Question

use the Laplace transform to solve each system

$\mathbf{44}\mathbf{.}{\mathbf{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{2}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{+}{\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$$\begin{gathered} x\left(t\right)=\frac{1}{2}-\frac{1}{2}{e}^{2t}+t{e}^{2t} \\ y\left(t\right)=-\frac{3}{4}-\frac{t}{2}+\frac{3}{4}{e}^{2t}-t{e}^{2t} \end{gathered}$$

Step-by-step solution

To Find the laplace transform on both sides of each given differential equation

We know that

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

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

$\begin{array}{rcl}\mathcal{L}{e}^{2t}& =& \mathcal{L}\left\{x^{\text{'}\text{'}}\right\}+\mathcal{L}\left\{y^{\text{'}\text{'}}\right\}\\ -\mathcal{L}{e}^{2t}& =& 2\mathcal{L}\left\{x^{\text{'}}\right\}+\mathcal{L}\left\{y^{\text{'}\text{'}}\right\}\end{array}$

$\mathcal{L}{e}^{2t}\stackrel{(*)}{=}{s}^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x^{\text{'}}\left(0\right)+s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)$

$-\mathcal{L}{e}^{2t}\stackrel{(*)}{=}2s\mathcal{L}\left\{x\right\}-2x\left(0\right)+s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)$

$\begin{array}{rcl}\mathcal{L}{e}^{2t}& =& s^2\mathcal{L}\left\{x\right\}+s^2\mathcal{L}\left\{y\right\}\left(x\left(0\right)=0,x^{\text{'}}\left(0\right)=0\right)\\ -\mathcal{L}{e}^{2t}& =& 2s\mathcal{L}\left\{x\right\}+s^2\mathcal{L}\left\{y\right\}\left(y\left(0\right)=0,y^{\text{'}}\left(0\right)=0\right)\end{array}$

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

$$\begin{gathered} \mathcal{L}{e}^{2t}=s^{2}X\left(s\right)+s^{2}Y\left(s\right)\mathcal{L}{e}^{at}=\frac{1}{s-a} \\ -\mathcal{L}{e}^{2t}=2sX\left(s\right)+s^{2}Y\left(s\right) \\ \frac{1}{s-2}=s^{2}X\left(s\right)+s^{2}Y\left(s\right) \\ -\frac{1}{s-2}=2sX\left(s\right)+s^{2}Y\left(s\right) \end{gathered}$$

System simplifies to the form.

$\begin{array}{rcl}{s}^{2}X\left(s\right)+s^{2}Y\left(s\right)& =& \frac{1}{s-2}\\ 2sX\left(s\right)+s^{2}Y\left(s\right)& =& -\frac{1}{s-2}\end{array}$

Multiply equation by -1

$\begin{array}{rcl}-s^{2}X\left(s\right)-s^{2}Y\left(s\right)& =& -\frac{1}{s-2}\\ 2sX\left(s\right)+s^{2}Y\left(s\right)& =& -\frac{1}{s-2}\end{array}$

Now add (3) and (4) equations to get

$\begin{array}{rcl}-s^{2}X\left(s\right)+2sX\left(s\right)& =& -\frac{1}{s-2}-\frac{1}{s-2}\\ \left(2s-s^2\right)X\left(s\right)& =& -\frac{2}{s-2}\\ X\left(s\right)& =& \frac{2}{s{(s-2)}^2}\end{array}$

Decompose the fraction$\frac{2}{s{(s-2)}^2}$using the partial fraction technique.

$\star \frac{2}{s{(s-2)}^2}=\frac{A}{s}+\frac{B}{s-2}+\frac{C}{{(s-2)}^2}$

$=\frac{A{(s-2)}^2+Bs(s-2)+Cs}{s{(s-2)}^2}$

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

$$\begin{gathered} 2=A(s-2{)}^2+Bs(s-2)+Cs \\ 2=A{s}^2-4As+4A+B{s}^2-2Bs+Cs \end{gathered}$$

If s=0 then

$2=4AA=\frac{1}{2}$

Comparing coefficient of$s^2$on both sides, we get

$A+B=0$

$B=-\frac{1}{2}$

If s=2 then

$$\begin{gathered} 2=2C \\ C=1 \end{gathered}$$

Substituting A, B and C in the equation (5), we get.

$\frac{2}{s{(s-2)}^2}=\frac{1}{2s}-\frac{1}{2(s-2)}+\frac{1}{{(s-2)}^2}$

The equation (*) becomes.

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

find the inverse laplace transform of both side

Take inverse Laplace transform of both side.

$\star {\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=\frac{1}{2}{\mathcal{L}}^{-1}\frac{1}{s}-\frac{1}{2}{\mathcal{L}}^{-1}\frac{1}{s-2}+{\mathcal{L}}^{-1}\frac{1}{{(s-2)}^2}$

$\stackrel{(**)}{=}\frac{1}{2}-\frac{1}{2}{e}^{2t}+t{e}^{2t}$

$e^{at}={\mathcal{L}}^{-1}\frac{1}{s-a}t{e}^{at}={\mathcal{L}}^{-1}\frac{1}{{(s-a)}^2}$

$1={\mathcal{L}}^{-1}\frac{1}{s}$so,

$x\left(t\right)={\mathcal{L}}^{-1}\left\{X\right(s\left)\right\}=\frac{1}{2}-\frac{1}{2}{e}^{2t}+t{e}^{2t}$

Now, multiply equation (1) by -2 and equation (2) by s.

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

Now add (6) and (7) equations to get

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

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

$\star \frac{2+s}{s^2{(s-2)}^2}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-2}+\frac{D}{{(s-2)}^2}$

$=\frac{As{(s-2)}^2+B{(s-2)}^2+C{s}^2(s-2)+D{s}^2}{s^2{(s-2)}^2}$

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

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

Comparing coefficient on both sides, we get

$\begin{array}{rcl}A+C& =& 0\\ -4A+B-2C+D& =& 0\\ 4A-4B& =& 1\\ B& =& \frac{1}{2}\end{array}$

Based on the previous system we get $A=\frac{3}{4},B=\frac{1}{2},$$C=-\frac{3}{4}$and D=1

Substituting and D in the equation (8), we get.

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

The equation (* *) becomes.

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

Take inverse Laplace transform of both side.

$\begin{array}{rcl}\star {\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}& =& -\frac{3}{4}{\mathcal{L}}^{-1}\frac{1}{s}-\frac{1}{2}{\mathcal{L}}^{-1}\frac{1}{s^2}+\frac{3}{4}{\mathcal{L}}^{-1}\frac{1}{s-2}\\ -{\mathcal{L}}^{-1}\frac{1}{{(s-2)}^2}\stackrel{(**)}{=}-\frac{3}{4}-\frac{t}{2}+\frac{3}{4}{e}^{2t}-t{e}^{2t}{e}^{at}& =& {\mathcal{L}}^{-1}\frac{1}{s-a}t{e}^{at}={\mathcal{L}}^{-1}\frac{1}{{(s-a)}^2}\text{(**)}\end{array}$

so,

$y\left(t\right)={\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}=-\frac{3}{4}-\frac{t}{2}+\frac{3}{4}{e}^{2t}-t{e}^{2t}$

Final proof

$$\begin{gathered} x\left(t\right)=\frac{1}{2}-\frac{1}{2}{e}^{2t}+t{e}^{2t} \\ y\left(t\right)=-\frac{3}{4}-\frac{t}{2}+\frac{3}{4}{e}^{2t}-t{e}^{2t} \end{gathered}$$