A seldom used formula for ODEs
Background
All too often a teacher of mathematics will encounter students who beg for a formula that
they can apply blindly and quickly to a narrow set of problems. If those students take the
time and effort to understand the process that leads to that formula, then they may gain
the ability to transfer that understanding beyond that narrow set of problems.
One example appears clearly in my teaching of ordinary differential equations to
engineering students in their second year of university studies. When the method of
variation of parameters is used to obtain the particular solution to a second order linear
ordinary differential equation, some students bypass the derivation and note only the
following formula.
The particular solution of the ODE y  p  x  y  q  x  y  r  x  is



 y2 r
yP  
dx  y1 
 y1 y2  y1 y2 


where the complementary function is yC  A y1



 y1r

dx  y2
 y1 y2  y1 y2 


 B y2 .
Much is lost in a blind reliance on this formula. Among other disadvantages, it can be
difficult to remember which of the two terms has the negative sign in the numerator of
the integral.
After I present the derivation of the method to my students, I rely upon formulas to a
lesser degree. The recipe that I use to find the particular solution is:
Find the complementary function yC  A y1  B y2 .
Evaluate the Wronskian W 
y1
y1
y2
 y1 y2  y1 y2 .
y2
Evaluate the related determinants W1 
0
r
y2
y 0
  y2 r and W2  1
  y1r .
y2
y1 r
W1
.
W
W
Find the function v(x) from v  2 .
W
Simplify yP  u y1  v y2 .
Find the function u(x) from u 
In addition to keeping the signs straight on  y2 r and  y1r , students are more likely to
recall the use of Cramer’s rule in the derivation of the method of variation of parameters
(and, indeed, to recall Cramer’s rule at all!).
Mathematical Gazette, vol. 92, pp. 344-348
Page 2
Example
As an example of the method, it can be shown that the complementary function for the
ordinary differential equation
x 2 y  5 x y  8 y  3x
or equivalently (for x ≠ 0),
5
8
3
y  y  2 y 
x
x
x
4
2
is
y  Ax  B x
Now find the particular solution by the method of variation of parameters.
y y2
x4 x2
W  1

 2 x5  4 x5   2 x5
y1 y2
4 x3 2 x
W1 
0
r
y2
3
  y2 r   x 2   3x
y2
x
y1 0
3
  y1r  x 4  3x3
y1 r
x
W
3 x
3
1
u  1 
 x 4
 u 
5
W
2 x
2
2 x3
W
3x3
3
3
v  2 
  x 2
 v 
5
W
2 x
2
2x
1
3 2
 yP  u y1  v y2   3 x 4 
x 
2x
2x
Therefore the general solution of the ODE x 2 y
y  A x4  B x2  x
W2 
 1 3
  x  x
 2 2
 5 x y  8 y  3x is
A general formula
However, once students have attained a comfort level with the solution of these ODEs, it
can be instructive to derive a formula for the general solution of any second order linear
ordinary differential equation with constant real coefficients,
y  p y  q y  r  x 
I have not noticed such a formula in the textbooks that I have seen, which is just as well.
Novice students of ordinary differential equations really should gain plenty of practice in
the methods of solving such problems before even considering such a shortcut.
The auxiliary (or characteristic) equation  2  p  q  0 produces the three cases
1. real distinct roots   a or b ;
2. real equal roots   a, a ; or
3. complex conjugate pair of roots   a  b j .
Mathematical Gazette, vol. 92, pp. 344-348
Page 3
1. Real, distinct roots:
Complementary function:
yC  A y1  B y2  Aeax  B ebx
Particular solution:
y
W  1
y1
y2
eax

y2
a eax
ebx
bx
be
  b  a  eaxebx
y2
W
ebx r
e  a x r
  y2 r   ebx r  u  1 

y2
W
ba
 b  a  ea xebx
W1 
0
r
W2 
y1 0
W
ea x r
ebx r
  y1r   ea x r  v  2 

y1 r
W
ba
 b  a  ea xebx

yP  u y1  v y2 
ea x  r e a x dx  ebx  r e bx dx
ba
Therefore the general solution (in the case b ≠ a) of the ODE
y   a  b  y  ab y  r  x 
is
y  x   Ae a x  B ebx 
ebx  r e bx dx  e a x  r e  a x dx
ba
2. Real, equal roots:
Complementary function:
yC  A y1  B y2  Ae a x  B x e a x
Particular solution:
y y2
ea x
W  1

y1 y2
a ea x
W1 
0
r
x ea x
ax 2
a x  e 
 ax  1 e
y2
W
 x eax r
  y2 r   x eax r  u  1  ax ax   x eax r
y2
W
e e
y1 0
W2
eax r
ax

W2 
  y1r   e r  v 
 ax ax  eax r
y1 r
W
e e

yP  u y1  v y2  e a x   x r e  a x dx  x e a x  r e  a x dx
Therefore the general solution of the ODE
y  2a y  a2 y  r  x 
is
Mathematical Gazette, vol. 92, pp. 344-348

y  x   A  Bx  x  r eax dx 
Page 4
 xre
ax

dx eax
3. Complex conjugate pair of roots:
This can be expressed in the same form as type (1), with the constants a, b being a pair of
complex conjugate numbers.
However, it is more convenient to express the
complementary function as a linear combination of a pair of real valued functions.
Complementary function:
yC  A y1  B y2  Aea x cos bx  B e a x sin bx
Particular solution:
y1 y2
ea x cos bx
e a x sin bx
W 
 ax
 b ea x ea x
a
x
y1 y2
e  a cos bx  b sin bx  e  a sin bx  b cos bx 
W1 
0
r
 u 
W2 
W1
ea x sin bx r
ea x sin bx r


W
b ea x ea x
b
y1 0
  y1r   ea x cos bx r
y1 r
 v 

y2
  y2 r   eax sin bx r
y2
W2
ea x cos bx r
ea x cos bx r


W
b ea x ea x
b
yP  u y1  v y2 
ea x cos bx  r e  a x sin bx dx  e a x sin bx  r e  a x cos bx dx
b
Therefore the general solution (in the case b ≠ 0) of the ODE
y  2a y   a 2  b 2  y  r  x 
is

sin bx  r e  a x cos bx dx  cos bx  r e  a x sin bx dx  a x
e
y  x    A cos bx  B sin bx 


b


and it is a nice exercise to show that this solution tends to the equal roots case in the limit
as b → 0.
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