allhvasstH

2021-01-25

Use the Laplace transform to solve the given system of differential equations.
$\frac{\left({d}^{2}x\right)}{\left(d{t}^{2}\right)}+\frac{\left({d}^{2}y\right)}{\left(d{t}^{2}\right)}=\frac{t}{2}$
$\frac{\left({d}^{2}x\right)}{\left(d{t}^{2}\right)}-\frac{\left({d}^{2}y\right)}{\left(d{t}^{2}\right)}=4t$
$x\left(0\right)=5,{x}^{\prime }\left(0\right)=0,$
$y\left(0\right)=0,{y}^{\prime }\left(0\right)=0$

Step 1
Given that
$\frac{\left({d}^{2}x\right)}{\left(d{t}^{2}\right)}+\frac{\left({d}^{2}y\right)}{\left(d{t}^{2}\right)}=\frac{t}{2}\dots \left(A\right)$
$\frac{\left({d}^{2}x\right)}{\left(d{t}^{2}\right)}-\frac{\left({d}^{2}y\right)}{\left(d{t}^{2}\right)}=4t\dots \left(B\right)$
$x\left(0\right)=5,{x}^{\prime }\left(0\right)=0,$
$y\left(0\right)=0,{y}^{\prime }\left(0\right)=0$
$⇒2\frac{\left({d}^{2}x\right)}{\left(d{t}^{2}\right)}=\left({t}^{2}+4t\right)$
$\frac{\left({d}^{2}x\right)}{\left(d{t}^{2}\right)}=\frac{\left({t}^{2}+4t\right)}{2}$
$\frac{\left({d}^{2}x\right)}{\left(d{t}^{2}\right)}=\frac{\left({t}^{2}\right)}{2}+2t$
Step 2
taking Laplace transform on both sides,
$s2X\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)=\frac{1}{2}+\frac{2}{{s}^{2}}+\frac{2}{{s}^{2}}$
$s2X\left(s\right)-5s=\frac{1}{\left({s}^{3}\right)}+\frac{2}{{s}^{2}}$
$s2X\left(s\right)=\frac{\left(1+2s+5{s}^{4}\right)}{{s}^{5}}$
$X\left(s\right)=\frac{1}{{s}^{5}}+\frac{2}{{s}^{4}}+\frac{5}{s}$
taking Inverse Laplace transform
$x\left(t\right)=\frac{{t}^{4}}{4!}+\frac{\left(2{t}^{3}\right)}{3!}+5u\left(t\right)$
Step 3
$u\left(t\right)=1,t\ge 0$
$=0,t<0$

And Subtracting equation (B) from (A)
$⇒2\frac{\left({d}^{2}y\right)}{\left(d{t}^{2}\right)}={t}^{2}-4t$
$\frac{\left({d}^{2}y\right)}{\left(d{t}^{2}\right)}=\frac{\left({t}^{2}-4t\right)}{2}=\frac{\left({t}^{2}\right)}{2}-2t$
Taking ILT
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)=\frac{1}{2}\frac{2}{{s}^{3}}-\frac{2}{{s}^{2}}$
${s}^{2}Y\left(s\right)=\frac{1}{\left({s}^{3}\right)}-\frac{2}{\left({s}^{2}\right)}=\frac{\left(1-2s\right)}{{s}^{3}}$
$Y\left(s\right)=\frac{\left(1-2s\right)}{{s}^{5}}$
Step 4
$Y\left(s\right)=\frac{1}{{s}^{5}}-\frac{2s}{\left({s}^{5}\right)}$
$Y\left(s\right)=\frac{1}{{s}^{5}}-\frac{2}{\left({s}^{4}\right)}$
Taking Inverse Laplace Transform
$y\left(t\right)=\frac{\left({t}^{4}\right)}{4!}-\frac{\left(2{t}^{3}\right)}{3!}$
$=\frac{\left({t}^{4}\right)}{24}-\frac{\left({t}^{3}\right)}{3}$

Do you have a similar question?