8.1.2 Solving First Order Differential Equations | Edexcel A Level Further Maths: Core Pure Revision Notes 2017

First Order Differential Equations

What is a differential equation?

A differential equation is simply an equation that contains derivatives

For example $\frac{d y}{d x} = 12 x y^{2}$ is a differential equation
And so is $\frac{d^{2} x}{d t^{2}} - 5 \frac{d x}{d t} + 7 x = 5 \sin t$

What is a first order differential equation?

A first order differential equation is a differential equation that contains first derivatives but no second (or higher) derivatives

For example $\frac{d y}{d x} = 12 x y^{2}$ is a first order differential equation
But $\frac{d^{2} x}{d t^{2}} - 5 \frac{d x}{d t} + 7 x = 5 \sin t$ is not a first order differential equation, because it contains the second derivative $\frac{d^{2} x}{d t^{2}}$

The general solution to a first order differential equation will have one unknown constant
To find the particular solution you will need to know an initial condition or a boundary condition

Wait – haven’t I seen first order differential equations before?

Yes you have!

For example $\frac{d y}{d x} = 3 x^{2}$ is also a first order differential equation, because it contains a first derivative and no second (or higher) derivatives
But for that equation you can just integrate to find the solution y = x³ + c (where c is a constant of integration)

In A Level Maths you will have solved some first order differential equations using the method of separation of variables

Integrating Factors

What is an integrating factor?

An integrating factor can be used to solve a differential equation that can be written in the standard form $\frac{d y}{d x} + p (x) y = q (x)$

Be careful – the ‘functions of x’ p(x) and q(x) may just be constants!

For example in $\frac{d y}{d x} + 6 y = e^{- 2 x}$ , p(x) = 6 and q(x) = e^-2x
While in $\frac{d y}{d x} + \frac{y}{2 x} = 12$ , $p (x) = \frac{1}{2 x}$ and q(x) = 12

For an equation in standard form, the integrating factor is $e^{\int p (x) d x}$

How do I use an integrating factor to solve a differential equation?

STEP 1: If necessary, rearrange the differential equation into standard form
STEP 2: Find the integrating factor

Note that you don’t need to include a constant of integration here when you integrate ∫p(x) dx

STEP 3: Multiply both sides of the differential equation by the integrating factor

This will turn the equation into an exact differential equation of the form $\frac{d}{d x} (y e^{\int p (x) d x}) = q (x) e^{\int p (x) d x}$

STEP 4: Integrate both sides of the equation with respect to x

The left side will automatically integrate to $y e^{\int p (x) d x}$
For the right side, integrate $\int q (x) e^{\int p (x) d x} d x$ using your usual techniques for integration
Don’t forget to include a constant of integration

Although there are two integrals, you only need to include one constant of integration

STEP 5: Rearrange your solution to get it in the form y = f(x)

What else should I know about using an integrating factor to solve differential equations?

After finding the general solution using the steps above you may be asked to do other things with the solution

For example you may be asked to find the solution corresponding to certain initial or boundary conditions

Worked example

Consider the differential equation $\frac{d y}{d x} = 2 x y + 5 e^{x^{2}}$ where y = 7 when x = 0.

Use an integrating factor to find the solution to the differential equation with the given boundary condition.

T8-Lavls_fm-8-1-2-integrating-factors-we-solution

Solving First Order Differential Equations (Edexcel A Level Further Maths: Core Pure)

Revision Note

First Order Differential Equations

What is a differential equation?

What is a first order differential equation?

Wait – haven’t I seen first order differential equations before?

Integrating Factors

What is an integrating factor?

How do I use an integrating factor to solve a differential equation?

What else should I know about using an integrating factor to solve differential equations?

Worked example

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Author: Roger