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277 lines
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277 lines
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Separation of Variables</h1>
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<p class="center">Separation of Variables is a special method to solve some Differential Equations</p>
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<div class="def">
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<p>A <a href="differential-equations.html">Differential Equation</a> is an equation with a <a href="../sets/function.html">function</a> and one or more of its <a href="derivatives-introduction.html">derivatives</a>:</p>
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<p class="center"><img src="images/diff-eq-sep-var.svg" alt="dy/dx = 5xy"><br>
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Example: an equation with the function <b>y</b> and its
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derivative<b> <span class="intbl"> <em>dy</em> <strong>dx</strong> </span></b> </p>
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</div>
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<h2>When Can I Use it?</h2>
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<div class="center80">
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/diff-eq-sep-var-a.svg" alt="Separation of Variables: dy/dx = 5xy becomes dy/y = 5x dx"></p>
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<p>Separation of Variables can be used when:</p>
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<p>All the y terms (including dy) can be moved to one side of the equation, and</p>
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<p>All the x terms (including dx) to the other side.</p>
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<div style="clear:both"></div>
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</div>
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<h2>Method</h2>
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<p><strong>Three Steps:</strong></p>
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<ul>
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<li><strong>Step 1</strong> Move all the y terms (including dy) to one side of the equation and all the x terms (including dx) to the other side.</li>
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<li><strong>Step 2 </strong>Integrate one side with respect to <b>y</b> and the other side with respect to <b>x</b>. Don't forget "+ C" (the constant of integration).</li>
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<li><b>Step 3</b> Simplify</li>
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</ul>
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<div class="example">
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<h3>Example: Solve this (k is a constant):</h3>
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<p class="center larger"> <span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = ky</p>
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<p><strong>Step 1</strong> Separate the variables by moving all the y terms to one side of the equation and all the x terms to the other side:</p>
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<div class="tbl">
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<div class="row"><span class="left">Multiply both sides by dx:</span><span class="right"> dy = ky dx</span></div>
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<div class="row"><span class="left">Divide both sides by y:</span><span class="right"> <span class="intbl"> <em>dy</em> <strong>y</strong> </span> = k dx</span></div>
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</div>
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<p><strong>Step 2 </strong><a href="integration-introduction.html">Integrate</a> both sides of the equation separately:</p>
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<div class="tbl">
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<div class="row"><span class="left">Put the integral sign in front:</span><span class="right"><span class="center"><span class="integral">∫</span> <span class="intbl"><em>dy</em> <strong>y</strong> </span> = <span class="integral">∫ </span> k dx </span></span></div>
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<div class="row"><span class="left">Integrate left side:</span><span class="right"> ln(y) + C = <span class="center"><span class="integral">∫</span> k dx </span></span></div>
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<div class="row"><span class="left">Integrate right side: </span><span class="right"> ln(y) + C = kx + D </span></div>
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</div>
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<p>C is the constant of integration. And we use D for the other, as it is a different constant.</p>
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<p> </p>
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<p><strong>Step 3 </strong> Simplify:</p>
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<div class="tbl">
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<div class="row"><span class="left">We can roll the two constants into one (a=D−C): </span><span class="right"><b> </b><span class="larger">ln(y) = kx + a </span></span></div>
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<div class="row"><span class="left"><b>e<sup>(ln(y))</sup> = y</b> , so let's <a href="../algebra/exponents-logarithms.html">take exponents</a> on both sides:</span><span class="right"><span class="larger">y = e<sup>kx + a</sup> </span></span></div>
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<div class="row"><span class="left"> And <b>e<sup>kx + a</sup> = e<sup>kx</sup> e<sup>a</sup></b> so we get:</span><span class="right"><span class="larger">y = e<sup>kx</sup> e<sup>a</sup></span></span></div>
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<div class="row"><span class="left"><span class="larger"> <b>e<sup>a</sup></b></span> is just a constant so we replace it with <b>c</b>:</span><span class="right"><span class="larger">y = ce<sup>kx</sup></span></span></div>
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</div>
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<p>We have solved it:</p>
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<p class="center"><span class="larger">y = ce<sup>kx</sup></span></p>
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<p>This is a general type of first order differential equation which turns up in all sorts of unexpected places in real world examples.</p>
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</div>
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<p>We used <b>y</b> and <b>x</b>, but the same method works for other variable names, like this:</p>
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<div class="example">
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/rabbits.jpg" alt="rabbits" height="147" width="200"></p>
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<h3>Example: Rabbits!</h3>
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<p>The more rabbits you have the more baby rabbits you will get. Then those rabbits grow up and have babies too! The population will grow faster and faster.</p>
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<p>The important parts of this are:</p>
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<ul>
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<li>the population <b>N</b> at any time <b>t</b></li>
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<li>the growth rate <b>r</b></li>
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<li>the population's rate of change <span class="larger"> <span class="center larger"> <span class="intbl"> <em>dN</em> <strong>dt</strong> </span> </span> </span></li>
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</ul>
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<p>The rate of change at any time equals the growth rate times the population:</p>
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<p class="center larger"><span class="intbl"> <em>dN</em> <strong> dt</strong> </span> = rN</p>
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<p>But hey! This is the same as the equation we just solved! It just has different letters:</p>
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<ul>
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<li>N instead of y</li>
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<li>t instead of x</li>
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<li>r instead of k</li>
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</ul>
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<p>So we can jump to a solution:</p>
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<p class="center"><span class="larger">N = ce<sup>rt</sup></span></p>
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<p> </p>
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<p>And here is an example, the graph of <span class="center"><span class="larger">N = 0.3e<sup>2t</sup></span></span>:</p>
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<p class="center"><img src="images/exponential-growth.gif" alt="exponential growth" height="213" width="233"><br>
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Exponential Growth</p>
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</div>
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<p>There are other equations that follow this pattern such as <a href="../money/compound-interest.html">continuous compound interest</a>.</p>
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<h2>More Examples</h2>
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<p>OK, on to some different examples of separating the variables:</p>
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<div class="example">
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<h3>Example: Solve this:</h3>
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<p class="center larger"><span class="intbl"><em>dy</em><strong>dx</strong></span> = <span class="intbl"><em>1</em><strong>y</strong></span></p>
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<p> </p>
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<p><strong>Step 1</strong> Separate the variables by moving all the y terms to one side of the equation and all the x terms to the other side:</p>
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<div class="tbl">
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<div class="row"><span class="left">Multiply both sides by dx:</span><span class="right">dy = (1/y) dx</span></div>
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<div class="row"><span class="left">Multiply both sides by y:</span><span class="right"> y dy = dx</span></div>
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</div>
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<p><strong>Step 2 </strong><a href="integration-introduction.html">Integrate</a> both sides of the equation separately:</p>
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<div class="tbl">
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<div class="row"><span class="left">Put the integral sign in front:</span><span class="right"><span class="center"><span class="integral">∫</span> y dy = <span class="integral">∫</span> dx </span></span></div>
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<div class="row"><span class="left">Integrate each side:</span><span class="right"> (y<sup>2</sup>)/2 = x + C<span class="center"> </span></span></div>
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</div>
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<p>We integrated both sides in the one line.</p>
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<p>We also used a shortcut of just one constant of integration <b>C.</b> This is perfectly OK as we could have +D on one, +E on the other and just say that C = E−D.</p>
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<p> </p>
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<p><strong>Step 3 </strong> Simplify:</p>
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<div class="tbl">
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<div class="row"><span class="left">Multiply both sides by 2: </span><span class="right"><span class="larger">y<sup>2</sup> = 2(x + C)<span class="center"> </span></span></span></div>
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<div class="row"><span class="left">Square root of both sides:</span><span class="right"><span class="larger">y = ±√(2(x + C))</span></span></div>
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</div>
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<p><i>Note: This is not the same as y = √(2x) + C, because the C was added <b>before</b> we took the square root. This happens a lot with differential equations. We cannot just add the C at the end of the process. It is added when doing the integration.</i></p>
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<p>We have solved it:</p>
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<p class="center"><span class="larger">y = ±√(2(x + C))</span></p>
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</div>
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<p>A harder example:</p>
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<div class="example">
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<h3>Example: Solve this:</h3>
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<p class="center larger"><span class="intbl"><em>dy</em><strong>dx</strong></span> = <span class="intbl"><em>2xy</em><strong>1+x<sup>2</sup></strong></span></p>
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<p> </p>
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<p><strong>Step 1</strong> Separate the variables:</p>
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<p>Multiply both sides by dx, divide both sides by y:</p>
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<p class="center larger"><span class="intbl"><em>1</em><strong>y</strong></span> dy = <span class="intbl"><em>2x</em><strong>1+x<sup>2</sup></strong></span>dx</p>
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<p><strong>Step 2 </strong><a href="integration-introduction.html">Integrate</a> both sides of the equation separately:</p>
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<p class="center larger"><span class="integral">∫</span><span class="intbl"><em>1</em><strong>y</strong></span> dy = <span class="integral">∫</span><span class="intbl"><em>2x</em><strong>1+x<sup>2</sup></strong></span>dx</p>
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<p>The left side is a simple logarithm, the right side can be integrated using substitution:</p>
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<div class="tbl">
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<div class="row"><span class="left">Let <b>u = 1 + x<sup>2</sup></b>, so <b>du = 2x dx</b>:</span><span class="right"><span class="integral">∫</span><span class="intbl"><em>1</em><strong>y</strong></span> dy = <span class="integral">∫</span><span class="intbl"><em>1</em><strong>u</strong></span>du</span></div>
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<div class="row"><span class="left">Integrate:</span><span class="right">ln(y) = ln(u) + C</span></div>
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<div class="row"><span class="left">Then we make <b>C = ln(k)</b>:</span><span class="right">ln(y) = ln(u) + ln(k)</span></div>
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<div class="row"><span class="left">So we can get this:</span><span class="right">y = uk</span></div>
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<div class="row"><span class="left">Now put u = 1 + x<sup>2</sup> back again:</span><span class="right">y = k(1 + x<sup>2</sup>) </span></div>
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</div>
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<p> </p>
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<p><strong>Step 3 </strong> Simplify:</p>
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<p>It is already as simple as can be. We have solved it:</p>
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<p class="center"><span class="larger">y = k(1 + x<sup>2</sup>) </span></p>
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</div>
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<p>An even harder example: the famous <i><b>Verhulst Equation</b></i></p>
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<div class="example">
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/rabbits.jpg" alt="rabbits" height="147" width="200"></p>
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<h3>Example: Rabbits Again!</h3>
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<p>Remember our growth Differential Equation:</p>
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<p class="center larger"><span class="intbl"><em>dN</em><strong>dt</strong></span> = rN</p>
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<p class="center larger"> </p>
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<p>Well, that growth can't go on forever as they will soon run out of available food.</p>
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<p>A guy called Verhulst included <b>k</b> (the maximum population the food can support) to get:</p>
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<p class="center larger"><span class="intbl"><em>dN</em><strong>dt</strong></span> = rN(1−N/k)</p>
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<p class="center"><i><b>The Verhulst Equation</b></i></p>
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<p>Can this be solved?</p>
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<p>Yes, with the help of one trick ...</p>
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<p> </p>
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<p><strong>Step 1</strong> Separate the variables:</p>
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<div class="tbl">
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<div class="row"><span class="left">Multiply both sides by dt:</span><span class="right"> dN = rN(1−N/k) dt</span></div>
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<div class="row"><span class="left">Divide both sides by <span class="larger">N(1-N/k)</span>:</span><span class="right"><span class="intbl"><em>1</em><strong>N(1−N/k)</strong></span>dN = r dt</span></div>
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</div>
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<p> </p>
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<p><strong>Step 2 </strong>Integrate:</p>
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<p class="center larger"><span class="integral">∫</span><span class="intbl"><em>1</em><strong>N(1−N/k)</strong></span>dN = <span class="integral">∫</span> r dt</p>
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<p>Hmmm... the left side looks hard to integrate. In fact it can be done with a little trick<span class="left"> from <a href="../algebra/partial-fractions.html">Partial Fractions</a></span> ... we rearrange it like this:</p>
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<div class="tbl">
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<div class="row"><span class="left">We start with this:</span><span class="right"><span class="intbl"><em>1</em><strong>N(1−N/k)</strong></span></span></div>
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<div class="row"><span class="left">Multiply top and bottom by k:</span><span class="right"><span class="intbl"><em>k</em><strong>N(k−N)</strong></span></span></div>
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<div class="row"><span class="left">Now here is the trick, add <b>N</b> and <b>−N</b> to the top:</span><span class="right"><span class="intbl"><em>N+k−N</em><strong>N(k−N)</strong></span></span></div>
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<div class="row"><span class="left">and split it into two fractions<a href="../algebra/partial-fractions.html"></a>:</span><span class="right"><span class="intbl"><em>N</em><strong>N(k−N)</strong></span> + <span class="intbl"><em>k−N</em><strong>N(k−N)</strong></span></span></div>
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<div class="row"><span class="left">Simplify each fraction:</span><span class="right"><span class="intbl"><em>1</em><strong>k−N</strong></span> + <span class="intbl"><em>1</em><strong>N</strong></span></span></div>
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</div>
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<p>Now it is a lot easier to solve. We can integrate each term separately, like this:</p>
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<div class="tbl">
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<div class="row"><span class="left">Our full equation is now:</span><span class="right"><span class="integral">∫</span><span class="intbl"><em>1</em><strong>k−N</strong></span>dN + <span class="integral">∫</span><span class="intbl"><em>1</em><strong>N</strong></span>dN = <span class="integral">∫</span> r dt</span></div>
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<div class="row"><span class="left">Integrate:</span><span class="right">−ln(k−N) + ln(N) = rt + C</span></div>
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</div>
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<p><i>(Why did that become <b>minus</b> ln(k−N)? Because we are integrating with respect to N.)</i></p>
|
||
<p> </p>
|
||
<p><strong>Step 3 </strong> Simplify:</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Negative of all terms:</span><span class="right">ln(k−N) − ln(N) = −rt − C</span></div>
|
||
<div class="row"><span class="left">Combine ln():</span><span class="right">ln((k−N)/N) = −rt − C</span></div>
|
||
<div class="row"><span class="left">Now <a href="../algebra/exponents-logarithms.html">take exponents</a> on both sides:</span><span class="right">(k−N)/N = e<sup>−rt−C</sup></span></div>
|
||
<div class="row"><span class="left"> Separate the powers of e:</span><span class="right"><span class="larger">(k−N)/N = e<sup>−rt</sup> e<sup>−C</sup></span></span></div>
|
||
<div class="row"><span class="left"><b>e<sup>−C</sup></b> is a constant, we can replace it with <b>A:</b></span><span class="right"><span class="larger">(k−N)/N = Ae<sup>−rt</sup></span></span></div>
|
||
</div>
|
||
<p> </p>
|
||
<p>We are getting close! Just a little more algebra to get N on its own:</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Separate the fraction terms:</span><span class="right">(k/N)−1 = Ae<sup>−rt</sup></span></div>
|
||
<div class="row"><span class="left">Add 1 to both sides:</span><span class="right">k/N = 1 + Ae<sup>−rt</sup></span></div>
|
||
<div class="row"><span class="left">Divide both by k:</span><span class="right">1/N = (1 + Ae<sup>−rt</sup>)/k</span></div>
|
||
<div class="row"><span class="left">Reciprocal of both sides:</span><span class="right">N = k/(1 + Ae<sup>−rt</sup>)</span></div>
|
||
</div>
|
||
|
||
<p>And we have our solution:</p>
|
||
<p class="center large">N = <span class="intbl"><em>k</em><strong>1 + Ae<sup>−rt</sup></strong></span></p>
|
||
<p> </p>
|
||
<p class="center">Here is an <b>example</b>, the graph of <span class="large"><span class="intbl"><em>40</em><strong>1 + 5e<sup>−2t</sup></strong></span></span></p>
|
||
<p class="center"><img src="images/verhulst.gif" alt="verhulst" height="216" width="297"><br>
|
||
It starts rising exponentially,<br>
|
||
then flattens out as it reaches k=40</p>
|
||
</div>
|
||
<p> </p>
|
||
|
||
<div class="questions">
|
||
<script>getQ(9409, 9410, 9411, 9412, 9413, 9414, 9415, 9416, 9417, 9418);</script>
|
||
</div>
|
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|
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<div class="related">
|
||
<a href="differential-equations.html">Differential Equation</a>
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<a href="index.html">Calculus Index</a>
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