MAP 2302
Differential Equations
Sanchez
Part II
Summary 3- Homogeneous Linear DE with Constant Coefficients
I. Theorem- If m is a double root of the characteristic polynomial of a Homogeneous Linear
Differential Equation with constant coefficients then
y1  e mx and y 2  xe mx are linearly independen t solutions of the differenti al equation .
Pr oof :
1. If a  m is a double root then y1  e mx is a solution of the DE ay   by   cy  0
 b  b 2  4ac
b

2a
2a
2. By Abel ' s formula a sec ond solution is given by
b
b
b
  dx
b
 x
  P( x)dx

x

x
e
e a
e a
y 2  y1
dx  e 2a
dx  e 2a
dx  e mx dx  xe mx
2
b
b
y1 
 x
 x
e a
e a
1
x
e mx
xe mx
Wronskian (e mx , xe mx ) 

 1  mx  mx  1  o
mx
mx
m 1  mx
me
(1  mx )e
and the discri min ant b 2  4ac  0 and m 




Therefore e mx and xe mx are linearly independen t solutions .
Pr oblem 1 :
Find the solution of the differenti al equation y   3y   4y  0
Solution :
The characteri stic equation is a 3  3a 2  4  0
 (a  1)(a  2)(a  2)  0
 a  1, a  2 and a  2
 y  c1e  x  c 2 e 2x  c 3 xe 2x is the general solution
Theorem . If m is a multiple root (k times ) of the characteri stic equation of a
homogeneous linear differential equation with constant coefficients, then
y  (c1  c 2 x  c3x 2  . . .  ck xk 1 )emx is a hom ogeneous solution of the DE.
Pr oblem 1. Find the solution of the DE y ( 4)  4y   6y   4y   y  0
Solution : a 4  4a 3  6a 2  4a  1  0  a  14  0
 a  1 is solution of multiplici ty four .
The solution is y  (c1  c 2 x  c 3 x 2  c 4 x 3 )e x
-1-
Problem 2. Find the general solution of y   2 y   y   2 y   x 3  3x 2  12 x  15 given the fact
that y= x 3  3x 2  5 is a particular solution
a) Step 1. Find the hom ogeneous solution:
y   2y   y   2y  0  a 3  2a 2  a  2  0 is the characteristic equation
2 |
1
1
-2
2
0
-1 2
0 -2
-1 0
 (a  2)( a 2  1)  0  (a  1) 2 (a  2)  0  a  1, a  1, a  2
are the characteri stic values
 y h  c1e  x  c 2e x  c 3e 2x is the general hom ogeneous solution
answer : y h  c1e  x  c 2 e x  c 3e 2x
b) The general solution is y  c1e x  c 2e x  c 3e 2x  x 3  3x 2  5
.
Theorem : If m1     i and m 2     i are conjugate complex roots of the characteri stic
polynomial of a linear hom ogeneous DE with cons tan t coefficients , then
y  e x (c1 cos  x  c 2 sin  x ) is a solution of the DE.
Pr oof : m1     i and m 2     i  y  k 1e    i x  k 2 e    i x is a solution of the DE

 y  e x k 1e i x  k 2 e  i x

Euler ' s formula e i x  cos  x  i sin  x
 e  i x  cos  x  i sin    e  i x  cos  x  i sin 
e i x  cos  x  i sin  x
  i x
e
 cos  x  i sin 


 y  e x k 1e i x  k 2 e  i x  e x k 1 cos  x  i sin  x   k 2 cos  x  i sin  
 y  e x (k 1  k 2 ) cos  x  ik 1  ik 2  sin  x
Let k 1  k 2  c1 and ik 1  ik 2  c 2
 y  e x (c1 cos  x  c 2 sin  x)
Theorem: if m and –m are solutions of a homogeneous linear differential equation with constant
coefficient, then the solution can be expressed in the form y=c1coshx + c2sinhx

ex  e x
cosh x 
e x  e  x  2 cosh x
 e x  cosh x  sinh x
2
Pr oof : 


 x
 x
x
x
 e  e  x  2 sinh x
e  cosh x  sinh x
 sinh x  e  e

2
a  m and a  m  y  k 1e mx  k 2 e  mc  y  k 1 cosh x  sinh x   k 2 cosh x  sinh x 
y  c1 cosh x  c 2 sinh x where c1  k 1  k 2 and c 2  k 1  k 2
Problem 3: Find the solution of the following differential equations
-2-
a) y   y   0
a) y   y  0
a 2  a  0  a(a  1)  0  a  0 or a  1
a 2  1  0  a 2  1  a  i or a   i
 y  c1  c 2 e  x
 y  c1 cos x  c 2 sin x
Problem 4. In each of the following cases write the general solution of the First Order
Homogeneous Differential Equation with constant coefficients.
1) Characteristic values m= 0, 2, -3
Answer : y  c1  c 2 e 2x  c 3e 3x
2) Characteristic m= 0, 0, 2, 2, 2
Answer : y  c1  c 2 x  c 3e 2x  c 4 xe 2x  c 5 x 2 e 2x
3) Characteristic m= -1, -2, 2, 3, 3
Answer : y  c1e  x  c 2 e 2x  c 3e 2x  c 4 e 3x  c 5 xe3x
4) Characteristic m= 0, 2+3i, 2-3i
Answer : y  c1  e 2x c 2 cos 3x  c 3 sin 3x 
5) i, i, -i, -i
Answer: y=cosx+sinx+xcosx +xsinx
6) 0, 2, 3  2i , 3  2i
Answer : y  c1  c 2 e 2x  e 3x c 3 cos 2x  c 4 sin 2x   xe 3x c 5 cos 2x  c 6 sin 2x 
7) Characteristic m= 0, 0, 0, 2, 2
Answer : y  c1  c 2 x  c 3 x 2  c 4 e 2x  c 5 xe 2x
Pr oblem 5. Find the solution of y ( 4)  y   y   0
Solution : a 4  a 3  a 2  0  a 2 (a 2  a  1)  0
 a  0, 0, a 
 y  c1  c 2
 1  1  4(1)(1)
1
3
 
i
2
2
2
1
 x
x  e 2 c


3 cos
3
3 
x  c 4 sin
x
2
2 
Pr oblem 6. Find the solution of y  4y  5y  0


Solution : a 3  4a 2  5a  0  a a 2  4a  5  0  a(a  5)(a  1)  0
 a  0, a  1 or a  5
 y  c1t  c 2 e  t  c 3e 5t
Definition: Boundary Value Problems (BVP): a linear differential equation of order two or
greater in which the dependent variable y or its derivatives are specified at different points.
A boundary value problem can have many, one, or no solutions.
-3-
Pr oblem 7. Solve the BVP y   10y   25y  0, y (0)  1, y (1)  0
Solution : a 2  10a  25  0  (a  5) 2  0  a  5, 5
 y  c1e 5x  c 2 xe5x
1  c1

y (0)  1, y (1)  0  
5
5  c1  1 and c 2  1
0  c1e  c 2 e
 y  e 5x  xe5x is the BVP solution
Pr oblem 8. Solve the BVP y   4y  0, y(0)  0 and y(  )  0
a 2  4  0  a  2i  y  c1 cos 2x  c 2 sin 2x
0  c1
y(0)  0 and y(  )  0  
 c1  0 and c 2 is any number
0  c1
There are infinitely many solution, they are of the form y=csin2x where c is any number.

Pr oblem 9. Solve the BVP y   y  0, y (0)  0, y    2
 2
a 2  1  0  a   i  y  c1 cos x  c 2 sin x and y   c1 sin x  c 2 cos x
 0  c2

y (0)  0, y    2  
 y  2 cos x is the solution of the BVP
 2
 2  c1
Problem 10. Find the solution of the following non-homogeneous DE.
y   y  2 sin x
Solution : a 2  1  0  a   i  y h  c1 cos x  c 2 sin x
Consider the hom ogeneous solution y1  cos x and let y  y1 v  v cos x
 y   v  cos x  v sin x and y   v  cos x  v  sin x  v  sin x  v cos x
 v  cos x  2 v  sin x  v cos x
 v  cos x  2 v  sin x  v cos x  v cos x  2x sin x
 v  cos x  2 v  sin x  2 sin x
 2 tan xdx
 v   2 v  tan x  2 tan x  w   2w tan x  2 tan x  e 
 e 2 ln cos x  cos 2 x
 cos 2 xw   2w cos 2 x tan x  2 tan x cos 2 x  cos 2 xw   2w sin x cos x  2 sin x cos x

 cos 2 x  w  2 sin x cos x  cos 2 x  w  sin 2 x  k 1


 w  v   tan 2 x  k 1 sec2 x  v   sec2 x  1  k 1 sec2 x  v  c1 tan x  x  c 2
 y  v cos x  cos xc1 tan x  x  c 2   c1 sin x  c 2 cos x  x cos x
Answer : y  c1 sin x  c 2 cos x  x cos x
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