Question

use the Laplace transform to solve each system

$$\begin{gathered} {\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{t} \\ \mathbf{4}\mathbf{x}\mathbf{+}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \end{gathered}$$

Short answer

$$\begin{gathered} x\left(t\right)=-\frac{1}{4}+\frac{1}{8}{e}^{2t}+\frac{9}{8}{e}^{-2t} \\ y\left(t\right)=t-\frac{1}{4}{e}^{2t}+\frac{9}{4}{e}^{-2t} \end{gathered}$$

Step-by-step solution

To Find the laplace transform equation.

We know that

$>\mathcal{L}\left\{x^{\text{'}}\right\}\mathcal{L}\left\{y^{\text{'}}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)\left(1\right)=s\mathcal{L}\left\{y\right\}-y\left(0\right)\left(2\right)$

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

$\mathcal{L}\left\{t\right\}=\mathcal{L}\left\{x^{\text{'}}\right\}+\mathcal{L}\left\{y\right\}$

$\mathcal{L}\left\{0\right\}=4\mathcal{L}\left\{x\right\}+\mathcal{L}\left\{y^{\text{'}}\right\}$

$\mathcal{L}\left\{t\right\}\stackrel{1}{=}s\mathcal{L}\left\{x\right\}-x\left(0\right)+\mathcal{L}\left\{y\right\}$

$0\stackrel{2}{=}4\mathcal{L}\left\{x\right\}+s\mathcal{L}\left\{y\right\}-y\left(0\right)$

$\begin{array}{rcl}\mathcal{L}\left\{t\right\}& =& s\mathcal{L}\left\{x\right\}-1+\mathcal{L}\left\{y\right\}\left(x\right(0)=1)\\ 0& =& 4\mathcal{L}\left\{x\right\}+s\mathcal{L}\left\{y\right\}-2\left(y\right(0)=2)\\ & & \end{array}$

Denote L{x}=X(s) and L{y}=Y(s) in the above system

$\begin{array}{rcl}3L\left\{t\right\}& =& sX\left(s\right)-1+Y\left(s\right)\frac{0=4X\left(s\right)+sY\left(s\right)-2}{s^2}=sX\left(s\right)-1+Y\left(s\right)\\ L\left\{t^n\right\}& =& \frac{n!}{s^{n+1}}0=4X\left(s\right)+sY\left(s\right)-2\end{array}$

System simplifies to the form.

$X\left(s\right)+Y\left(s\right)=\frac{1+s^2}{s^2}4X\left(s\right)+sY\left(s\right)=2$

Multiply equation (3) by 4 and equation (4) by -s

$\begin{array}{rcl}4sX\left(s\right)+4Y\left(s\right)& =& \frac{4+4s^2}{s^2}\\ -4sX\left(s\right)-s^{2}Y\left(s\right)& =& -2s\end{array}$

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

Decompose the fraction$\frac{4+4s^2-2s^3}{s^2\left(4-s^2\right)}$using the partial fraction technique.

$$\begin{gathered} a\frac{4+4s^2-2s^3}{s^2\left(4-s^2\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{2-s}+\frac{D}{2+s} \\ =\frac{As\left(4-s^2\right)+B\left(4-s^2\right)+C(2+s)s^2+D(2-s)s^2}{s^2\left(4-s^2\right)} \end{gathered}$$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$$\begin{gathered} 4+4s^2-2s^3=As\left(4-s^2\right)+B\left(4-s^2\right)+C(2+s)s^2+D(2-s)s^2 \\ 4+4s^2-2s^3=4As-A{s}^3+4B-B{s}^2+2C{s}^2+C{s}^3+2D{s}^2-D{s}^3 \end{gathered}$$

If s=0 then

$36=16D$

$D=\frac{9}{4}$

If s=2 then

$4=16CC=\frac{1}{4}$

Comparing the coefficient of $s^3$on both sides, we get

$-2=-A+C-DA=\frac{1}{4}-\frac{9}{4}+2A=0$

Final proof

Substituting A,B,C and D in the equation (7), we get.

$\frac{4+4s^2-2s^3}{s^2\left(4-s^2\right)}=\frac{1}{s^2}+\frac{1}{4}\frac{1}{2-s}+\frac{9}{4}\frac{1}{2+s}$

The equation (*) becomes.

$Y\left(s\right)=\frac{1}{s^2}-\frac{1}{4}\frac{1}{s-2}+\frac{9}{4}\frac{1}{2+s}$

Apply inverse Laplace transform to each side.

${\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}={\mathcal{L}}^{-1}\frac{1}{s^2}-\frac{1}{4}\frac{1}{s-2}+\frac{9}{4}\frac{1}{2+s}={\mathcal{L}}^{-1}\frac{1}{s^2}-\frac{1}{4}{\mathcal{L}}^{-1}\frac{1}{s-2}+\frac{9}{4}{\mathcal{L}}^{-1}\frac{1}{s+2}\stackrel{(**)}{=}t-\frac{1}{4}{e}^{2t}+\frac{9}{4}{e}^{-2t}$$e^{at}={\mathcal{L}}^{-1}\frac{1}{s-a},{\mathcal{L}}^{-1}\frac{1}{s^2}=t$

Therefore,

$y\left(t\right)=t-\frac{1}{4}{e}^{2t}+\frac{9}{4}{e}^{-2t}$

Substituting$y\left(t\right)=t-\frac{1}{4}{e}^{2t}+\frac{9}{4}{e}^{-2t}$in the given equation $4x+\frac{dy}{dt}=0$, we get.

$\begin{array}{rcl}4x+\frac{d}{dt}t-\frac{1}{4}{e}^{2t}+\frac{9}{4}{e}^{-2t}& =& 0\\ 4x+1-\frac{1}{2}{e}^{2t}-\frac{9}{2}{e}^{-2t}& =& 0\\ 4x& =& -1+\frac{1}{2}{e}^{2t}+\frac{9}{2}{e}^{-2t}\\ x& =& -\frac{1}{4}+\frac{1}{8}{e}^{2t}+\frac{9}{8}{e}^{-2t}\end{array}$

Therefore,

$\begin{array}{rcl}x\left(t\right)& =& -\frac{1}{4}+\frac{1}{8}{e}^{2t}+\frac{9}{8}{e}^{-2t}\\ x\left(t\right)& =& -\frac{1}{4}+\frac{1}{8}{e}^{2t}+\frac{9}{8}{e}^{-2t}\\ y\left(t\right)& =& t-\frac{1}{4}{e}^{2t}+\frac{9}{4}{e}^{-2t}\end{array}$