15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS
y=xλ. This leads to an algebraic equation whose solution gives the allowed values
ofλ; the general solution is then the linear superposition of these functions.
15.2.2 Exact equationsSometimes an ODE may be merely the derivative of another ODE of one order
lower. If this is the case then the ODE is called exact. Thenth-order linear ODE
an(x)dny
dxn+···+a 1 (x)dy
dx+a 0 (x)y=f(x), (15.41)is exact if the LHS can be written as a simple derivative, i.e. if
an(x)dny
dxn+···+a 0 (x)y=d
dx[
bn− 1 (x)dn−^1 y
dxn−^1+···+b 0 (x)y]. (15.42)
It may be shown that, for (15.42) to hold, we require
a 0 (x)−a′ 1 (x)+a′′ 2 (x)−···+(−1)na(nn)(x)=0, (15.43)where the prime again denotes differentiation with respect tox. If (15.43) is
satisfied then straightforward integration leads to a new equation of one order
lower. If this simpler equation can be solved then a solution to the original
equation is obtained. Of course, if the above process leads to an equation that is
itself exact then the analysis can be repeated to reduce the order still further.
Solve
(1−x^2 )d^2 y
dx^2− 3 xdy
dx−y=1. (15.44)Comparing with (15.41), we havea 2 =1−x^2 ,a 1 =− 3 xanda 0 =−1. It is easily shown
thata 0 −a′ 1 +a′′ 2 = 0, so (15.44) is exact and can therefore be written in the form
d
dx[
b 1 (x)dy
dx+b 0 (x)y]
=1. (15.45)
Expanding the LHS of (15.45) we find
d
dx(
b 1dy
dx+b 0 y)
=b 1d^2 y
dx^2+(b′ 1 +b 0 )dy
dx+b′ 0 y. (15.46)Comparing (15.44) and (15.46) we find
b 1 =1−x^2 ,b′ 1 +b 0 =− 3 x, b′ 0 =− 1.These relations integrate consistently to giveb 1 =1−x^2 andb 0 =−x, so (15.44) can be
written as
d
dx
[
(1−x^2 )dy
dx−xy]
=1. (15.47)
Integrating (15.47) gives us directly the first-order linear ODE
dy
dx−
( x1 −x^2)
y=x+c 1
1 −x^2,
which can be solved by the method of subsection 14.2.4 and has the solution
y=c 1 sin−^1 x+c 2
√
1 −x^2