1 Introduction

You have already studied ordinary differential equations (ODEs) and have learnt how to obtain the solution of certain types. Since a knowledge of the solution of certain ODEs (i.e. those with constant coefficients) will be required in solving partial differential equations (PDEs), we will begin this unit reminding you of some important results.

Key Point 1

The first order ODE

d y d x = k y
has general solution
y = A e k x
Here k is a constant which can be positive or negative and A is an arbitrary constant.

In Key Point 1 the quantity A in the general solution is a constant. To obtain the value of A we have to know the value of y at some value of x , perhaps x = 0 . In other words, we need to know an initial condition .

Task!

Find y as a function of x if

d y d x = − 2 y

and the initial condition is y ( 0 ) = 3.

From Key Point 1 with k = − 2 we have the general solution

y = A e − 2 x

Putting x = 0 and y = 3 into this we obtain 3 = A e 0 i.e. A = 3 so the solution to the given initial value problem is

y = 3 e − 2 x

We shall also need to be familiar with solutions to second order, homogeneous, constant coefficient ODEs, summarised in Key Point 2.

Key Point 2

A second order ODE of the form

a d 2 y d x 2 + b d y d x + c y = 0 (1)

where a , b , c are constants, has an auxiliary equation

a m 2 + b m + c = 0 (2)

obtained by inserting the trial solution y = e m x in (1).

The general solution of (1) then depends on the solutions (or roots) of the quadratic equation (2).

  1. If (2) has real, distinct roots m = m 1 and m = m 2 then

    y = A e m 1 x + B e m 2 x

  2. If (2) has a repeated root m = m 1 then

    y = ( A + B x ) e m 1 x

  3. If (2) has complex roots (which will be a conjugate pair) m = α ± j β then

    y = e α x ( A cos β x + B sin β x )

Note that in each of these cases 1. to 3. the general solution is a linear combination of two particular solutions:

For 1. they are e m 1 x and e m 2 x .

For 2. they are e m 1 x and x e m 1 x .

For 3. they are e α x cos β x and e α x sin β x .

Task!

Use Key Point 2 to find the general solution of d 2 y d x 2 − 4 y = 0.

First write down the auxiliary equation:

m 2 − 4 = 0

Now find the roots of the auxiliary equation:

m = ± 2

Finally give the general solution to the ODE:

y = A e 2 x + B e − 2 x (Since the roots of the auxiliary equation are real and distinct.)

Task!

Find the general solution of d 2 y d x 2 + 9 y = 0

First write down the auxiliary equation:

m 2 + 9 = 0

Now Find the roots of this auxiliary equation:

m = ± 3 ß

Finally give the general solution to the ODE:

y = A cos 3 x + B sin 3 x

(Since the roots of the auxiliary equation are complex conjugates with real part α = 0 and imaginary part β = 3 .)

The two Tasks above can be generalised as in Key Point 3.

Key Point 3
  1. The general solution to: d 2 y d x 2 − n 2 y = 0 is
    y = A e n x + B e − n x
    or, equivalently using hyperbolic functions,
    y = C cosh n x + D sinh n x
  2. The general solution to: d 2 y d x 2 + n 2 y = 0 is

    y = A cos n x + B sin n x

Those of you who are familiar with elementary dynamics will recognise the second differential equation in Key Point 3 as modelling simple harmonic motion .