2y''' + 5y'' − 22y' + 15y = 0.
2007-09-01 07:49:30 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let y = e^(rx).
Plug in y = e^(rx) and solve for r,
2r^3+5r^2-22r+15 = 0
(r-1)((2r-3)(r+5) = 0
r = 1, 3/2, -5
The general solution is,
y = Ae^x + Be^(1.5x) + Ce^(-5x), where A, B and C are all constants.
2007-09-01 08:00:22 · answer #1 · answered by sahsjing 7 · 1⤊ 0⤋
2y''' + 5y'' - 22y' + 15y = 0
Find the solution to the characteristic equation:
2r^3 + 5r^2 - 22r + 15 = 0
Notice by examining coefficients that r=1 is a solution.
So use synthetic division to divide by r - 1
1|.....2.....5.....-22.....15
...............2.......7...... -15
........2.....7......-15.....0
2r^2 + 7r - 15 = 0
Factor:
(2r - 3)(r + 5) = 0
r = 3/2, r = -5
So there are 3 roots: 1, 3/2, and -5
So the general solution is:
y = Ae^(x) + Be^(3x/2) + Ce^(-5x)
where A, B, and C are constants.
2007-09-01 15:05:36 · answer #2 · answered by whitesox09 7 · 0⤊ 0⤋
Use LaPlace transforms
2007-09-01 14:59:53 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋
is the answer 22/15
2007-09-01 14:55:39 · answer #4 · answered by ? 3 · 0⤊ 2⤋
dont know
2007-09-01 14:57:38 · answer #5 · answered by J.T Rebel 2 · 0⤊ 2⤋