = \ln x + c$
$\int e^x dx = e^x + c$
Standard trigonometric integrals

Constants of integration: Always include the constant c when integrating, and use initial conditions to find its specific value.

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Calculator mode: When dealing with inverse trigonometric functions, ensure your calculator is in the correct mode (radians for most cases).

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Key Points to Remember:

Differential equations contain derivatives - they show relationships between functions and their rates of change
Separate variables first - get all y terms on one side and all x terms on the other before integrating
Always integrate both sides and include the constant of integration c
Use initial conditions to find the specific value of c for particular solutions
For second-order equations, use substitution (let $v = \frac{dy}{dx}$ ) to reduce them to first-order problems

Solving Differential Equations (Leaving Cert Applied Maths): Revision Notes