Lecture 8. Second-Order Homogeneous Equations I. July 15, 2013
8.1. Example
Let y 00 + y 0 − 2y = 0. Find the solution in the form y = eλt . Plugging into the equation, we get:
λ2 eλt + λeλt − 2eλt = 0 ⇒ λ2 + λ − 2 = 0.
Its roots are λ1 = 1 and λ2 = −2. Therefore, et and e−2t are solutions. Their linear combinations
are also solutions:
y(t) = C1 et + C2 e−2t
for any constants C1 , C2 .
Solve the initial value problem: y(0) = 1, y 0 (0) = 0. Then
y(0) = C1 + C2 = 1, y 0 (0) = C1 − 2C2 = 0.
Solving this system, we get:
C1 = 2/3, C1 = 1/3,
So the solution is
2
1
y(t) = et + e−2t .
3
3
8.2. Real Distinct Roots
Consider the equation
ay 00 + by 0 + cy = 0,
where a, b, c are real numbers. The equation
aλ2 + bλ + c = 0
is called the characteristic equation. If b2 − 4ac > 0, then it has distinct roots λ1 and λ2 , so eλ1 t and
eλ2 t are its solutions. The general solution is
C1 eλ1 t + C2 eλ2 t
An initial value problem is when you fix the values of y(t0 ) and y 0 (t0 ) for some t0 . The argument
should be the same for y and y 0 .
8.3. Repeated Root
Assume b2 − 4ac = 0, then we have repeated root: −b/2a. Consider an example: y 00 − 2y 0 + y = 0.
The characteristic equation is λ2 − 2λ + 1 = 0, so λ = 1 is the only root. The function y(t) = et , and
therefore y(t) = Cet for any constant C, is a solution. But we need another solution! Let us use the
method of variation of parameters: let C = C(t) be not a constant, but a function. Then
y 0 (t) = C 0 (t)et + C(t)et , y 00 (t) = C 00 (t)et + 2C 0 (t)et + C(t)et .
Plugging these into y 00 − 2y 0 + y = 0, we get:
C 00 (t)et = 0 ⇒ C 00 (t) = 0 ⇒ C 0 (t) = C1 ⇒ C(t) = C1 t + C2
1
for some constants C1 and C2 . Plugging this back into the expression for y(t), we get:
y(t) = (C1 t + C2 )et = C1 tet + C2 et .
In general, if you have a repeated root λ, the general solution is
C1 teλt + C2 eλt
2