General ordinary differential equations

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
15.3 General ordinary differential equations
In this section, we discuss miscellaneous methods for simplifying general ODEs.
These methods are applicable to both linear and non-linear equations and in
some cases may lead to a solution. More often than not, however, finding a
closed-form solution to a general non-linear ODE proves impossible.
15.3.1 Dependent variable absent
If an ODE does not contain the dependent variable y explicitly, but only its
derivatives, then the change of variable p = dy/dx leads to an equation of one
order lower.
Solve
d2 y
dy
+2
= 4x
dx2
dx
(15.76)
This is transformed by the substitution p = dy/dx to the first-order equation
dp
+ 2p = 4x.
dx
(15.77)
The solution to (15.77) is then found by the method of subsection 14.2.4 and reads
p=
dy
= ae−2x + 2x − 1,
dx
where a is a constant. Thus by direct integration the solution to the original equation,
(15.76), is
y(x) = c1 e−2x + x2 − x + c2 . An extension to the above method is appropriate if an ODE contains only
derivatives of y that are of order m and greater. Then the substitution p = dm y/dxm
reduces the order of the ODE by m.
Solution method. If the ODE contains only derivatives of y that are of order m and
greater then the substitution p = dm y/dxm reduces the order of the equation by m.
15.3.2 Independent variable absent
If an ODE does not contain the independent variable x explicitly, except in d/dx,
d2 /dx2 etc., then as in the previous subsection we make the substitution p = dy/dx
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15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONS
but also write
d2 y
dy dp
dp
dp
=
=p
=
dx2
dx
dx dy
dy
2
dp
d2 p
d3 y
d
dp
dy d
dp
=
,
p
=
p
= p2 2 + p
3
dx
dx
dy
dx dy
dy
dy
dy
(15.78)
and so on for higher-order derivatives. This leads to an equation of one order
lower.
Solve
1+y
d2 y
+
dx2
dy
dx
2
= 0.
(15.79)
Making the substitutions dy/dx = p and d2 y/dx2 = p(dp/dy) we obtain the first-order
ODE
dp
1 + yp
+ p2 = 0,
dy
which is separable and may be solved as in subsection 14.2.1 to obtain
(1 + p2 )y 2 = c1 .
Using p = dy/dx we therefore have
p=
dy
=±
dx
c21 − y 2
,
y2
which may be integrated to give the general solution of (15.79); after squaring this reads
(x + c2 )2 + y 2 = c21 . Solution method. If the ODE does not contain x explicitly then substitute p =
dy/dx, along with the relations for higher derivatives given in (15.78), to obtain an
equation of one order lower, which may prove easier to solve.
15.3.3 Non-linear exact equations
As discussed in subsection 15.2.2, an exact ODE is one that can be obtained by
straightforward differentiation of an equation of one order lower. Moreover, the
notion of exact equations is useful for both linear and non-linear equations, since
an exact equation can be immediately integrated. It is possible, of course, that
the resulting equation may itself be exact, so that the process can be repeated.
In the non-linear case, however, there is no simple relation (such as (15.43) for
the linear case) by which an equation can be shown to be exact. Nevertheless, a
general procedure does exist and is illustrated in the following example.
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HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
Solve
2y
d3 y
dy d2 y
+6
= x.
dx3
dx dx2
(15.80)
Directing our attention to the term on the LHS of (15.80) that contains the highest-order
derivative, i.e. 2y d3 y/dx3 , we see that it can be obtained by differentiating 2y d2 y/dx2 since
d2 y
d3 y
d
dy d2 y
2y 2 = 2y 3 + 2
.
(15.81)
dx
dx
dx
dx dx2
Rewriting the LHS of (15.80) using (15.81), we are left with 4(dy/dx)(d2 y/dy 2 ), which may
itself be written as a derivative, i.e.
2
dy d2 y
d
dy
4
=
.
(15.82)
2
dx dx2
dx
dx
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 to x, and using (15.81) and
(15.82), now gives
2 d2 y
x2
dy
2y 2 + 2
= x dx =
(15.83)
+ c1 .
dx
dx
2
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. 2y d2 y/dx2 , we note
that we obtain this by differentiating 2y dy/dx, as follows:
2
dy
d2 y
dy
d
2y
= 2y 2 + 2
.
dx
dx
dx
dx
The above expression already contains all the terms on the LHS of (15.83), so we can
integrate (15.83) to give
x3
dy
=
+ c1 x + c2 .
dx
6
Integrating once more we obtain the solution
2y
y2 =
x4
c1 x2
+
+ c2 x + c3 . 24
2
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 containing y or 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.
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15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONS
15.3.4 Isobaric or homogeneous equations
It is straightforward to generalise the discussion of first-order isobaric equations
given in subsection 14.2.6 to equations of general order n. An nth-order isobaric
equation is one in which every term can be made dimensionally consistent upon
giving y and dy each a weight m, and x and dx each a weight 1. Then the nth
derivative of y with respect to x, for example, would have dimensions m in y
and −n in x. In the special case m = 1, for which the equation is dimensionally
consistent, the equation is called homogeneous (not to be confused with linear
equations with a zero RHS). If an equation is isobaric or homogeneous then the
change in dependent variable y = vxm (y = vx in the homogeneous case) followed
by the change in independent variable x = et leads to an equation in which the
new independent variable t is absent except in the form d/dt.
Solve
x3
dy
d2 y
− (x2 + xy)
+ (y 2 + xy) = 0.
dx2
dx
(15.84)
Assigning y and dy the weight m, and x and dx the weight 1, the weights of the five terms
on the LHS of (15.84) are, from left to right: m + 1, m + 1, 2m, 2m, m + 1. For these
weights all to be equal we require m = 1; thus (15.84) is a homogeneous equation. Since it
is homogeneous we now make the substitution y = vx, which, after dividing the resulting
equation through by x3 , gives
dv
d2 v
+ (1 − v)
= 0.
dx2
dx
t
Now substituting x = e into (15.85) we obtain (after some working)
x
d2 v
dv
−v
= 0,
dt2
dt
which can be integrated directly to give
dv
= 12 v 2 + c1 .
dt
Equation (15.87) is separable, and integrates to give
dv
1
t + d2 =
2
v 2 + d21
1
v
.
=
tan−1
d1
d1
(15.85)
(15.86)
(15.87)
Rearranging and using x = et and y = vx we finally obtain the solution to (15.84) as
y = d1 x tan 12 d1 ln x + d1 d2 . Solution method. Assume that y and dy have weight m, and x and dx weight 1,
and write down the combined weights of each term in the ODE. If these weights can
be made equal by assuming a particular value for m then the equation is isobaric
(or homogeneous if m = 1). Making the substitution y = vxm followed by x = et
leads to an equation in which the new independent variable t is absent except in the
form d/dt.
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HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
15.3.5 Equations homogeneous in x or y alone
It will be seen that the intermediate equation (15.85) in the example of the
previous subsection was simplified by the substitution x = et , in that this led to
an equation in which the new independent variable t occurred only in the form
d/dt; see (15.86). A closer examination of (15.85) reveals that it is dimensionally
consistent in the independent variable x taken alone; this is equivalent to giving
the dependent variable and its differential a weight m = 0. For any equation that
is homogeneous in x alone, the substitution x = et will lead to an equation that
does not contain the new independent variable t except as d/dt. Note that the
Euler equation of subsection 15.2.1 is a special, linear example of an equation
homogeneous in x alone. Similarly, if an equation is homogeneous in y alone, then
substituting y = ev leads to an equation in which the new dependent variable, v,
occurs only in the form d/dv.
Solve
x2
dy
d2 y
2
+x
= 0.
+
dx2
dx y 3
This equation is homogeneous in x alone, and on substituting x = et we obtain
2
d2 y
+ 3 = 0,
dt2
y
which does not contain the new independent variable t except as d/dt. Such equations
may often be solved by the method of subsection 15.3.2, but in this case we can integrate
directly to obtain
dy
= 2(c1 + 1/y 2 ).
dt
This equation is separable, and we find
dy
= t + c2 .
2(c1 + 1/y 2 )
By multiplying the numerator and denominator of the integrand on the LHS by y, we find
the solution
c1 y 2 + 1
√
= t + c2 .
2c1
Remembering that t = ln x, we finally obtain
c1 y 2 + 1
√
= ln x + c2 . 2c1
Solution method. If the weight of x taken alone is the same in every term in the
ODE then the substitution x = et leads to an equation in which the new independent
variable t is absent except in the form d/dt. If the weight of y taken alone is the
same in every term then the substitution y = ev leads to an equation in which the
new dependent variable v is absent except in the form d/dv.
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