University Maths Notes: Calculus – Series Solutions of Differential Equations
Many differential equations do not have analytical solutions – solutions expressed in terms of elementary functions. In these circumstances, a numerical solution must be found. One possible numerical solution is in terms of a series expansion.
To find the series expansion we assume a solution of
the form
and
substitute into the differential equation, collecting together and
equating powers of
If
there are any elementary functions in the equation, these are also
expanded in powers ofThe
solution is then found iteratively using the boundary or initial
condition(s). If there is one condition,
or
then
or
respectively
is found first, then the other
found
by iteration. For higher order differential equations, equations in
themust
be formed and solved then the other
found
by iteration.
Example: Solve
with
and
Assume a series solution of the form
then
and
The
equation to be solved becomes
We have to collect like powers of
but
must first re - index the summations so that
and
become
To
do this write
The equation becomes
We can take all the summations inside a single summation to give
Equate each coefficient of
to
zero, since the right hand side is zero.
But
and
so
and
The
solution is
with
higher order terms defined with a recurrence relation in terms of
coefficients of
Example
withand
Assume a series solution of the formthenandThe
equation to be solved becomes
We have to collect like powers ofbut
must first re - index the summations so thatand
becomeTo
do this write
The equation becomes
We can take all the summations inside a single summation to give
Equate each coefficient ofto
zero, since the right hand side is zero.
Butandso
and
The
solution is
with
higher order terms defined with a recurrence relation in terms of
coefficients of