Lecture 23, Solving nthorder, homogeneous, linear ODEs with

Lecture 23, Solving nth order, homogeneous, linear ODEs with constant coefficients
BONUS EXAMPLES
Example: Find the general solution of
y(7) + 7y(6) + 23y(5) + 45y(4) + 56y000 + 44y00 + 20y0 + 4y = 0
SOLUTION: This is a linear, seventh order linear eq.
Use the substitution y = ert to recover the characteristic equation
r7 + 7r6 + 23r5 + 45r4 + 56r3 + 44r2 + 20r + 4 = 0.
which factors to
(r + 1)3 (r2 + 2r + 2)2 = 0
How? Note that r = −1 is a root, then perform polynomial long division to get
(r + 1)(r6 + 6r5 + 17r4 + 28r3 + 28r2 + 16r + 4) = 0,
note that r = −1 is a root of the 6th order polynomial, and again with the long division to get
(r + 1)2 (r5 + 5r4 + 12r3 + 16r2 + 12r + 4) = 0,
etc.
Our characteristic equation (r + 1)3 (r2 + 2r + 2)2 = 0 has roots:
• r = −1 with multiplicity 3, i.e., repeated three times (the (r + 1) term is cubed). Therefore,
y1 (t) = e−t , y2 (t) = te−t , and y3 (t) = t 2 e−t
are solutions of the ODE.
• r = −1 ± i with multiplicity 2, i.e., repeated twice (the (r2 + 2r + 2) term is squared). Therefore,
y4 (t) = e−t sin(t), y5 (t) = e−t cos(t), y6 (t) = te−t sin(t), and y7 (t) = te−t cos(t)
are solutions of the ODE.
The 7th order homogeneous ODE has 7 solutions. The fundamental solution set of y(7) + 7y(6) +
23y(5) + 45y(4) + 56y000 + 44y00 + 20y0 + 4y = 0 is
−t
e , te−t , t 2 e−t , e−t sin(t), e−t cos(t), te−t sin(t), te−t cos(t) .
Therefore the general solution is
y(t) = C1 e−t +C2te−t +C3t 2 e−t +C4 e−t sin(t) +C5 e−t cos(t) +C6te−t sin(t) +C7te−t cos(t).