Find a general solution for the given linear system using th

Find a general solution for the given linear system using the elimination method of Section 5.2. d^2x/dt^2 - x + 5y = 0, 2x + d^2y/dt^2 + 2y = 0

Solution

Consider operator D = d/dt

d²x/dt² - x + 5y = 0 => (D² -1)x +5y = 0..(1)

d²y/dt² + 2x + 2y = 0 => 2x + (D² + 2)y = 0..(2)

Multiply (1) by 2 and (2) by (D²-1)

2(D²-1)x + 10y = 0

2(D²-1)x + (D²-1)(D²+2)y = 0

Subtracting :

10y - (D²-1)(D²+2)y = 0

=> [D⁴+2D²-D² -2-10]y = 0

=> (D⁴ + D² -12)y = 0

The Auxiliory equation is given by :

m⁴ + m² -12 = 0

(m²)² + m² -12 =0

m² = t

Then t² + t -12 = 0

=> t² + 4t -3t -12 = 0

=> t(t + 4) -3(t+4) = 0

=> (t+4)(t-3) = 0

t = -4, 3

=> m² = -4, 3

=> m = +2i, -2i, +sqrt(3), -sqrt(3)

y = A cos(2t) + Bsin(2t) + Ce^sqrt(3)t + D e^-sqrt(3)t

Now substitute this in (2)

2x + (D² + 2)( A cos(2t) + Bsin(2t) + Ce^sqrt(3)t + D e^-sqrt(3)t) = 0

=> 2x + D(-2Asin(2t) +2Bcos(2t) +sqrt(3)Ce^sqrt(3)t -sqrt(3)De^-sqrt(3)t)+2( A cos(2t) + Bsin(2t) + Ce^sqrt(3)t + D e^-sqrt(3)t) = 0

=> 2x + (-4Acos(2t) -4Bsin(2t)+3Ce^sqrt(3)t +3De^-sqrt(3)t) + 2Acos(2t)+2Bsin(2t)+2Ce^sqrt(3)t+2De^-sqrt(3)t = 0

=> x -Acos(2t)-Bsin(2t)+5C/2e^sqrt(3)t +5/2De^-sqrt(3)t = 0

=> x = Acos(2t) + Bsin(2t) - 5C/2e^sqrt(3)t - 5/2De^-sqrt(3)t