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Solution of the Linear Differential Equation with Nonconstant Coefficients. Reduction of Order Method.

Introductory Remarks

We now examine the general linear differential equation

fn(x)y(n)+f(n1)(x)y(n1)++f1(x)y+f0(x)y=Q(x),(23.1)

and its related homogeneous equation

fn(x)y(n)+f(n1)(x)y(n1)++f1(x)y+f0(x)y=0,(23.11)

where f0(x),f1(x),,fn(x),Q(x) are each continuous functions of x on a common interval I and fn(x)0 when x is in I.

There are no standard methods of finding solutions of (23.1) in the general case. For an unrestricted (23.11) that has a solution expressible in terms of elementary functions, you can use a standard method to find one independent solution, if the other n1 independent solutions are known. For (23.1), the best you can obtain from a standard method is to find one independent solution of (23.11) and a particular solution of (23.1), provided the other n1 independent solutions of (23.11) are known.

Therefore, even when following a standard method for finding a general solution of only the second order equation:

f2(x)y+f1(x)y+f0(x)y=Q(x),(23.12)

it is essential that f0(x),f1(x),f2(x) be of such a character that the needed first solution of its related homogenous equation can be discovered.

Reduction of Order Method

Assuming we have been able to find a non-trivial solution y1 (a solution other than y0, which is always true) to the homogenous equation

f2(x)y+f1(x)y+f0(x)y=0,(23.14)

There is a method, called the reduction of order method, to obtain a second independent solution of (23.14) as well as a particular solution of the related nonhomogenous equation:

f2(x)y+f1(x)y+f0(x)y=Q(x).(23.15)

Let y2(x) be a second solution of (23.14) and assume that it will have the form:

y2(x)=y1(x)u(x)dx(23.2)

where u(x) is an unknown function to be determined.

(jmh why do we/why can we make this assumption? the book does not explain).

We can then find y2 and y2 via differntiation and substitute them into (23.14) and simplify, which will result in a first order differential equation of u. We can then solve for u and substitute it back into (23.2) to solve for y2.

Now that we know y1 and y2, we can write the general solution of (23.14) as y=c1y1+c2y2.

The same approach can be used for finding a general solution to (23.15) using the substitution:

y(x)=y1(x)u(x)dx(23.3)

See the book for details...