Question

Solve system (1) when

${\mathit{k}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{,}{\mathit{k}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{,}{\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathit{a}\mathit{n}\mathit{d}{\mathit{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{x}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$$\begin{gathered} X_1\left(t\right)=\frac{2}{5}\cos t+\frac{1}{5}\sin t+\frac{4}{5\sqrt{6}}\sin \sqrt{6}t-\frac{2}{5}\cos \sqrt{6}t \\ {X}_2\left(t\right)=\frac{4}{5}\cos t+\frac{2}{5}\sin t-\frac{2}{5\sqrt{6}}\sin \sqrt{6}t+\frac{1}{5}\cos \sqrt{6}t \end{gathered}$$

Step-by-step solution

Given Information.

$$\begin{gathered} thesystemis \\ \begin{array}{l}=-k_1{x}_1+k_2\left(x_2-x_1\right)\\ m_1{x}_1^{\text{'}\text{'}}{m}_2{x}_2^{\text{'}\text{'}}=-k_2\left(x_2-x_1\right)\end{array} \end{gathered}$$

Apply Laplace Transform

Replace the given values to get

$\begin{array}{l}-3x_1+2\left(x_2-x_1\right)=x_1^{\text{'}\text{'}}\\ \underset{¯}{-2\left(x_2-x_1\right)=x_2^{\text{'}\text{'}}}\\ -5x_1+2x_2=x_1^{\text{'}\text{'}}\\ \underset{¯}{-2x_2+2x_1=x_2^{\text{'}\text{'}}}\end{array}$

Take Laplace to transform to get

$\begin{array}{l}-5L\left\{x_1\right\}+2L\left\{x_2\right\}=L\left\{x_1^{\text{'}\text{'}}\right\}\\ -2L\left\{x_2\right\}+2L\left\{x_1\right\}=L\left\{x_2^{\text{'}\text{'}}\right\}\end{array}$

We know that

\(\begin{aligned}{*{20}{c}}{L\left\{ {x_1^{\prime \prime }} \right\} = {s^2}L\left\{ {{x_1}} \right\} - s{x_1}(0) - x_1^\prime (0)}\\{L\left\{ {x_2^{\prime \prime }} \right\} = {s^2}L\left\{ {{x_2}} \right\} - s{x_2}(0) - x_2^\prime (0)}\end{aligned}\)

\( - 5L\left\{ {{x_1}} \right\} + 2L\left\{ {{x_2}} \right\} = {s^2}L\left\{ {{x_1}} \right\} - s{x_1}(0) - x_1^\prime (0)\;\;\;{\rm{ }}\left( {{x_1}(0) = 0,x_1^\prime (0) = 1} \right)\)

\( - 2L\left\{ {{x_2}} \right\} + 2L\left\{ {{x_1}} \right\} = {s^2}L\left\{ {{x_2}} \right\} - s{x_2}(0) - x_2^\prime (0)\;\;\;{\rm{ }}\left( {{x_2}(0) = 1,x_2^\prime (0) = 0} \right)\)

\( - 5L\left\{ {{x_1}} \right\} + 2L\left\{ {{x_2}} \right\} = {s^2}L\left\{ {{x_1}} \right\} - 1\)

\( - 2L\left\{ {{x_2}} \right\} + 2L\left\{ {{x_1}} \right\} = {s^2}L\left\{ {{x_2}} \right\} - s\)

Substitute

\(\begin{aligned}{*{20}{c}}{ - 5{X_1}(s) + 2{X_2}(s) = {s^2}{X_1}(s) - 1}\\{ - 2{X_2}(s) + 2{X_1}(s) = {s^2}{X_2}(s) - s}\end{aligned}\)
System simplifies to the form

\(\left( {5 + {s^2}} \right){X_1}(s) - 2{X_2}(s) = 1.....(3)\)

\( - 2{X_1}(s) + \left( {{s^2} + 2} \right){X_2}(s) = s......(4)\)

Multiply equation (3) by \(\left( {{s^2} + 2} \right)\)and equation (4) by 2 .

\(\left( {5 + {s^2}} \right)\left( {{s^2} + 2} \right){X_1}(s) - 2\left( {{s^2} + 2} \right){X_2}(s) = {s^2} + 2....(5)\)

\( - 4{X_1}(s) + 2\left( {{s^2} + 2} \right){X_2}(s) = 2s......(6)\)

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

\(\left( {5 + {s^2}} \right)\left( {{s^2} + 2} \right){X_1}(s) - 4{X_1}(s) = {s^2} + 2 + 2s\)

\(\left( {5{s^2} + 10 + {s^4} + 2{s^2} - 4} \right){X_1}(s) = {s^2} + 2 + 2s\)

\(\left( {{s^4} + 7{s^2} + 6} \right){X_1}(s) = {s^2} + 2 + 2s\)

\(\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right){X_1}(s) = {s^2} + 2 + 2s\)

\({X_1}(s) = \frac{{{s^2} + 2 + 2s}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}}\)

Apply Partial Fraction 

Decompose the fraction \(\frac{{{s^2} + 2 + 2s}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}}\)using the partial fraction technique.

\(\frac{{{s^2} + 2 + 2s}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}} = \frac{{As + B}}{{{s^2} + 1}} + \frac{{Cs + D}}{{{s^2} + 6}}\)

\( = \frac{{(As + B)\left( {{s^2} + 6} \right) + (Cs + D)\left( {{s^2} + 1} \right)}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}}\)

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

\(\begin{aligned}{l}{s^2} + 2 + 2s = A{s^3} + 6As + B{s^2} + 6B + C{s^3} + Cs + D{s^2} + D\\{s^2} + 2 + 2s = (A + C){s^3} + (B + D){s^2} + (6A + C)s + 6B + D\end{aligned}\)

Comparing coefficient on both sides

we get

\(A + C = 0\)

\(6A + C = 2\)

\(B + D = 1\)

\(6B + D = 2\)

Based on the previous two systems, we get

From the first equation \(A = - C\)

The second equation is \( - 6C + C = 2\)

\( - 5C = 2\)

\(C = - \frac{2}{5}\)

From the third equation as \(B = 1 - D\)

Substitute the fourth equation as \(6 - 6D + D = 2\)

\( - 5D = 2 - 6\)

\( - 5D = - 4\)

\(D = \frac{4}{5}\)

\(A = \frac{2}{5}\)

\(B = 1 - \frac{4}{5}\)

\(B = \frac{{5 - 4}}{5}\)

\(B = \frac{1}{5}\)

\(\frac{{{s^2} + 2 + 2s}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}} = \frac{{2s + 1}}{{5\left( {{s^2} + 1} \right)}} + \frac{{4 - 2s}}{{5\left( {{s^2} + 6} \right)}}\)

\({X_1}(s) = \frac{{2s + 1}}{{5\left( {{s^2} + 1} \right)}} + \frac{{4 - 2s}}{{5\left( {{s^2} + 6} \right)}}\)

Take inverse Laplace transform of both sides

\({L^{ - 1}}\left\{ {{X_1}(s)} \right\} = {L^{ - 1}}\left\{ {\frac{{2s + 1}}{{5\left( {{s^2} + 1} \right)}} + \frac{{4 - 2s}}{{5\left( {{s^2} + 6} \right)}}} \right\}\)

\( = \frac{2}{5}{L^{ - 1}}\left\{ {\frac{s}{{{s^2} + 1}}} \right\} + \frac{1}{5}{L^{ - 1}}\left\{ {\frac{1}{{{s^2} + 1}}} \right\} + \frac{4}{{5\sqrt 6 }}{L^{ - 1}}\left\{ {\frac{{\sqrt 6 }}{{{s^2} + {{(\sqrt 6 )}^2}}}} \right\} - \frac{2}{5}{L^{ - 1}}\left\{ {\frac{s}{{{s^2} + {{(\sqrt 6 )}^2}}}} \right\}\)

\({x_1}(t) = {L^{ - 1}}\left\{ {{X_1}(s)} \right\} = \frac{2}{5}\cos t + \frac{1}{5}\sin t + \frac{4}{{5\sqrt 6 }}\sin \sqrt 6 t - \frac{2}{5}\cos \sqrt 6 t\)

Multiply equation (3) by 2 and equation (4) by \(\left( {5 + {s^2}} \right).\)

\(2\left( {5 + {s^2}} \right){X_1}(s) - 4{X_2}(s) = 2....(8)\)

\( - 2\left( {5 + {s^2}} \right){X_1}(s) + \left( {5 + {s^2}} \right)\left( {{s^2} + 2} \right){X_2}(s) = s\left( {5 + {s^2}} \right)...(9)\)

Now add (8) and (9) equations to get.

\(\left( {5 + {s^2}} \right)\left( {{s^2} + 2} \right){X_2}(s) - 4{X_2}(s) = 2 + s\left( {5 + {s^2}} \right)\)

\(\left( {5{s^2} + 10 + {s^4} + 2{s^2} - 4} \right){X_2}(s) = 2 + 5s + {s^3}\)

\(\left( {{s^4} + 7{s^2} + 6} \right){X_2}(s) = 2 + 5s + {s^3}\)

\(\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right){X_2}(s) = 2 + 5s + {s^3}\)

\({X_2}(s) = \frac{{2 + 5s + {s^3}}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}}\)

Decompose the fraction

Decompose the fraction \(\frac{{2 + 5s + {s^3}}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}}\)using the partial fraction technique.

\(\frac{{2 + 5s + {s^3}}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}} = \frac{{As + B}}{{{s^2} + 1}} + \frac{{Cs + D}}{{{s^2} + 6}}.....(10)\)

\( = \frac{{(As + B)\left( {{s^2} + 6} \right) + (Cs + D)\left( {{s^2} + 1} \right)}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}}\)

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

\(2 + 5s + {s^3} = A{s^3} + 6As + B{s^2} + 6B + C{s^3} + Cs + D{s^2} + D\)

\(2 + 5s + {s^3} = (A + C){s^3} + (B + D){s^2} + (6A + C)s + 6B + D\)

Comparing coefficient on both sides, we get

\(A + C = 1\)

\(6A + C = 5\)

\(B + D = 0\)

\(6B + D = 2\)

Based on the previous two systems, we get

Substituting A, B, C and D in the equation (10), we get.

\(\frac{{2 + 5s + {s^3}}}{{\left( {{s^2} + 1} \right)\left( {{s^2} + 6} \right)}} = \frac{{4s + 2}}{{5\left( {{s^2} + 1} \right)}} + \frac{{s - 2}}{{5\left( {{s^2} + 6} \right)}}\)

\({X_2}(s) = \frac{{4s + 2}}{{5\left( {{s^2} + 1} \right)}} + \frac{{s - 2}}{{5\left( {{s^2} + 6} \right)}}\)

Take inverse Laplace transform of both sides.

\({L^{ - 1}}\left\{ {{X_2}(s)} \right\} = {L^{ - 1}}\left\{ {\frac{{4s + 2}}{{5\left( {{s^2} + 1} \right)}} + \frac{{s - 2}}{{5\left( {{s^2} + 6} \right)}}} \right\}\)

\({L^{ - 1}}\left\{ {\frac{s}{{{s^2} + {k^2}}}} \right\} = \cos kt,\;\;\;{L^{ - 1}}\left\{ {\frac{k}{{{s^2} + {k^2}}}} \right\} = \sin kt\)

\( = \frac{4}{5}{L^{ - 1}}\left\{ {\frac{s}{{{s^2} + 1}}} \right\} + \frac{2}{5}{L^{ - 1}}\left\{ {\frac{1}{{{s^2} + 1}}} \right\} - \frac{2}{{5\sqrt 6 }}{L^{ - 1}}\left\{ {\frac{{\sqrt 6 }}{{{s^2} + {{(\sqrt 6 )}^2}}}} \right\} + \frac{1}{5}{L^{ - 1}}\left\{ {\frac{s}{{{s^2} + {{(\sqrt 6 )}^2}}}} \right\}\)

\({x_2}(t) = {L^{ - 1}}\left\{ {{X_2}(s)} \right\}\)

\( = \frac{4}{5}\cos t + \frac{2}{5}\sin t - \frac{2}{{5\sqrt 6 }}\sin \sqrt 6 t + \frac{1}{5}\cos \sqrt 6 t\)