Second order matrix differential equation - 2

Calculus Level 3

A $2 \times 2$ matrix $B(t)$ satisfies the differential equation,

$\dfrac{d^2 B }{d t^2} + A^2 B = 0$

where $A = \begin{bmatrix} 1 && 2 \\ 2 && 4 \end{bmatrix}$

subject to the initial conditions,

$B(0) = \begin{bmatrix} 1 && 1 \\ 2 && 3 \end{bmatrix}$

$\dfrac{dB}{dt}(0) = \begin{bmatrix} 1 && 0 \\ 3 && 1 \end{bmatrix}$

Apply the Laplace transform to the above differential equation, and find $B_{12}(t)$ . You'll find that,

$B_{12}(t) = a_0 + a_1 t + a_2 \cos( \omega t ) + a_3 \sin( \omega t )$ .

Report the value of $\lfloor 100 ( a_0 + a_1 + a_2 + a_3 + \omega) \rfloor$

1 solution

Hosam Hajjir
May 21, 2019

Taking the Laplace transform of the given differential equation, yields,

$s^2 B - s B(0) - B'(0) + A^2 B = 0$

so that,

$B(s) = (s^2 I + A^2)^{-1} ( s B(0) + B'(0) )$

The rest is straight forward matrix evaluation, inversion, and multiplication. The resulting $B_{12}(s)$ is given by,

$B_{12}(s) = \dfrac{ s^3 - 10 s - 10 }{ s^2 (s^2 + 25) } = \dfrac{(-0.4)}{s} + \dfrac{(-0.4)}{s^2} + \dfrac{1.4 s + 0.4}{s^2 + 25}$

Applying the inverse Laplace transform, we get,

$B_{12}(t) = -0.4 - 0.4 t + 1.4 \cos( 5 t ) + 0.08 \sin( 5 t )$

Therefore, the answer is $\lfloor 100 ( -0.4 -0.4 + 1.4 + 0.08 + 5 ) \rfloor = \boxed{568}$