Second Order Differential Equations
Equivalent Forms; Initial Value Problems
second order differential equations take the form
In general,
dy
d2 y
=
f
x,
y,
,
dx2
dx
!
where f (x, y, z) is a continuous function of the variables x, y
and z in some region R ⊂ R3 and is continuously differentiable
with respect to y and z throughout that region.
dy
If we define z = dx
, then we can write the above equation
equivalently as a system of two (coupled) first order equations:
dy
dz
= z;
= f (x, y, z).
dx
dx
Written in this form and drawing on experience with first order
equations, we would expect to have to give initial values
y(x0 ) = y0, z(x0 ) = z0 ,
in order to specify an initial value problem. Going back to the
second order equation, this means we need to give
dy
dy
(x0 ) = z0 ; equivalently : (x0 ) = y1,
dx
dx
in order to specify an initial value problem for it.
y(x0 ) = y0 ,
General Solutions We say that an expression y(x, c1 , c2 ) is a
general solution for the second order differential equation above
if, as a function of x, it satisfies the differential equation, and,
given initial values as above at a point x0 , with (x0 , y0 , y1) in
1
the region R where f (x, y, z) has the required properties, we
can solve the equations
y(x0 , c1 , c2 ) = y0,
dy
(x0, c1 , c2 ) = y1
dx
for c1 and c2 to satisfy that initial value problem.
Example 1
Consider the second order differential equation
d2 y
2 dy
2
=
−
y
dx2
x dx
x2
in the region x > 0, and let us try to find solutions of the form
y = xr . Substituting, we have
r(r − 1) xr−2 = 2r xr−2 − 2xr−2
or
xr−2 (r2 − 3r + 2) = 0.
Since xr−2 is not identically zero, we need to solve the so-called
indicial equation
r2 − 3r + 2 = 0 −→ r = 2 or r = 1.
So we conclude that y(x) = x and y(x) = x2 are solutions,
which is easily checked out. Since the equation is linear with
respect to y (if y and ŷ are solutions, so is any linear combination
of the two) it then makes sense to try a general solution of the
form
y(x, c1 , c2 ) = c1 x + c2 x2 .
To satisfy initial conditions as specified above at x0 > 0 we
need
c1 x0 + c2 x0 2 = y0
,
c1 + 2 c2 x0 = y1
a system of two linear algebraic equations. The determinant of
the coefficients of c1 and c2 on the left hand side is 2 x0 2 − x0 2 =
x0 2 6= 0 for x0 6= 0 so we conclude we can always obtain a unique
2
solution of these equations if x0 6= 0. We conclude therefore that
y(x, c1 , c2 ) = c1 x + c2 x2 is the general solution of the second
d2 y
2 dy
2
order differential equation dx
2 = x dx − x2 y for x 6= 0.
A General Method of Solution in the “Autonomous”
Case
If our second order differential equation does not depend explicitly on x, which is technically referred to as the autonomous,
i.e., ”self-governing”, case, we can employ a transformation of
variables to reduce the second order equation to two first order
differential equations. Assuming the equation is
d2 y
dy
=
f
(y,
),
dx2
dx
we let
dy
= z.
dx
Substituting this into the second derivative we have
d2 y
dz
dz dy
dz
=
=
=
z
dx2
dx
dy dx
dy
and thus the original second order equation becomes
z
dz
= f (y, z).
dy
If we can solve this last equation to get z = z(y, c1 ), then the
equation used to define z becomes
dy
= z(y, c1 )
dx
which, being in separable form
1
dy = dx
z(y, c1)
can be integrated to give
Z y
1
dη = x + c2 .
z(η, c1 )
3
Then, at least in principle, we can solve this to get y = y(x, c1 , c2 ).
Success of the project ultimately depends on being able to get a
closed form solution for the transformed second order equation,
z dz
dy = f (y, z), which has now become a first order equation in
z and y.
Example 2 We consider the autonomous second order differential equation
d2 y
dy
.
=
y
dx2
dx
Following the plan indicated above, we arrive at
z
dz
dz
= y z or
= y.
dy
dy
This is solved immediately to obtain
y 2 + c1
z = z(y, c1 ) =
.
2
Then we have
dy
y 2 + c1
=
.
dx
2
There are three cases here, depending on whether c1 is 0, positive, or negative. Taking just the case where c1 is positive, we
write c1 = c2 and we have
2
dy = dx.
c2 + y 2
Setting y = c w we have
2
1
dw = dx
c 1 + w2
!
and we integrate to obtain
2
tan−1 (w) = x + c2 .
c
4
Then w = tan
c
(x
2
+ c2 ) and we have as the general solution
c
y = y(x, c, c2 ) = c tan (x + c2 ) .
2
!
The cases corresponding to c1 = 0 and c1 < 0 are handled in
a similar way but different expressions are obtained. (Try those
cases as an exercise.)
5