Lecture Notes in Differential Equations - Bruce E. Shapiro

138 LESSON 16. LINEAR EQS. W/ CONST. COEFFICENTS
When a root is repeated, the square root is zero, hence
r = − b
2a
(16.33)
Rearranging gives
2ar + b = 0 (16.34)
whenever r is a repeated root. Substituting equation (16.34) into equation
(16.31) gives L(te rt ) = 0.
Example 16.3. We can use theorem (16.4) to find two solutions of the
homogeneous linear differential equation
The characteristic equation is
y ′′ − 7y ′ + 12y = 0 (16.35)
r 2 − 7r + 12 = 0 (16.36)
Factoring gives
(r − 3)(r − 4) = 0 (16.37)
Since the roots are r = 3 and r = 4, two solutions of the differential equation
(16.35) are
Thus for any real numbers A and B,
is also a solution.
y H1 = e 3t (16.38)
y H2 = e 4t (16.39)
y = Ae 3t + Be 4t (16.40)
Because equation (16.14) involves a second-order derivative the solution
will in general include two constants of integration rather than the single
arbitrary constant that we had when we were solving first order equations.
These initial conditions are expressed as the values of both the function
and its derivative at the same point, e.g.,
}
y(t 0 ) = y 0
y ′ (t 0 ) = y 1
(16.41)