Math240
Elementary Differential Equations
Fall 2003
Kansas State University
Extra Credit Assignment #1:
Reducible Second-Order Equations
Due in recitation Thursday, October 2, 2003
You are encouraged to collaborate with your colleagues. For credit, however, your
final write-up must be done individually. Show all your work and make your presentation comprehensible.
A second-order differential equation has the general form
F (x, y, y 0 , y 00 ) = 0,
where F is a funtion relating x, y, y 0 , and y 00 . If the independent vairable x or the
dependent variable y is not present in the differential equation, then the order of
the equation can be reduced using a substitution.
Dependent variable y missing: The differential equation has the form
F (x, y 0 , y 00 ) = 0.
We can substitute v = y 0 , so v 0 = y 00 . The differential equation becomes
F (x, v, v 0 ) = 0,
which is a first-order equation. If this can be solved for the unknown function v,
then the unknown function y is
Z
Z
0
y(x) = y (x) dx = v(x) dx.
The solution y will have two arbitrary constants (as is typical for solutions to
second-order equations).
Example. Solve xy 00 + 2y 0 = 6x (y is not explicitly present in this equation).
Substituting v = y 0 leads to
2
xv 0 + 2v = 6x ⇒ v 0 + v = 6.
x
This is a first-order linear equation; it’s solutions are
v = 2x +
C1
.
x2
Now v = y 0 , so
y(x) =
Z
0
y (x) dx =
Z µ
C1
2x +
x
where C1 and C2 are arbitrary constants.
1
¶
dx = x2 +
C1
+ C2 ,
x
Independent variable x missing: The differential equation has the form
F (y, y 0 , y 00 ) = 0.
For these types of equations, it’s useful to think of x as a function of y, and thus
y 0 (x) can also be thought of as a function of y (since x is). We substitute
v = y0 =
dv
dv dy
dv
dy
⇒ y 00 =
=
=v ;
dx
dx
dy dx
dy
we used the chain rule, since we are thinking of y 0 as a function of y. The differential
equation becomes
¶
µ
dv
= 0,
F y, v, v
dy
where p is a function of y. If this can be solved for v, then we can find x as a
function y by
Z
Z
Z µ ¶−1
dx
1
dy
x(y) =
dx =
dy =
dy.
dy
dx
v(y)
This yields an implicit formula for y wich will include two arbitrary constants.
Example. Solve yy 00 = (y 0 )2 (x is not present in this equation). Substituting v = y 0
leads to
dv
= v2.
yv
dy
This is a first-order separable equation; it’s solutions are
v = C1 y.
Now
1
= x0 , so
v
x(y) =
Z
0
x (y) dy =
Z
1
1
dy =
ln |y| + C2 .
C1 y
C1
We solve this solution for y explicitly
ln |y| = C1 x − C2 ⇒ y(x) = C2 eC1 x ,
where C1 and C2 are arbitrary constants.
2
Solve the following problems.
1. Find all solutions to xy 00 = y 0 .
2
2. Find all solutions to y 00 = (y 0 ) .
3. Suppose that a uniform flexible cable is suspended between two points (−L, H)
and (L, H) at equal heights. Physical principles can be used to show that
the shape of the hanging cable can be represented by the graph of a function
y = y(x) satisfying the differential equation
q
00
ay = 1 + (y 0 )2 ,
where the constant a is determined by the properties of the cable. Find the
shape function y for the hanging cable. (Note: the graph of the solution is
called a catenary curve, and it does match very closely with a real hanging
cable.)
4. In your calculus courses, you learned that the curvature κ of a curve y = y(x)
at the point (x, y) is given by
κ= £
|y 00 (x)|
1 + (y 0 (x))2
¤ 32 .
Assuming that y 00 ≥ 0, a curve with constant curvature r will satisfy
i3
1 00 h
0 2 2
y = 1 + (y )
.
r
Find the general solution to this differential equation. Give a geometric description of a curve with constant curvature r?
3