where M, N are homogeneous polynomials in x and y, of the same order;
by putting v = y/x and eliminating y, the equation becomes of the
first type considered above, in v and x. An equation (aB <> bA)
(ax + by + c)dy/dx = Ax + By + C
may be reduced to this rule by first putting x + h, y + k for x and y,
and determining h, k so that ah + bk + c = 0, Ah + Bk + C = 0.
(2) An equation in which y does not explicitly occur,
f(x, dy/dx) = 0,
may, theoretically, be reduced to the type dy/dx = F(x); similarly an
equation F(y, dy/dx) = 0.
(3) An equation
f(dy/dx, x, y) = 0,
which is an integral polynomial in dy/dx, may, theoretically, be
solved for dy/dx, as an algebraic equation; to any root dy/dx = F1(x,
y) corresponds, suppose, a solution [phi]1(x, y, c) = 0, where c is an
arbitrary constant; the product equation [phi]1(x, y, c)[phi]2(x, y,
c) ... = 0, consisting of as many factors as there were values of
dy/dx, is effectively as general as if we wrote [phi]1(x, y, c1)
[phi]2(x, y, c2) ... = 0; for, to evaluate the first form, we must
necessarily consider the factors separately, and nothing is then
gained by the multiple notation for the various arbitrary constants.
The equation [phi]1(x, y, c)[phi]2(x, y, c) ... = 0 is thus the
solution of the given differential equation.
In all these cases there is, except for cases of singular solutions,
one and only one arbitrary constant in the most general solution of
the differential equation; that this must necessarily be so we may
take as obvious, the differential equation being supposed to arise by
elimination of this constant from the equation expressing its solution
and the equation obtainable from this by differentiation in regard to
x.
A further type of differential equation of the first order, of the
form
dy/dx = A + By + Cy^2
in which A, B, C are functions of x, will be briefly considered below
under differential equations of the second order.
When we pass to ordinary differential equations of the second order,
that is, those expressing a relation between x, y, dy/dx and d^2y/dx^2,
the number of types for which the solution can be found by a known
procedure is very considerably reduced. Consider the general linear
equation
d^2y dy
--- + P-- + Qy = R,
dx^2 dx
where P, Q, R are functions of x only. There is no method always
effective; the main general result for such a linear equation is that
if any particular function of x, say y1, can be discovered, for which
d^2y1 dy1
---- + P--- + Qy1 = 0,
dx^2 dx
then the substitution y = y1[eta] in the original equation, with R on
the right side, reduces this to a linear equation of the first order
with the dependent variable d[eta]/dx. In fact, if y = y1[eta] we have
dy d[eta] dy1 d^2y d^2[eta] dy1 d[eta] d^2y1
-- = y1------ + [eta]--- and --- = y1------- + 2--- ------ + [eta]-----,
dx dx dx dx^2 dx^2 dx dx dx^2
and thus
d^2y dy d^2[eta] / dy1 \ d[eta] /d^2y1 dy1 \
--- + P -- + Qy = y1------- + ( 2--- + Py1) ------ + ( ----- + P--- + Qy1)[eta];
dx^2 dx dx^2 \ dx / dx \ dx^2 dx /
if then
d^2y1 dy1
---- + P --- + Qy1 = 0,
dx^2 dx
and z denote d[eta]/dx, the original differential equation becomes
dz / dy1 \
y1-- + ( 2--- + Py1)z = R.
dx \ dx /
From this equation z can be found by the rule given above for the
linear equation of the first order, and will involve one arbitrary
constant; thence y = y1 [eta] = y1 [int] zdx + Ay1, where A is another
arbitrary constant, will be the general solution of the original
equation, and, as was to be expected, involves two arbitrary
constants.
The case of most frequent occurrence is that in which the coefficients
P, Q are constants; we consider this case in some detail. If [t]*
be a root of the quadratic equation [t]^2 + [t]P + Q = 0, it
can be at once seen that a particular integral of the differential
equation with zero on the right side is y1 = e^[theta]x. Supposing
first the roots of the quadratic equation to be different, and [phi]
to be the other root, so that [p] + [t] = -P, the auxiliary
differential equation for z, referred to above, becomes dz/dx +
([t] - [p])z = Re^(-[t]^x), which leads to
ze^{([t]-[p])^x} = B + [int] Re^(-[p]^x)dx, where B is an
arbitrary constant, and hence to
(*) [t] = [theta]; [p] = [phi].
_ _ _
/ / /
y = Ae^([t]^x) + e^([t]^x)| Be^([p]-[t])^x dx + e^[t]^x | e^([p]-[t])^x | Re^-[p]^x dxdx,
_/ _/ _/
or say to y = Ae^[t]^x + Ce^[p]^x + U, where A, C are arbitrary
constants and U is a function of x, not present at all when R = 0. If
the quadratic equation [t]^2 + P[t] + Q = 0 has equal roots, so that
2[t] = -P, the auxiliary equation in z becomes dz/dx = Re^-[t]^x,
giving z = B + [int] Re^-[t]^x dx, where B is an arbitrary constant,
and hence
_ _
/ /
y = (A + Bx)e^[t]^x + e^[t]^x | | Re^-[t]^x dxdx,
_/ _/
or, say, y = (A + Bx)e^[t]^x + U, where A, B are arbitrary constants,
and U is a function of x not present at all when R = 0. The portion
Ae^[t]^x + Be^[p]^x or (A + Bx)e^[t]^x of the solution, which is known
as the _complementary function_, can clearly be written down at once
by inspection of the given differential equation. The remaining
portion U may, by taking the constants in the complementary function
properly, be replaced by any particular solution whatever of the
differential equation
d^2v dy
---- + P -- + Qy = R;
dx^2 dx
for if u be any particular solution, this has a form
u = A0 e^[t]^x + B0 e^[p]^x + U,
or a form
u = (A0 + B0x)e^[t]^x + U;
thus the general solution can be written
(A - A0)e^[t]^x + (B - B0)e^[p]^x + u,
or
{A - A0 + (B - B0)x}e^[t]^x + u,
where A - A0, B - B0, like A, B, are arbitrary constants.
A similar result holds for a linear differential equation of any
order, say
d^n y d^n-1 y
----- + P1 ------- + ... + P_n y = R,
dx_n dx^n-1
where P1, P2, ... Pn are constants, and R is a function of x. If we
form the algebraic equation [t]^n + P1[t]^n-1 + ... + P_n = 0, and all
the roots of this equation be different, say they are [t]1, [t]2, ...
[t]n, the general solution of the differential equation is
y = A1 e^[t]1^x + A2 e^[t]2^x + ... + A_n e^[t]_n^x + u,
where A1, A2, ... An are arbitrary constants, and u is any particular
solution whatever; but if there be one root [t]1 repeated r times, the
terms A1 e^[t]1^x + ... + A_r e^[t]_r^x must be replaced by (A1 + A2x
+ ... + A_r x^r-1)e^[t]1x where A1, ... An are arbitrary constants;
the remaining terms in the complementary function will similarly need
alteration of form if there be other repeated roots.
To complete the solution of the differential equation we need some
method of determining a particular integral u; we explain a procedure
which is effective for this purpose in the cases in which R is a sum
of terms of the form e^ax[p](x), where [p](x) is an integral
polynomial in x; this includes cases in which R contains terms of the
form cos bx.[p](x) or sin bx.[p](x). Denote d/dx by D; it is clear
that if u be any function of x, D(e^ax u) = e^ax Du + ae^ax u, or say,
D(e^ax u) = e^ax (D + a)u; hence D^2(e^ax u), i.e. d^2/dx^2 (e^ax u),
being equal to D(e^ax v), where v=(D + a)u, is equal to e^ax(D + a)v,
that is to e^ax(D + a)^2u. In this way we find D^n(e^ax u) = e^ax(D +
a)^n u, where n is any positive integer. Hence if [psi](D) be any
polynomial in D with constant coefficients, [psi](D)(e^ax u) = e^ax
[psi](D + a)u. Next, denoting [int] udx by D^-1 u, and any solution of
the differential equation dz/dx + az = u by z = (d + a)^-1 u, we have
D[e^ax(D + a)^-1 u] = D(e^ax z) = e^ax(D + a)z = e^ax u, so that we
may write D^-1(e^ax u) = e^ax(D+a)^-1 u, where the meaning is that one
value of the left side is equal to one value of the right side; from
this, the expression D^-2(e^axu), which means D^-1[D^-1(e^ax u)], is
equal to D^-1(e^ax z) and hence to e^ax(D + a)^-1 z, which we write
e^ax(D + a)^-2 u; proceeding thus we obtain
D^-n(e^ax u) = e^ax(D + a)^-n u,
where n is any positive integer, and the meaning, as before, is that
one value of the first expression is equal to one value of the second.
More generally, if [psi](D) be any polynomial in D with constant
coefficients, and we agree to denote by 1/[psi](D) u any solution z of
the differential equation [psi](D)z = u, we have, if v = 1/[psi](D +
a) u, the identity [psi](D)(e^ax v) = e^ax [psi](D + a)v = e^ax u,
which we write in the form
1 1
--------(e^ax u) = e^ax ------------ u.
[psi](D) [psi](D + a)
This gives us the first step in the method we are explaining, namely
that a solution of the differential equation [psi](D)y = e^ax u + e^bx
v + ... where u, v, ... are any functions of x, is any function
denoted by the expression
1 1
e^ax ------------ u + e^ax ------------ v + ....
[psi](D + a) [psi](D + b)
It is now to be shown how to obtain one value of 1/[psi](D + a) u,
when u is a polynomial in x, namely one solution of the differential
equation [psi](D + a)z = u. Let the highest power of x entering in u
be x^m; if t were a variable quantity, the rational fraction in t,
1/[psi](t + a), by first writing it as a sum of partial fractions, or
otherwise, could be identically written in the form
K_r t^-r + K_r-1 t^-r+1 + ... + K1 t^-1 + H + H1t + ... + H_m t^m + t^m+1 [p](t)/[psi](t + a),
where [p](t) is a polynomial in t; this shows that there exists an
identity of the form
1 = [psi](t + a)(K_r t^-r + ... + K1t^-1 + H + H1t + ... + H_m t^m) + [p](t)t^m+1,
and hence an identity
u = [psi](D + a)[K_r D^-r + ... + K1D^-1 + H + H1D + ... + H_m D^m]u + [p](D)D^m+1 u;
in this, since u contains no power of x higher than x^m, the second
term on the right may be omitted. We thus reach the conclusion that a
solution of the differential equation [psi](D + a)z = u is given by
z = (K_r D^-r + ... + K1D^-1 + H + H1D + ... + H_m D^m)u,
of which the operator on the right is obtained simply by expanding
1/[psi](D + a) in ascending powers of D, as if D were a numerical
quantity, the expansion being carried as far as the highest power of D
which, operating upon u, does not give zero. In this form every term
in z is capable of immediate calculation.
_Example._--For the equation
d^4v d^2y
---- + 2--- + y = x^3 cos x or (D^2 + 1)^2y = x^3 cos x,
dx^4 dx^3
the roots of the associated algebraic equation ([t]^2+1)^2 = 0 are [t] =
[+-]i, each repeated; the complementary function is thus
(A + Bx)e^ix + (C + Dx)e^ix,
where A, B, C, D are arbitrary constants; this is the same as
(H + Kx) cos x + (M + Nx) sin x,
where H, K, M, N are arbitrary constants. To obtain a particular
integral we must find a value of (1 + D^2)^-2 x^3 cos x; this is the
real part of (1 + D^2)^-2 e^ix x^3 and hence of e^ix [1 + (D + i)^2]^-2
x^3
or e^ix [2iD(1 + 1/2iD)]^-2 x^3,
or -1/4e^ix D^-2 (1 + iD - 3/4D^2 - 1/2iD^3 + 5/16 D^4 + 3/16 iD^5 ...)x^3,
or -1/4e^ix(1/20 x^5 + 1/4ix^4 - 3/4x^3 - 3/2 ix^2 + 15/8 x + 9/8 i);
the real part of this is
-1/4(1/20 x^5 - 3/4x^2 + 15/8 x) cos x + 1/4(1/4x^4 - 3/2 x^2 + 9/8) sin x.
This expression added to the complementary function found above gives
the complete integral; and no generality is lost by omitting from the
particular integral the terms -15/32 x cos x + 9/32 sin x, which are
of the types of terms already occurring in the complementary function.
The symbolical method which has been explained has wider applications
than that to which we have, for simplicity of explanation, restricted
it. For example, if [psi](x) be any function of x, and a1, a2, ... an
be different constants, and [(t + a1) (t + a2) ... (t + an)]^-1 when
expressed in partial fractions be written [Sigma]c_m(t + a_m)^-1, a
particular integral of the differential equation (D + a1)(D + a2) ...
(D + a_n)y = [psi](x) is given by
y = [Sigma]c_m(D + a_m)^-1 [psi](x) = [Sigma]c_m(D + a_m)^-1 e^-a m^x e^a m^x [psi](x) =
[Sigma]c_m e^-a m^x D^-1 (e^a m^x [psi](x)) = [Sigma]c_m e^-a m^x [int] e^a m^x [psi](x)dx.
The particular integral is thus expressed as a sum of n integrals.