Use the Laplace transform to solve the given system of differential equations.(d2x)(dt2)+(d2y)(dt2)=t2 (d2x)(dt2)−(d2y)(dt2)=4t x(0)=5,x′(0)=0,y(0)=0,y′(0)=0

Step 1
Given that
$\frac{(d^{2}x)}{(d{t}^2)}+\frac{(d^{2}y)}{(d{t}^2)}=\frac{t}{2}\dots (A)$
$\frac{(d^{2}x)}{(d{t}^2)}-\frac{(d^{2}y)}{(d{t}^2)}=4t\dots (B)$
$x(0)=5,x^{\prime }(0)=0,$
$y(0)=0,y^{\prime }(0)=0$
adding equation( A) and (B)
$\Rightarrow 2\frac{(d^{2}x)}{(d{t}^2)}=(t^2+4t)$
$\frac{(d^{2}x)}{(d{t}^2)}=\frac{(t^2+4t)}{2}$
$\frac{(d^{2}x)}{(d{t}^2)}=\frac{(t^2)}{2}+2t$
Step 2
taking Laplace transform on both sides,
$s2X(s)-sx(0)-x^{\prime }(0)=\frac{1}{2}+\frac{2}{s^2}+\frac{2}{s^2}$
$s2X(s)-5s=\frac{1}{(s^3)}+\frac{2}{s^2}$
$s2X(s)=\frac{(1+2s+5s^4)}{s^5}$
$X(s)=\frac{1}{s^5}+\frac{2}{s^4}+\frac{5}{s}$
taking Inverse Laplace transform
$x(t)=\frac{t^4}{4!}+\frac{(2t^3)}{3!}+5u(t)$
Step 3
$u(t)=1,t\ge 0$
$=0,t<0$
$\therefore x(t)=\frac{(t^4)}{24}+\frac{(t^3)}{3}+5\text{ for }t\ge 0$
And Subtracting equation (B) from (A)
$\Rightarrow 2\frac{(d^{2}y)}{(d{t}^2)}=t^2-4t$
$\frac{(d^{2}y)}{(d{t}^2)}=\frac{(t^2-4t)}{2}=\frac{(t^2)}{2}-2t$
Taking ILT
$s^{2}Y(s)-sy(0)-y^{\prime }(0)=\frac{1}{2}\frac{2}{s^3}-\frac{2}{s^2}$
$s^{2}Y(s)=\frac{1}{(s^3)}-\frac{2}{(s^2)}=\frac{(1-2s)}{s^3}$
$Y(s)=\frac{(1-2s)}{s^5}$
Step 4
$Y(s)=\frac{1}{s^5}-\frac{2s}{(s^5)}$
$Y(s)=\frac{1}{s^5}-\frac{2}{(s^4)}$
Taking Inverse Laplace Transform
$y(t)=\frac{(t^4)}{4!}-\frac{(2t^3)}{3!}$
$=\frac{(t^4)}{24}-\frac{(t^3)}{3}$