Question

In Problems 1-12 Use the Laplace transform to solve the given system of differential equations.

$\begin{array}{l}\mathbf{8}\mathbf{.}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{0}\\ \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathbf{4}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{0}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\end{array}$

Short answer

$\begin{array}{l}x\left(t\right)=e^{-t}\cos 2t+\frac{1}{2}{e}^{-t}\sin 2t\\ y\left(t\right)=2e^{-t}\sin 2t-e^{-t}\cos 2t\end{array}$

Step-by-step solution

Definition of Laplace Transform

LetF be afunction defined for. Then the integral$t\ge 0$

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of f, provided that the integral converges.

Use Laplace Transform

We know that

$\begin{array}{l}\mathcal{L}\left\{x^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x^{\text{'}}\left(0\right)\\ \mathcal{L}\left\{x^{\text{'}}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)\\ \mathcal{L}\left\{y^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)\\ \mathcal{L}\left\{y^{\text{'}}\right\}=s\mathcal{L}\left\{y\right\}-y\left(0\right)\end{array}$

Take the Laplace to transform on both sides of each given differential equation

$\begin{array}{l}\mathcal{L}\left\{0\right\}=\mathcal{L}\left\{x^{\text{'}\text{'}}\right\}+\mathcal{L}\left\{x^{\text{'}}\right\}+\mathcal{L}\left\{y^{\text{'}}\right\}----\left(1\right)\\ \mathcal{L}\left\{0\right\}=\mathcal{L}\left\{y^{\text{'}\text{'}}\right\}+\mathcal{L}\left\{y^{\text{'}}\right\}-4\mathcal{L}\left\{x^{\text{'}}\right\}---\left(2\right)\end{array}$

$$\begin{gathered} 0^{(*)}=s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x^{\text{'}}\left(0\right)+s\mathcal{L}\left\{x\right\}-x\left(0\right)+s\mathcal{L}\left\{y\right\}-y\left(0\right) \\ {0}^{(*)}=s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)+s\mathcal{L}\left\{y\right\}-y\left(0\right)-4s\mathcal{L}\left\{x\right\}+4x\left(0\right) \\ \begin{array}{l}0=s^2\mathcal{L}\left\{x\right\}-s+s\mathcal{L}\left\{x\right\}-1+s\mathcal{L}\left\{y\right\}+1\left(x\left(0\right)=1,x^{\text{'}}\left(0\right)=0\right)\\ \\ 0=s^2\mathcal{L}\left\{y\right\}+s-5+s\mathcal{L}\left\{y\right\}+1-4s\mathcal{L}\left\{x\right\}+4\left(y\left(0\right)=-1,y^{\text{'}}\left(0\right)=5\right)\end{array} \\ Denote\mathcal{L}\left\{x\right\}=X\left(s\right)and\mathcal{L}\left\{y\right\}=Y\left(s\right)intheabovesyst \\ \begin{array}{l}0=s^{2}X\left(s\right)-s+sX\left(s\right)-1+sY\left(s\right)+1\\ 0=s^{2}Y\left(s\right)+s-5+sY\left(s\right)+1-4sX\left(s\right)+4\\ s=s^{2}X\left(s\right)+sX\left(s\right)+sY\left(s\right)\\ -s=s^{2}Y\left(s\right)+sY\left(s\right)-4sX\left(s\right)\end{array} \end{gathered}$$

Simplify the equations

The system simplifies the form.

$\begin{array}{l}\left(s+1\right)X\left(s\right)+Y\left(s\right)=1---\left(3\right)\\ -4X\left(s\right)+\left(s+1\right)Y\left(s\right)=-1--\left(4\right)\\ Multiplyequation\left(3\right)by4andequation\left(4\right)by(s+1)\\ 4\left(s+1\right)X\left(s\right)+4Y\left(s\right)=4----\left(5\right)\\ -4(s+1)X\left(s\right)+(s+1{)}^{2}Y\left(s\right)=-(s+1)---\left(6\right)\end{array}$

Now add (5) and (6) equations to get.

$\begin{array}{l}4Y\left(s\right)+(s+1{)}^{2}Y\left(s\right)=4-(s+1)\\ \left(s^2+2s+5\right)Y\left(s\right)=3-s\\ Y\left(s\right)=\frac{3-s}{s^2+2s+5}\\ Y\left(s\right)=\frac{3-s+1-1}{{(s+1)}^2+2^2}\\ Y\left(s\right)=\frac{4}{{(s+1)}^2+2^2}-\frac{s+1}{{(s+1)}^2+2^2}\end{array}$

Use Inverse Laplace Transform

Apply inverse Laplace transform to each side.

$\begin{array}{l}{\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\frac{4}{{(s+1)}^2+2^2}-\frac{s+1}{{(s+1)}^2+2^2}\right\}\\ =2{\mathcal{L}}^{-1}\left\{\frac{2}{{(s-(-1\left)\right)}^2+2^2}\right\}-{\mathcal{L}}^{-1}\left\{\frac{s-(-1)}{{(s-(-1\left)\right)}^2+2^2}\right\}\end{array}$

$$\begin{gathered} {}^{(**)}=2e^{-t}\sin 2t-e^{-t}\cos 2t \\ \left({\mathcal{L}}^{-1}\left\{\frac{k}{{(s-a)}^2+k^2}\right\}=e^{at}\sin kt,{\mathcal{L}}^{-1}\left\{\frac{s-a}{{(s-a)}^2+k^2}\right\}=e^{at}\cos kt\right)(**) \end{gathered}$$

Therefore,

$y\left(t\right)=2e^{-t}\sin 2t-e^{-t}\cos 2t$

Now, multiply equation (3) by and equation (4) by -1

$\begin{array}{l}(s+1{)}^{2}X\left(s\right)+(s+1)Y\left(s\right)=s+1---\left(7\right)\\ 4X\left(s\right)-\left(s+1\right)Y\left(s\right)=1---\left(8\right)\end{array}$

Now add (7) and (8) equations to get.
$\begin{array}{l}(s+1{)}^{2}X\left(s\right)+4X\left(s\right)=s+1+1\\ \left(s^2+2s+5\right)X\left(s\right)=s+1+1\\ X\left(s\right)=\frac{s+1+1}{s^2+2s+5}\\ X\left(s\right)=\frac{s+1}{{(s+1)}^2+2^2}+\frac{1}{{(s+1)}^2+2^2}\end{array}$

Apply inverse Laplace transform to each side.

$\begin{array}{l}{\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\frac{s+1}{{(s+1)}^2+2^2}+\frac{1}{{(s+1)}^2+2^2}\right\}\\ ={\mathcal{L}}^{-1}\left\{\frac{s-(-1)}{{(s-(-1\left)\right)}^2+2^2}\right\}+\frac{1}{2}{\mathcal{L}}^{-1}\left\{\frac{2}{{(s-(-1\left)\right)}^2+2^2}\right\}\end{array}$

$\begin{array}{l}{}^{(**)}=e^{-t}\cos 2t+\frac{1}{2}{e}^{-t}\sin 2t\\ \left({\mathcal{L}}^{-1}\left\{\frac{k}{{(s-a)}^2+k^2}\right\}=e^{at}\sin kt,{\mathcal{L}}^{-1}\left\{\frac{s-a}{{(s-a)}^2+k^2}\right\}=e^{at}\cos kt\right)(**)\\ therefore,\\ x\left(t\right)=e^{-t}\cos 2t+\frac{1}{2}{e}^{-t}\sin 2t\\ result,\\ x\left(t\right)=e^{-t}\cos 2t+\frac{1}{2}{e}^{-t}\sin 2t\\ y\left(t\right)=2e^{-t}\sin 2t-e^{-t}\cos 2t\end{array}$