HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
Solve2 yd^3 y
dx^3+6
dy
dxd^2 y
dx^2=x. (15.80)Directing our attention to the term on the LHS of (15.80) that contains the highest-order
derivative, i.e. 2yd^3 y/dx^3 , we see that it can be obtained by differentiating 2yd^2 y/dx^2 since
d
dx(
2 yd^2 y
dx^2)
=2yd^3 y
dx^3+2
dy
dxd^2 y
dx^2. (15.81)
Rewriting the LHS of (15.80) using (15.81), we are left with 4(dy/dx)(d^2 y/dy^2 ), which may
itself be written as a derivative, i.e.
4
dy
dxd^2 y
dx^2=
d
dx[
2
(
dy
dx) 2 ]
. (15.82)
Since, therefore, we can write the LHS of (15.80) as a sum of simple derivatives of other
functions, (15.80) is exact. Integrating (15.80) with respect tox, and using (15.81) and
(15.82), now gives
2 yd^2 y
dx^2+2
(
dy
dx) 2
=
∫
xdx=x^2
2+c 1. (15.83)Now we can repeat the process to find whether (15.83) is itself exact. Considering the term
on the LHS of (15.83) that contains the highest-order derivative, i.e. 2yd^2 y/dx^2 , we note
that we obtain this by differentiating 2ydy/dx, as follows:
d
dx(
2 ydy
dx)
=2yd^2 y
dx^2+2
(
dy
dx) 2
.
The above expression already contains all the terms on the LHS of (15.83), so we can
integrate (15.83) to give
2 ydy
dx=
x^3
6+c 1 x+c 2.Integrating once more we obtain the solution
y^2 =x^4
24+
c 1 x^2
2+c 2 x+c 3 .It is worth noting that both linear equations (as discussed in subsection 15.2.2)and non-linear equations may sometimes be made exact by multiplying through
by an appropriate integrating factor. Although no general method exists for
finding such a factor, one may sometimes be found by inspection or inspired
guesswork.
Solution method.Rearrange the equation so that all the terms containingyor its
derivatives are on the LHS, then check to see whether the equation is exact by
attempting to write the LHS as a simple derivative. If this is possible then the
equation is exact and may be integrated directly to give an equation of one order
lower. If the new equation is itself exact the process can be repeated.