This equation is of form am^2+bm+c = 0
a = 1 b = 1 c = 1
m=[-b+/-sqrt(b^2-4ac)]/2a]
m=[-1 +/-sqrt(1^2-4(1)(1)]/(2)(1)
discriminant is b^2-4ac =-3
i^2 = -1, so √i^2 = i
No real roots: The complex roots are
m=[-1 +i √(3)] / (2)(1)
m=[-1 -i √(3)] / (2)(1)
y(c) = C1 e^(-x/2) sin ( √3 x /2) + C2 e^(-x/2) cos( √3 x /2)
Particular solution:
y(p) = Ax^3+Bx^2+Cx+D
y' = 3Ax^2+2Bx + C
y'' = 6Ax+2B
y''+y'+y = 6Ax+2B +3Ax^2+2Bx + C + Ax^3+Bx^2+Cx+D
6Ax+2B +3Ax^2+2Bx + C + Ax^3+Bx^2+Cx+D = x^3+3x^2+3x-1
Match the coefficient of x^3
A = 1
Match the coefficient of x^2
3A +B = 3
3(1)+ B = 3
B = 0
Match the coeficient of x
6A+2B+C = 3
6(1)+2(0) + C=3
C = -3
Match the constants
C+ D = -1
-3 + D = -1
D = 2
y(p) = (1)x^3+(0)x^2-3x+2
y(p) = x^3-3x+2
y = y(c)+y(p)
y = C1 e^(-x/2) sin ( √3 x /2) + C2 e^(-x/2) cos( √3 x /2) + x^3-3x+2
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http://www.wolframalpha.com/input/?i=F(x...
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ted s say: Sorry , we don't know about the meaning of F(x)...
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Paris say: No I am NOT doing your homework.
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alex say: what's F(x) +f (x)+f"(x) ?
assume it's F(x) +f'(x)+f"(x)
then
let f(x) = x^3+bx^2+cx+d
set up equation , then solve for b , c , d
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