Lecture Notes in Differential Equations - Bruce E. Shapiro

141
The general solution to
is given by
y =
ay ′′ + by ′ + cy = 0 (16.60)
{
Ae r1t + Be r2t r 1 ≠ r 2 (distinct roots)
(A + Bt)e rt r = r 1 = r 2 (repeated root)
(16.61)
where r 1 and r 2 are roots of the characteristic equation ar 2 + br + c = 0.
Example 16.4. Solve the initial value problem
⎫
y ′′ − 6y ′ + 8y = 0⎪⎬
y(0) = 1
⎪⎭
y ′ (0) = 1
(16.62)
The characteristic equation is
0 = r 2 − 6r + 8 = (r − 4)(r − 2) (16.63)
The roots are r = 2 and r = 4, hence
y = Ae 2t + Be 4t (16.64)
From the first initial condition,
1 = A + B (16.65)
Differentiating (16.64)
y ′ = 2Ae 2t + 4Be 4t (16.66)
From the second initial condition
1 = 2A + 4B (16.67)
From (16.65), B = 1 − A, hence
hence
1 = 2A + 4(1 − A) = 4 − 2A =⇒ 2A = 3 =⇒ A = 3 2
B = 1 − A = 1 − 3 2 = −1 2
(16.68)
(16.69)
The solution to the IVP is
y = 3 2 e2t − 1 2 e4t (16.70)