Find a particular solution yp of the given equation.
2006-10-27 06:19:51 · 3 answers · asked by spyderkiley 1 in Science & Mathematics ➔ Mathematics
The roots of the characteristic equation are +2 and -3, none of which lead to a solution of the form sin(3x). Therefore the particular solution has the form y = A cos(3x) + B sin(3x).
Plug this into the equation:
y' = -3A sin(3x) + 3B cos(3x)
y'' = -9A cos(3x) - 9B sin(3x)
y'' - y' - 6y =
(-9A-3B-6A)cos(3x) + (-9B+3A-6B)sin(3x) =
(-15A-3B)cos(3x) + (-15B+3A)sin(3x).
Therefore -15A-3B=0, and -15B+3A=2.
From the first equation, B=-5A, so plug that into the second equation, and you get 78A=2, so A=1/39, and B=-5/39.
2006-10-27 06:33:03 · answer #1 · answered by James L 5 · 0⤊ 0⤋
The roots of the characteristic equation are +2 and -3, none of which lead to a solution of the form sin(3x). Therefore the particular solution has the form y = A cos(3x) + B sin(3x).
Plug this into the equation:
y' = -3A sin(3x) + 3B cos(3x)
y'' = -9A cos(3x) - 9B sin(3x)
y'' - y' - 6y =
(-9A-3B-6A)cos(3x) + (-9B+3A-6B)sin(3x) =
(-15A-3B)cos(3x) + (-15B+3A)sin(3x).
Therefore -15A-3B=0, and -15B+3A=2.
From the first equation, B=-5A, so plug that into the second equation, and you get 78A=2, so A=1/39, and B=-5/39.
2006-10-27 06:35:45 · answer #2 · answered by craftyboy 2 · 0⤊ 0⤋
I am a math idiot and my son isin't home from school yet so I'll tell you at about 4:45 this evening...
2006-10-27 06:27:48 · answer #3 · answered by Anonymous · 1⤊ 0⤋