Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

y(0) =y(π/2) = 0 is given by

y(x)=

∫π/ 2

0

G(x, z)coseczdz

=−cosx

∫x

0

sinzcoseczdz−sinx

∫π/ 2

x

coszcoseczdz

=−xcosx+sinxln(sinx),

which agrees with the result obtained in the previous subsections.

As mentioned earlier, once a Green’s function has been obtained for a given

LHS and boundary conditions, it can be used to find a general solution for any

RHS; thus, the solution ofd^2 y/dx^2 +y=f(x), withy(0) =y(π/2) = 0, is given

immediately by

y(x)=

∫π/ 2

0

G(x, z)f(z)dz

=−cosx

∫x

0

sinzf(z)dz−sinx

∫π/ 2

x

coszf(z)dz. (15.68)

As an example, the reader may wish to verify that iff(x)=sin2xthen (15.68)

givesy(x)=(−sin 2x)/3, a solution easily verified by direct substitution. In

general, analytic integration of (15.68) for arbitraryf(x) will prove intractable;

then the integrals must be evaluated numerically.

Another important point is that although the Green’s function method above

has provided a general solution, it is also useful for finding a particular integral

if the complementary function is known. This is easily seen since in (15.68) the

constant integration limits 0 andπ/2 lead merely to constant values by which

the factors sinxand cosxare multiplied; thus the complementary function is

reconstructed. The rest of the general solution, i.e. the particular integral, comes

from the variable integration limitx. Therefore by changing

∫π/ 2

x to−

∫x

,andso

dropping the constant integration limits, we can find just the particular integral.

For example, a particular integral ofd^2 y/dx^2 +y=f(x) that satisfies the above

boundary conditions is given by

yp(x)=−cosx

∫x

sinzf(z)dz+sinx

∫x

coszf(z)dz.

A very important point to realise about the Green’s function method is that a

particularG(x, z) applies to a given LHS of an ODEandthe imposed boundary

conditions, i.e.the same equation with different boundary conditions will have a

different Green’s function. To illustrate this point, let us consider again the ODE

solved in (15.68), but with different boundary conditions.