Question

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

Short answer

The solution of the given system of differential equations is,

\(x(t) = - \frac{1}{2}t - \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\)

\(y(t) = - \frac{1}{2}t + \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\)

Step-by-step solution

Definition of Laplace Transform

Let \(f\)be a function defined for\(t \ge 0\)Then the integral

\(\mathcal{L}\{ f(t)\} = \int_0^x {{e^{ - xt}}} f(t)dt\)

is said to be the Laplace transform of\(f\), provided that the integral converges.

Use Laplace Transform

We know that

\(\mathcal{L}\left\{ {{x^{\prime \prime }}} \right\} = {s^2}\mathcal{L}\{ x\} - sx(0) - {x^\prime }(0)\)

\(\mathcal{L}\left\{ {{x^\prime }} \right\} = s\mathcal{L}\{ x\} - x(0){\rm{ }}\)

\(\mathcal{L}\left\{ {{y^{\prime \prime }}} \right\} = {s^2}\mathcal{L}\{ y\} - sy(0) - {y^\prime }(0){\rm{ }}\)

\(\mathcal{L}\left\{ {{y^\prime }} \right\} = s\mathcal{L}\{ y\} - y(0)\)

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

\(\mathcal{L}\left\{ y \right\} - \mathcal{L}\left\{ x \right\} = \mathcal{L}\left\{ {{x^{\prime \prime }}} \right\}\)------------ (1)

\(\mathcal{L}\left\{ x \right\} - \mathcal{L}\left\{ y \right\} = \mathcal{L}\left\{ {{y^{\prime \prime }}} \right\}\)

------------ (2)

Substitute the known values,

\(\mathcal{L}\left\{ y \right\} - \mathcal{L}\left\{ x \right\} = {s^2}\mathcal{L}\{ x\} - sx(0) - {x^\prime }(0)\)

\(\mathcal{L}\left\{ x \right\} - \mathcal{L}\left\{ y \right\} = {s^2}\mathcal{L}\{ y\} - sy(0) - {y^\prime }(0){\rm{ }}\)

\(\mathcal{L}\left\{ y \right\} - \mathcal{L}\left\{ x \right\} = {s^2}\mathcal{L}\{ x\} + 2\) \(\left( {x(0) = 0,{x^\prime }(0) = - 2} \right)\)

\(\mathcal{L}\left\{ x \right\} - \mathcal{L}\left\{ y \right\} = {s^2}\mathcal{L}\{ y\} - 1{\rm{ }}\) \(\left( {y(0) = 0,{y^\prime }(0) = 1} \right)\)

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

\(\begin{aligned}{l}Y(s) - X(s) = {s^2}X(s) + 2\\X(s) - Y(s) = {s^2}Y(s) - 1\end{aligned}\)

Simplify the equations

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

\( - {\rm{ }}X\left( s \right) + \left( {{s^2} + 1} \right){\rm{ }}Y\left( s \right){\rm{ }} = 1\)------------ (4)

Multiply equation (4) by\(({s^2} + 1)\).

\(\left( {{s^2} + 1} \right){\rm{ }}X\left( s \right) - Y\left( s \right){\rm{ }} = - 2{\rm{ }}\)------------ (5)

\( - {\rm{ }}({s^2} + 1)X\left( s \right) + {\left( {{s^2} + 1} \right)^2}{\rm{ }}Y\left( s \right){\rm{ }} = ({s^2} + 1)\)------------ (6)

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

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

Use the partial fraction technique.

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

\(\begin{aligned}{l}\frac{{{s^2} - 1}}{{{s^2}({s^2} + 2)}} = \frac{{As + B}}{{{s^2}}} + \frac{{Cs + D}}{{{s^2} + 2}}......(7)\\ = \frac{{(As + B)({s^2} + 2) + (Cs + D){s^2}}}{{{s^2}({s^2} + 2)}}\end{aligned}\)

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

\(\begin{aligned}{l}{s^2} - 1 = (As + B)({s^2} + 2) + (Cs + D){s^2}\\{s^2} - 1 = A{s^3} + 2As + B{s^2} + 2B + C{s^3} + D{s^2}\\{s^2} - 1 = (A + C){s^3} + (B + D){s^2} + 2As + 2B\end{aligned}\)

Comparing the coefficient on both sides, we get,

\(\begin{aligned}{l}A + C = 0\\B + D = 1\\2A = 0\\2B = - 1\\C = 0\\D = \frac{3}{2}\\A = 0\\B = - \frac{1}{2}\end{aligned}\)

Substitute A , B,C, and D in equation (7), and we get,

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

It becomes,

\(Y(s) = - \frac{1}{2}\frac{1}{{{s^2}}} + \frac{3}{2}\frac{1}{{{s^2} + 2}}.\)

Use Inverse Laplace Transform

Apply inverse Laplace transform to each side.

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

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

\(\mathop = \limits^{(**)} - \frac{1}{2}t + \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\)

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

Therefore,

\(y(t) = - \frac{1}{2}t + \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\)

Now, multiply equation (3) by\(({s^2} + 1)\),

\({({s^2} + 1)^2}X(s) - ({s^2} + 1)Y(s) = - 2({s^2} + 1)\)

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

\[{\left( {{s^2} + 1} \right)^2}X\left( s \right) - \left( {{s^2} + 1} \right)Y(s) + \left[ { - X\left( s \right) + \left( {{s^2} + 1} \right)Y\left( s \right)} \right] = 1\]

\[{s^2}\left( {{s^2} + 2} \right)X(s) = - 2{s^2} - 1\]

\[X(s) = \frac{{ - 2{s^2} - 1}}{{{s^2}({s^2} + 2)}}\]

Use the partial fraction technique.

Decompose the fraction \[\frac{{ - 2{s^2} - 1}}{{{s^2}({s^2} + 2)}}\] using the partial fraction technique.

\[\begin{array}{l}\frac{{ - 2{s^2} - 1}}{{{s^2}({s^2} + 2)}} = \frac{{As + B}}{{{s^2}}} + \frac{{Cs + D}}{{{s^2} + 2}}......(10)\\ = \frac{{(As + B)({s^2} + 2) + (Cs + D){s^2}}}{{{s^2}({s^2} + 2)}}\end{array}\]

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

\[\begin{array}{l} - 2{s^2} - 1 = (As + B)({s^2} + 2) + (Cs + D){s^2}\\ - 2{s^2} - 1 = A{s^3} + 2As + B{s^2} + 2B + C{s^3} + D{s^2}\\ - 2{s^2} - 1 = (A + C){s^3} + (B + D){s^2} + 2As + 2B\end{array}\]

Comparing the coefficient on both sides, we get,

\[\begin{array}{l}A + C = 0\\B + D = - 2\\2A = 0\\2B = - 1\\C = 0\\D = - \frac{3}{2}\\A = 0\\B = - \frac{1}{2}\end{array}\]

Substitute A , B,C, and D in equation (10), and we get,

\[\frac{{ - 2{s^2} - 1}}{{{s^2}({s^2} + 2)}} = - \frac{1}{2}\frac{1}{{{s^2}}} - \frac{3}{2}\frac{1}{{{s^2} + 2}}.\]

It becomes,

\[X(s) = - \frac{1}{2}\frac{1}{{{s^2}}} - \frac{3}{2}\frac{1}{{{s^2} + 2}}.\]

Use Inverse Laplace Transform

Apply inverse Laplace transform to each side.

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

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

\[\mathop = \limits^{(**)} - \frac{1}{2}t - \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\]

\[\left( {{\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{{s^2}}}} \right\} = t,\;\;\;{\mathcal{L}^{ - 1}}\left\{ {\frac{k}{{{s^2} + {k^2}}}} \right\} = \sin kt} \right)\;\;\;(**)\]

Therefore,

\[x(t) = - \frac{1}{2}t - \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\]

Result

\[x(t) = - \frac{1}{2}t - \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\]

\[y(t) = - \frac{1}{2}t + \frac{3}{{2\sqrt 2 }}\sin \sqrt 2 t\]