HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
Substituting equations (15.37) into the original equation (15.36), the latter becomes
a linear ODE with constant coefficients, i.e.
anαn
d
dt
(
d
dt
− 1
)
···
(
d
dt
−n+1
)
y+···+a 1 α
dy
dt
+a 0 y=f
(
et−β
α
)
,
which can be solved by the methods of section 15.1.
A special case of Legendre’s linear equation, for whichα= 1 andβ=0,is
Euler’s equation,
anxn
dny
dxn
+···+a 1 x
dy
dx
+a 0 y=f(x); (15.38)
it may be solved in a similar manner to the above by substitutingx=et.If
f(x) = 0 in (15.38) then substitutingy=xλleads to a simple algebraic equation
inλ, which can be solved to yield the solution to (15.38). In the event that the
algebraic equation forλhas repeated roots, extra care is needed. Ifλ 1 is ak-fold
root (k>1) then theklinearly independent solutions corresponding to this root
arexλ^1 ,xλ^1 lnx,...,xλ^1 (lnx)k−^1.
Solve
x^2
d^2 y
dx^2
+x
dy
dx
− 4 y= 0 (15.39)
by both of the methods discussed above.
First we make the substitutionx=et, which, after cancellinget, gives an equation with
constant coefficients, i.e.
d
dt
(
d
dt
− 1
)
y+
dy
dt
− 4 y=0 ⇒
d^2 y
dt^2
− 4 y=0. (15.40)
Using the methods of section 15.1, the general solution of (15.40), and therefore of (15.39),
is given by
y=c 1 e^2 t+c 2 e−^2 t=c 1 x^2 +c 2 x−^2.
Since the RHS of (15.39) is zero, we can reach the same solution by substitutingy=xλ
into (15.39). This gives
λ(λ−1)xλ+λxλ− 4 xλ=0,
which reduces to
(λ^2 −4)xλ=0.
This has the solutionsλ=±2, so we obtain again the general solution
y=c 1 x^2 +c 2 x−^2 .
Solution method. If the ODE is of the Legendre form (15.36) then substituteαx+
β=et. This results in an equation of the same order but with constant coefficients,
which can be solved by the methods of section 15.1. If the ODE is of the Euler
form (15.38) with a non-zero RHS then substitutex=et; this again leads to an
equation of the same order but with constant coefficients. If, however,f(x)=0in
the Euler equation (15.38) then the equation may also be solved by substituting