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      <text>20.4.4.1 Applications of systems of differential equations</text>
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      <text><![CDATA[<h2>Applications of systems of differential equations</h2>
<p>Coupled electrical circuits and mechanical vibrating systems involving several masses in springs offer examples of engineering systems modelled by systems of differential equations.</p>
<h3>Electrical circuits</h3>
<p>Consider the RL (resistance/inductance) circuit with a voltage \(v(t)\) applied as shown in the figure below.</p>
<p><center><img src="@@PLUGINFILE@@/20-4-fig-17-circuit-exercise.svg?time=1691407549410" alt="Diagram showing a RL (resistance/inductance) circuit." width="317" height="169" class="img-responsive atto_image_button_text-bottom"></center></p>
<p>If \(i_1\) and \(i_2\) denote the currents in each loop we obtain, using Kirchhoff's voltage law:</p>
<p>(i) in the right loop:  \(L_1\frac{\mathrm{d}i_1}{\mathrm{d}t}+R_2(i_1-i_2)+R_1i_1=v(t)\)</p>
<p> (ii) in the left loop: \(L_2\frac{\mathrm{d}i_2}{\mathrm{d}t}+R_2(i_2-i_1)=0\) </p>
<p></p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Suppose, in the above circuit, that
\(L_1=0.8\) henry,    \(L_2=1\) henry,    \(R_1=1.4\, \Omega \quad R_2=1\, \Omega\)</p>
<p>Assume zero initial conditions: \(i_1(0)=i_2(0)=0\).</p>
<p> Suppose that the applied voltage is constant: \( v(t)=100\) volts    \(t\ge 0\). </p>
<p>Solve the problem by Laplace transforms.</p>
<h4 class="HELM_solution">Solution</h4>
<p>Begin by obtaining \(V(s)\), the Laplace transform of \(v(t)\):</p>
<p>We have, from the definition of the Laplace transform:
\[ V(s)=\int_0^{\infty}100\text{e}^{-st}dt=100\left[\dfrac{\text{e}^{-st}}{-s}\right]_0^{\infty}=\dfrac{100}{s}\] This is simply the Laplace transform of the step function of height 100.</p>
<p>Now insert the parameter values into the differential equations and obtain the Laplace transform of each equation. Denote by \(I_1(s), I_2(s)\) the Laplace transforms of the unknown currents. (These are equivalent to \(X(s)\) and \(Y(s)\) of the theory.):</p>
<p>\[\begin{align} 0.8\dfrac{di_1}{dt}+i_1-i_2+1.4i_1&amp;=v(t) \\  \dfrac{di_2}{dt}+i_2-i_1&amp;=0 \end{align}\] Rearranging and dividing the first equation by 0.8: \[\begin{align} \dfrac{di_1}{dt}+3i_1-1.25i_2&amp;=1.25v(t) \\  \dfrac{di_2}{dt}-i_1+i_2&amp;=0 \end{align} \] Taking Laplace transforms and inserting the initial conditions \(i_1(0)=0,\ i_2(0)=0\): \[\begin{align} (s+3)I_1(s)-1.25I_2(s)&amp;=\dfrac{125}{s} \\ -I_1(s)+(s+1)I_2(s)&amp;=0 \end{align}\] </p>
<p>Now solve these equations for \(I_1(s)\) and \(I_2(s)\). Put each expression into partial
fractions and finally take the inverse Laplace transform to obtain \(i_1(t)\) and \(i_2(t)\):</p>
<p>
We find
\[ I_1(s)=\dfrac{125(s+1)}{s(s+1/2)(s+7/2)}=\dfrac{500}{7s}-\dfrac{125}{3(s+1/2)}-\dfrac{625}{21(s+7/2)}\] in partial fractions.</p>
<p> Hence    \(\displaystyle i_1(t)=\dfrac{500}{7}-\dfrac{125}{3}\text{e}^{-t/2}-\dfrac{625}{21}\text{e}^{-7t/2}\)</p>
<p>Similarly \[ I_2(s)=\dfrac{125}{s(s+1/2)(s+7/2)}=\dfrac{500}{7s}-\dfrac{250}{3(s+1/2)} +\dfrac{250}{21(s+7/2)}\] which has inverse Laplace transform: \[ i_2(t)=\dfrac{500}{7}-\dfrac{250}{3}\text{e}^{-t/2}+\dfrac{250}{21}\text{e}^{-7t/2}\]</p>
<p> Notice in both cases that \(i_1(t)\) and \(i_2(t)\)  tend to the steady state value \(\dfrac{500}{7}\) as \(t\) increases. </p>]]></text>
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