Question

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

$\begin{array}{l}\mathbf{5}\mathbf{.}\mathbf{2}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{1}\\ \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{-}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{2}\\ \mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\end{array}$

Short answer

The solution of the given system of differential equations is,

$\begin{array}{l}x\left(t\right)=-\frac{1}{2}-2e^{3t}+\frac{5}{2}{e}^{2t}\\ y\left(t\right)=-\frac{1}{6}+\frac{8}{3}{e}^{3t}-\frac{5}{2}{e}^{2t}\end{array}$

Step-by-step solution

Definition of Laplace Transform

Let be a function defined for$t\ge 0$. 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 f, provided that the integral converges.

Use Laplace Transform

We know that

$\begin{array}{l}\mathcal{L}\left\{x^{\text{'}}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)----\left(1\right)\\ \mathcal{L}\left\{y^{\text{'}}\right\}=s\mathcal{L}\left\{y\right\}-y\left(0\right)----\left(2\right)\\ TaketheLaplacetransformonbothsidesofeachgivendifferentialequation\\ \mathcal{L}\left\{1\right\}=2\mathcal{L}\{x\text{'}\}+\mathcal{L}\{y\text{'}\}-2\mathcal{L}\left\{x\right\}\\ 2\mathcal{L}\left\{1\right\}=\mathcal{L}\{x\text{'}\}+\mathcal{L}\{y\text{'}\}-3\mathcal{L}\left\{x\right\}-3\mathcal{L}\left\{y\right\}\\ Substitutetheequation\left(1\right)and\left(2\right),weget\\ \mathcal{L}\left\{1\right\}=2\left(s\mathcal{L}\right\{x\}-x(0\left)\right)+s\mathcal{L}\left\{y\right\}-y\left(0\right)-2\mathcal{L}\left\{x\right\}\\ 2\mathcal{L}\left\{1\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)+s\mathcal{L}\left\{y\right\}-y\left(0\right)-3\mathcal{L}\left\{x\right\}-3\mathcal{L}\left\{y\right\}\\ Wehave,X\left(0\right)=0,Y\left(0\right)=0,so\\ \mathcal{L}\left\{1\right\}=2\left(s\mathcal{L}\right\{x\left\}\right)+s\mathcal{L}\left\{y\right\}-2\mathcal{L}\left\{x\right\}\\ 2\mathcal{L}\left\{1\right\}=s\mathcal{L}\left\{x\right\}+s\mathcal{L}\left\{y\right\}-3\mathcal{L}\left\{x\right\}-3\mathcal{L}\left\{y\right\}\\ Denote\mathcal{L}\left\{x\right\}=X\left(s\right)and\mathcal{L}\left\{y\right\}=Y\left(s\right)intheabovesystem.\\ \frac{1}{s}=(2s-2)L\left\{x\right\}+sL\left\{y\right\}.......\left(3\right)\\ \frac{2}{s}=(s-3)L\left\{x\right\}+(s-3)L\left\{y\right\}......\left(4\right)\end{array}$

Simplify the equations

Multiply equation (3) by S-3 and equation (4) by S.

$\begin{array}{l}\mathcal{L}\left\{x\right\}=\frac{-s-3}{s(s-3)(s-2)}\\ \mathcal{L}\left\{y\right\}=\frac{3s-1}{s(s-3)(s-2)}\\ SO,\\ X\left(s\right)=\frac{-s-3}{s(s-3)(s-2)}..........\left(5\right)\\ Y\left(s\right)=\frac{3s-1}{s(s-3)(s-2)}..........\left(6\right)\\ Withaidofpartialfractionweget,\\ \frac{-s-3}{s(s-3)(s-2)}=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s-2}\\ \frac{3s-1}{s(s-3)(s-2)}=\frac{A_1}{s}+\frac{B_1}{s-3}+\frac{C_1}{s-2}\\ Evaluatingconstantcoefficientasusualmethod\\ A=\frac{1}{2},B=-2,C=\frac{5}{2}\\ A_1=-\frac{1}{6},B_1=\frac{8}{3},C_1=-\frac{5}{2}\end{array}$

Use Inverse Laplace Transform

Apply inverse Laplace transform to each side.

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

\( = - \frac{1}{6}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}} \right\} + \frac{8}{3}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{(s - 3)}}} \right\} - \frac{5}{2}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{s - 2}}} \right\}\)

\( = - \frac{1}{6} + \frac{8}{3}{e^{3t}} - \frac{5}{2}{e^{2t}}\)

Therefore,

\(y(t) = - \frac{1}{6} + \frac{8}{3}{e^{3t}} - \frac{5}{2}{e^{2t}}\)

Apply inverse Laplace transform to each side.

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

\( = - \frac{1}{2}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{s}} \right\} - 2{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{(s - 3)}}} \right\} + \frac{5}{2}{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{s - 2}}} \right\}\)

\( = - \frac{1}{2} - 2{e^{3t}} + \frac{5}{2}{e^{2t}}\)

Therefore,

\(x(t) = - \frac{1}{2} - 2{e^{3t}} + \frac{5}{2}{e^{2t}}\)

Result

\(x(t) = - \frac{1}{2} - 2{e^{3t}} + \frac{5}{2}{e^{2t}}\)

\(y(t) = - \frac{1}{6} + \frac{8}{3}{e^{3t}} - \frac{5}{2}{e^{2t}}\)