I'm going to assume that "−" is supposed to be some kind of minus character, so that the given system of DEs is supposed to be
[tex]\begin{cases}\dfrac{\mathrm d^2x}{\mathrm dt^2} + x - y = 0 \\ \dfrac{\mathrm d^2y}{\mathrm dt^2} + y - x = 0 \\ x(0) = 0, x'(0) = -6, y(0) = 0, y'(0) = 1\end{cases}[/tex]
Take the Laplace transform of both sides of both equations. Recall the transform for a second-order derivative,
[tex]L_s\left\{\dfrac{\mathrm d^2f(t)}{\mathrm dt^2}\right\} = s^2 F(s) - sf(0) - f'(0)[/tex]
where F(s) denotes the transform of f(t). You end up with
[tex]\begin{cases}s^2X(s)-sx(0)-x'(0) + X(s) - Y(s) = 0 \\ s^2Y(s) - sy(0) - y'(0) + Y(s) - X(s) = 0\end{cases} \\\\ \begin{cases}(s^2+1)X(s) - Y(s) = -6 \\ (s^2+1)Y(s) - X(s) = 1\end{cases}[/tex]
and solving for X(s) and Y(s) (nothing tricky here, just two linear equations) gives
[tex]X(s) = -\dfrac{6s^2+5}{s^2(s^2+2)} \text{ and } Y(s) = \dfrac{s^2-5}{s^2(s^2+2)}[/tex]
Now solve for x(t) and y(t) by computing the inverse transforms. To start, split up both X(s) and Y(s) into partial fractions.
• Solving for x(t) :
[tex]-\dfrac{6s^2+5}{s^2(s^2+2)} = \dfrac as + \dfrac b{s^2} + \dfrac{cs+d}{s^2+2} \\\\ -6s^2-5 = as(s^2+2) + b(s^2+2) + (cs+d)s^2 \\\\ -6s^2-5 = (a+c)s^3 + (b+d)s^2 + 2as + 2b \\\\ \implies \begin{cases}a+c=0\\b+d=-6\\2a=0\\2b=-5\end{cases} \implies a=c=0, b=-\dfrac52, d=-\dfrac72[/tex]
[tex]\implies X(s) = -\dfrac5{2s^2} - \dfrac7{2(s^2+2)}[/tex]
Taking the inverse transform, you get
[tex]L^{-1}_t\left\{X(s)\right\} = -\dfrac52 L^{-1}_t\left\{\dfrac{1!}{s^{1+1}}\right\} - \dfrac7{2\sqrt2} L^{-1}_t\left\{\dfrac{\sqrt2}{s^2+(\sqrt2)^2}\right\} \\\\ \implies \boxed{x(t) = -\dfrac52 t - \dfrac7{2\sqrt2}\sin(\sqrt2\,t)}[/tex]
• Solving for y(t) :
[tex]\dfrac{s^2-5}{s^2(s^2+2)} = \dfrac as + \dfrac b{s^2} + \dfrac{cs+d}{s^2+2} \\\\ s^2 - 5 = as(s^2+2) + b(s^2+2) + (cs+d)s^2 \\\\ s^2 - 5 = (a+c)s^3 + (b+d)s^2 + 2as + 2b \\\\ \implies \begin{cases}a+c=0\\b+d=1\\2a=0\\2b=-5\end{cases} \implies a=c=0, b=-\dfrac52, d=\dfrac72[/tex]
[tex]\implies Y(s) = -\dfrac5{2s^2} + \dfrac7{2(s^2+2)}[/tex]
Inverse transform:
[tex]L^{-1}_t\left\{Y(s)\right\} = -\dfrac52 L^{-1}_t\left\{\dfrac{1!}{s^{1+1}}\right\} + \dfrac7{2\sqrt2} L^{-1}_t\left\{\dfrac{\sqrt 2}{s^2+(\sqrt2)^2}\right\} \\\\ \implies \boxed{y(t) = -\dfrac52 t + \dfrac7{\sqrt2} \sin(\sqrt2\,t)}[/tex]
