Lecture Notes in Differential Equations - Bruce E. Shapiro

190 LESSON 22. NON-HOMOG. EQS.W/ CONST. COEFF.
Algorithm for 2nd Order, Linear, Constant Coefficients IVP
To solve the initial value problem
ay ′′ + by ′ + cy = f(t)
y(t 0 ) = y 0
y ′ (t 0 ) = y 1
⎫
⎪⎬
⎪ ⎭
(22.21)
1. Find the roots of the characteristic polynomial ar 2 + br + c = 0.
2. Factor the differential equation as a(D − r 1 ) (D − r 2 )y = f(t).
} {{ }
z(t)
3. Substitute z = (D − r 2 )y = y ′ − r 2 y to get a(D − r 1 )z = f(t).
4. Solve the resulting differential equation z ′ − r 1 z = 1 f(t) for z(t).
a
5. Solve the first order differential equation y ′ − r 2 y = z(t) where z(t) is
the solution you found in step (4).
6. Solve for arbitrary constants using the initial conditions.
Example 22.1. Solve the initial value problem
⎫
y ′′ − 10y ′ + 21y = 3 sin t⎪⎬
y(0) = 1
⎪⎭
y ′ (0) = 0
The characteristic polynomial is
(22.22)
r 2 − 10r + 21 = (r − 3)(r − 7) (22.23)
Thus the roots are r = 3, 7 and since a = 1 (the coefficient of y ′′ ), the
differential equation can be factored as
(D − 3) (D − 7)y = 3 sin t (22.24)
} {{ }
z
To solve equation (22.24) we make the substitution
Then (22.24) becomes
z = (D − 7)y = y ′ − 7y (22.25)
(D − 3)z = 3 sin t (22.26)
z ′ − 3z = 3 sin t (22.27)