ODE Lecture Notes
Section 3.1
Page 1 of 6
Section 3.1: Homogeneous Equations with Constant Coefficients
Big idea: Linear second-order differential equations with constant coefficients can be solved by
assuming a solution with an exponential form: i.e., ay  t   by  t   cy t   0 has solution
y  t   ert or y  t   tert for some values r.
Big skill: You should be able to solve second-order differential equations with constant
coefficients by plugging in the correct form of the solution, then solving the characteristic
equation.
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A second-order ordinary differential equation has the form:
d2y
dy 

 f  t , y, 
2
dt
dt 

dy 
dy

A second-order ODE is linear if f has the form: f  t , y,   g  t   p  t   q  t  y .
dt 
dt

Otherwise, it is nonlinear.
Usual form of a linear second-order ODE:
y  p  t  y  q t  y  g t 
OR
Q t 
R t 
G t 
P  t  y  Q  t  y  R  t  y  G  t   y  
y 
y
P t 
P t 
P t 
A second-order ODE is homogeneous if g(t) = 0 for all t: y  p  t  y  q t  y  0 .
Otherwise, it is called nonhomogeneous.
This section examines linear, homogeneous second-order ODEs with constant
coefficients.
ay  by  cy  0
ODE Lecture Notes
Section 3.1
Page 2 of 6
Here are four fundamental second-order differential equations and their solutions:
y  t   0
y  t   k 2 y  t 
y  t    2 y  t 
OR
OR
y  t   k 2 y  t   0
y  t    2 y  t   0
ODE Lecture Notes
Section 3.1
Page 3 of 6
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Note that two constants of integration arise when solving y  t   0 by direct integration.
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Note that the most general solution y  t   k 2 y t  and y  t    2 y t  is a linear
combination of each of the two solutions for each problem. This is analogous to there
being two constants of integration.
This is a general result: if you can find two linearly independent solutions of a secondorder equation, then all possible solutions can be written as a linear combination of those
two solutions.
The constants in the linear combination can only be explicitly determined given two
initial conditions, usually: y  t0   y0 , y  t0   y0 , but sometimes
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y  t0   y0 , y  t1   y1 (or other combos)
To solve a second-order differential equation with constant coefficients:
ay  t   by t   cy t   0
Assume a solution of the form y  t   ert , substitute it and its derivatives y  t   rert and
y  t   r 2ert into the equation.
ar 2 e rt  brert  ce rt  0
 ar
2
 br  c  e rt  0
Then find the values r1 and r2 that satisfy the resulting characteristic equation:
 ar 2  br  c   0
The general solution is a linear combination of the two solutions:
y  t   c1er1t  c2er2t
Practice:
1. Solve y  4 y  0 subject to the initial conditions y  0  1, y  0  1 . Then re-write the
solution as sum of cosh and sinh functions.
ODE Lecture Notes
Section 3.1
Page 4 of 6
2. Solve y  9 y  0 subject to the initial conditions y  0  1, y  0  1 . Use Euler’s
Formula: eix  cos  x   i sin  x  . Then re-write the solution as a single sinusoid using the
identity A sin  x   B cos  x   A2  B2 sin  x    ,
 B   0 if a  0
where   tan 1    
. We will explore complex roots more in section 3.3.
 A   if a  0
ODE Lecture Notes
Section 3.1
Page 5 of 6
3. Solve y  0 subject to the initial conditions y  0  1, y  0  1 using the characteristic
equation. We will explore the case of repeated roots more in section 3.4.
ODE Lecture Notes
Section 3.1
Page 6 of 6
4. Show that if there are two distinct roots to the characteristic equation, then a linear
combination of the solutions satisfies the original differential equation, and find
expressions for the constants of integration given the initial conditions
y  t0   y0 , y  t0   y0 .