Question

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

$\begin{array}{l}\mathbf{4}\mathbf{.}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{1}\\ \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathbf{x}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\\ \mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\end{array}$

Short answer

The solution of the given system of differential equations is,

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

Step-by-step solution

Definition of Laplace Transform

Let be a function defined for. Then the integral

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

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

Use Laplace Transform

We know that

$\begin{array}{l}\mathcal{L}\left\{x^{\text{'}}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)\\ \mathcal{L}\left\{y^{\text{'}}\right\}=s\mathcal{L}\left\{y\right\}-y\left(0\right)\\ TaketheLaplacetransformonbothsidesofeachgivendifferentialequation\\ \mathcal{L}\left\{1\right\}=\mathcal{L}\{x\text{'}\}+3\mathcal{L}\left\{x\right\}+\mathcal{L}\{y\text{'}\}\\ \mathcal{L}\left\{e^t\right\}=\mathcal{L}\{x\text{'}\}-\mathcal{L}\left\{x\right\}+\mathcal{L}\{y\text{'}\}-\mathcal{L}\left\{y\right\}\\ Substitutetheequation\left(1\right)and\left(2\right),weget\\ \mathcal{L}\left\{1\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)+3\mathcal{L}\left\{x\right\}+s\mathcal{L}\left\{y\right\}-y\left(0\right)\\ \mathcal{L}\left\{e^t\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)-\mathcal{L}\left\{x\right\}+s\mathcal{L}\left\{y\right\}-y\left(0\right)-\mathcal{L}\left\{y\right\}\\ Wehave,X\left(0\right)=0,Y\left(0\right)=0,so\\ \mathcal{L}\left\{1\right\}=s\mathcal{L}\left\{x\right\}+3\mathcal{L}\left\{x\right\}+s\mathcal{L}\left\{y\right\}\\ \mathcal{L}\left\{e^t\right\}=s\mathcal{L}\left\{x\right\}-\mathcal{L}\left\{x\right\}+s\mathcal{L}\left\{y\right\}-\mathcal{L}\left\{y\right\}\\ Denote\mathcal{L}\left\{x\right\}=X\left(s\right)and\mathcal{L}\left\{y\right\}=Y\left(s\right)intheabovesystem.\\ \frac{1}{s}=sX\left(s\right)+3X\left(s\right)+sY\left(s\right)\\ \frac{1}{s-1}=sX\left(s\right)-X\left(s\right)+sY\left(s\right)-Y\left(s\right)\end{array}$

Simplify the equations

The system simplifies the form.

$\begin{array}{l}\left(s+3\right)X\left(s\right)+sY\left(s\right)=\frac{1}{s}---\left(3\right)\\ X\left(s\right)+Y\left(s\right)=\frac{1}{{(s-1)}^2}---\left(4\right)\\ Multiplyequation\left(3\right)by-1andequation\left(4\right)byS+3.\\ -\left(s+3\right)X\left(s\right)-sY\left(s\right)=-\frac{1}{s}---\left(5\right)\\ (s+3)X\left(s\right)+(s+3)Y\left(s\right)=\frac{(s+3)}{{(s-1)}^2}---\left(6\right)\\ Nowadd\left(5\right)and\left(6\right)equationstoget.\\ -\left(s+3\right)X\left(s\right)-sY\left(s\right)+(s+3)X\left(s\right)+(s+3)Y\left(s\right)=-\frac{1}{s}+\frac{(s+3)}{{(s-1)}^2}\\ 3Y\left(s\right)=\frac{(s+3)}{{(s-1)}^2}-\frac{1}{s}\\ Y\left(s\right)=\frac{(s+3)}{3{(s-1)}^2}-\frac{1}{3s}\\ Y\left(s\right)=\frac{s}{3{(s-1)}^2}+\frac{1}{{(s-1)}^2}-\frac{1}{3s}\end{array}$

Use Inverse Laplace Transform

Apply inverse Laplace transform to each side.

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

\( = \frac{1}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{s}{{{{(s - 1)}^2}}}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{{{(s - 1)}^2}}}} \right\} - \frac{1}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}} \right\}\)

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

\(\mathop = \limits^{(**)} \frac{1}{3}{e^t} + \frac{4}{3}t{e^t} - \frac{1}{3}\)

\(\left( {{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}} \right\} = 1,\;\;\;{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{s - a}}} \right\} = {e^{at}},{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{{{(s - a)}^2}}}} \right\} = t{e^{at}}} \right)\;\;\;(**)\)

Therefore,

\(y(t) = \frac{1}{3}{e^t} + \frac{4}{3}t{e^t} - \frac{1}{3}\)

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

\( - \left( {s + 3} \right){\rm{ }}X\left( s \right) - sY\left( s \right){\rm{ }} = - \frac{1}{s}{\rm{ }}\)------------ (7)

\(sX\left( s \right){\rm{ + s}}Y\left( s \right){\rm{ }} = \frac{s}{{{{(s - 1)}^2}}}\)------------ (8)

Now add (7) and (8) equations to get.

\(\begin{aligned}{l} - \left( {s + 3} \right){\rm{ }}X\left( s \right) - sY\left( s \right){\rm{ + }}sX\left( s \right){\rm{ + s}}Y\left( s \right){\rm{ }} = \frac{s}{{{{(s - 1)}^2}}} - \frac{1}{s}{\rm{ }}\\{\rm{ - 3X(s) = }}\frac{s}{{{{(s - 1)}^2}}} - \frac{1}{s}{\rm{ }}\\X(s) = - \frac{s}{{3{{(s - 1)}^2}}} + \frac{1}{{3s}}{\rm{ }}\end{aligned}\)

Apply inverse Laplace transform to each side.

\({\mathcal{L}^{ - 1}}\{ X(s)\} {\rm{ }} = {\mathcal{L}^{ - 1}}\left\{ { - \frac{s}{{3{{(s - 1)}^2}}} + \frac{1}{{3s}}{\rm{ }}} \right\}\)

\( = - \frac{1}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{s}{{{{(s - 1)}^2}}}} \right\} + \frac{1}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}{\rm{ }}} \right\}\)

\( = - \frac{1}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{(s - 1)}}} \right\} - \frac{1}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{{{(s - 1)}^2}}}} \right\} + \frac{1}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}{\rm{ }}} \right\}\)

\(\mathop = \limits^{(**)} - \frac{1}{3}({e^t} + t{e^t}) + \frac{1}{3}\)

\(\left( {{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}} \right\} = 1,{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{(s - a)}}} \right\} = {e^{at}},\;\;\;{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{{{(s - a)}^2}}}} \right\} = t{e^{at}}} \right)\;\;\;(**)\)

$\begin{array}{l}Therefore,\\ x\left(t\right)=-\frac{1}{3}(e^t+t{e}^t)+\frac{1}{3}\\ Result\\ x\left(t\right)=-\frac{1}{3}(e^t+t{e}^t)+\frac{1}{3}\\ y\left(t\right)=\frac{1}{3}{e}^t+\frac{4}{3}t{e}^t-\frac{1}{3}\end{array}$