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665 lines
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<h1 class="center">First Order Linear Differential Equations</h1>
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<p class="center">You might like to read about <a href="differential-equations.html">Differential Equations</a><br>
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and <a href="separation-variables.html">Separation of Variables</a> first!</p>
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<div class="def">
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<p>A Differential Equation 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-1.svg" alt="y + dy/dx = 5x" height="123" width="188"><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">
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<em>dy</em>
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<strong>dx</strong>
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</span></b> </p>
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</div>
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<p>Here we will look at solving a special class of Differential Equations called <b>First Order Linear Differential Equations</b></p>
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<h2>First Order</h2>
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<p>They are "First Order" when there is only <b> <span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span></b> , not <b> <span class="intbl">
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<em>d<sup>2</sup>y</em>
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<strong>dx<sup>2</sup></strong>
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</span></b> or <b> <span class="intbl">
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<em>d<sup>3</sup>y</em>
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<strong>dx<sup>3</sup></strong>
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</span></b> etc</p>
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<h2>Linear</h2>
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<p>A <b>first order differential equation</b> is <b>linear</b> when it can be made to look like this:</p>
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<p class="center large"><span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> + P(x)y = Q(x)</p>
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<p>Where <b>P(x)</b> and <b>Q(x)</b> are functions of x.</p>
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<p>To solve it there is a special method:</p>
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<ul>
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<li>We invent two new functions of x, call them <b>u</b> and <b>v</b>, and say that <b>y=uv</b>.</li>
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<li>We then solve to find <b>u</b>, and then find <b>v</b>, and tidy up and we are done!</li>
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</ul>
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<p>And we also use the derivative of <b>y=uv</b> (see <a href="derivatives-rules.html"> <span class="center">Derivative Rules</span> (Product Rule) </a>):</p>
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<p class="center large"><span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> = u<span class="intbl">
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<em>dv</em>
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<strong>dx</strong>
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</span> + v<span class="intbl">
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<em>du</em>
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<strong>dx</strong>
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</span></p>
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<h2>Steps</h2>
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<p>Here is a step-by-step method for solving them:</p>
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<ul>
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<li>1. Substitute <b>y = uv</b>, and
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<p class="center large"><span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> = u<span class="intbl">
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<em>dv</em>
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<strong>dx</strong>
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</span> + v<span class="intbl">
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<em>du</em>
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<strong>dx</strong>
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</span></p>
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into
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<p class="center large"><span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> + P(x)y = Q(x)</p></li>
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<li>2. Factor the parts involving <b>v</b></li>
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<li>3. Put the <b>v</b> term equal to zero (this gives a differential equation in <b>u</b> and <b>x</b> which can be solved in the next step)</li>
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<li>4. Solve using <a href="separation-variables.html">separation of variables</a> to find <b>u</b></li>
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<li>5. Substitute <b>u</b> back into the equation we got at step 2</li>
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<li>6. Solve that to find <b>v</b></li>
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<li>7. Finally, substitute <b>u</b> and <b>v</b> into <b>y = uv</b> to get our solution!</li>
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</ul>
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<p>Let's try an example to see:</p>
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<div class="example">
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<h3>Example 1: Solve this:
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<p class="center larger"> <span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> − <span class="intbl">
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<em>y</em>
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<strong>x</strong>
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</span> = 1</p></h3>
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<p>First, is this linear? Yes, as it is in the form</p>
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<p class="center"><span class="larger"><span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> + P(x)y = Q(x)</span><br>
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where <b>P(x) = −<span class="intbl">
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<em>1</em>
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<strong>x</strong>
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</span></b> and <b>Q(x) = 1</b></p>
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<p>So let's follow the steps:</p>
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<p>Step 1:
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Substitute <b>y = uv</b>, and <span class="center"> <b> <span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> = u <span class="intbl">
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<em>dv</em>
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<strong>dx</strong>
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</span> + v <span class="intbl">
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<em>du</em>
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<strong>dx</strong>
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</span></b> </span></p>
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<div class="tbl">
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<div class="row"><span class="left">So this:</span><span class="right"><span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> − <span class="intbl">
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<em>y</em>
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<strong>x</strong>
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</span> = 1</span></div>
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<div class="row"><span class="left">Becomes this:</span><span class="right">u<span class="intbl">
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<em>dv</em>
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<strong>dx</strong>
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</span> + v<span class="intbl">
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<em>du</em>
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<strong>dx</strong>
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</span> − <span class="intbl">
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<em>uv</em>
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<strong>x</strong>
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</span> = 1</span></div>
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</div>
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<p>Step 2: Factor the parts involving <b>v</b></p>
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<div class="tbl">
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<div class="row"><span class="left">Factor <b>v</b>:</span><span class="right">u <span class="intbl">
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<em>dv</em>
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<strong>dx</strong>
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</span> + v( <span class="intbl">
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<em>du</em>
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<strong>dx</strong>
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</span> − <span class="intbl">
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<em>u</em>
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<strong>x</strong>
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</span> ) = 1</span></div>
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</div>
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<p>Step 3: Put the <b>v</b> term equal to zero</p>
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<div class="tbl">
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<div class="row"><span class="left"><b>v</b> term equal to zero:</span><span class="right"> <span class="intbl">
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<em>du</em>
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<strong>dx</strong>
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</span> − <span class="intbl">
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<em>u</em>
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<strong>x</strong>
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</span> = 0</span></div>
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<div class="row"><span class="left">So:</span><span class="right"> <span class="intbl">
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<em>du</em>
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<strong>dx</strong>
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</span> = <span class="intbl">
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<em>u</em>
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<strong>x</strong>
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</span></span></div>
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</div>
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<p>Step 4: Solve using <a href="separation-variables.html">separation of variables</a> to find <b>u</b></p>
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<div class="tbl">
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<div class="row"><span class="left">Separate variables:</span><span class="right"> <span class="intbl">
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<em>du</em>
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<strong>u</strong>
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</span> = <span class="intbl">
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<em>dx</em>
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<strong>x</strong>
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</span> </span></div>
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<div class="row"><span class="left">Put integral sign:</span><span class="right"> <span class="integral">∫</span><span class="intbl">
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<em>du</em>
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<strong>u</strong>
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</span> = <span class="integral">∫</span><span class="intbl">
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<em>dx</em>
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<strong>x</strong>
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</span> </span></div>
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<div class="row"><span class="left">Integrate:</span><span class="right">ln(u) = ln(x) + C</span></div>
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<div class="row"><span class="left">Make C = ln(k):</span><span class="right">ln(u) = ln(x) + ln(k)</span></div>
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<div class="row"><span class="left">And so:</span><span class="right">u = kx</span></div>
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</div>
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<p>Step 5: Substitute <b>u</b> back into the equation at Step 2</p>
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<div class="tbl">
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<div class="row"><span class="left">(Remember <b>v</b> term equals 0 so can be ignored):</span><span class="right">kx <span class="intbl">
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<em>dv</em>
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<strong>dx</strong>
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</span> = 1</span></div>
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</div>
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<p>Step 6: Solve this to find <b>v</b></p>
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<div class="tbl">
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<div class="row"><span class="left">Separate variables:</span><span class="right">k dv = <span class="intbl">
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<em>dx</em>
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<strong>x</strong>
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</span> </span></div>
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<div class="row"><span class="left">Put integral sign:</span><span class="right"> <span class="integral">∫</span>k dv
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<strong></strong>
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= <span class="integral">∫</span><span class="intbl">
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<em>dx</em>
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<strong>x</strong>
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</span> </span></div>
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<div class="row"><span class="left">Integrate:</span><span class="right">kv = ln(x) + C</span></div>
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<div class="row"><span class="left">Make C = ln(c):</span><span class="right">kv = ln(x) + ln(c)</span></div>
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<div class="row"><span class="left">And so:</span><span class="right">kv = ln(cx)</span></div>
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<div class="row"><span class="left">And so:</span><span class="right">v = <span class="intbl">
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<em>1</em>
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<strong>k</strong>
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</span> ln(cx)</span></div>
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</div>
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<p>Step 7: Substitute into <b>y = uv</b> to find the solution to the original equation.</p>
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<div class="tbl">
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<div class="row"><span class="left">y = uv:</span><span class="right">y = kx <span class="intbl">
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<em>1</em>
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<strong>k</strong>
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</span> ln(cx)</span></div>
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<div class="row"><span class="left">Simplify:</span><span class="right">y = x ln(cx)</span></div>
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</div>
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<p>And it produces this nice family of curves:</p>
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<p class="center"><img src="images/diff-eq-lin-b.svg" alt="differential equation at 0.2, 0.4, 0.6, 0.8 and 1.0" height="207" width="178"> <span class="larger"><br>
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y = x ln(cx)</span> for various values of <b>c</b></p>
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</div>
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<p>What is the meaning of those curves?</p>
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<p class="center">They are the solution to the equation <span class="center"> <b> <span class="intbl">
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<em>dy</em>
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<strong>dx</strong>
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</span> − <span class="intbl">
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<em>y</em>
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<strong>x</strong>
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</span> = 1</b></span></p>
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<p>In other words:</p>
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<p class="center"><b>Anywhere on any of those curves<br>
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the slope minus <span class="intbl">
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<em>y</em>
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<strong>x</strong>
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</span> equals 1</b></p>
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<p>Let's check a few points on the <b>c=0.6</b> curve:</p>
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<p class="center"><img src="images/diff-eq-lin-c.svg" alt="differential equation graph and points" height="200" width="227"></p>
|
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<div class="simple">
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<p>Estmating off the graph (to 1 decimal place):</p>
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<table style="border: 0; margin:auto;">
|
||
<tbody>
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<tr style="text-align:center;">
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<th>Point</th>
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<th>x</th>
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<th>y</th>
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<th>Slope (<span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
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</span>)</th>
|
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<th><span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> − <span class="intbl">
|
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<em>y</em>
|
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<strong>x</strong>
|
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</span> </th>
|
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</tr>
|
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<tr style="text-align:center;">
|
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<th>A</th>
|
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<td>0.6</td>
|
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<td>−0.6</td>
|
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<td>0</td>
|
||
<td>0 − <span class="intbl">
|
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<em>−0.6</em>
|
||
<strong>0.6</strong>
|
||
</span> = 0 + 1 = <b>1</b></td>
|
||
</tr>
|
||
<tr style="text-align:center;">
|
||
<th>B</th>
|
||
<td>1.6</td>
|
||
<td>0</td>
|
||
<td>1</td>
|
||
<td>1 − <span class="intbl">
|
||
<em>0</em>
|
||
<strong>1.6</strong>
|
||
</span> = 1 − 0 = <b>1</b></td>
|
||
</tr>
|
||
<tr style="text-align:center;">
|
||
<th>C</th>
|
||
<td>2.5</td>
|
||
<td>1</td>
|
||
<td> 1.4</td>
|
||
<td>1.4 − <span class="intbl">
|
||
<em>1</em>
|
||
<strong>2.5</strong>
|
||
</span> = 1.4 − 0.4 = <b>1</b></td>
|
||
</tr>
|
||
</tbody></table>
|
||
</div>
|
||
<p>Why not test a few points yourself? You can <a href="../data/function-grapher6e0c.html?func1=x*ln%280.6*x%29&xmin=-3.840&xmax=3.480&ymin=-2.380&ymax=2.500&aval=0.2480&uni=1">plot the curve here</a>.</p>
|
||
<p> </p>
|
||
<p>Perhaps another example to help you? Maybe a little harder?</p>
|
||
<div class="example">
|
||
|
||
<h3>Example 2: Solve this:
|
||
|
||
|
||
<p class="center larger"> <span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> − <span class="intbl">
|
||
<em>3y</em>
|
||
<strong>x</strong>
|
||
</span> = x</p></h3>
|
||
<p>First, is this linear? Yes, as it is in the form</p>
|
||
<p class="center"><span class="larger"><span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> + P(x)y = Q(x)</span><br>
|
||
where <b>P(x) = − <span class="intbl">
|
||
<em>3</em>
|
||
<strong>x</strong>
|
||
</span></b> and <b>Q(x) = x</b></p>
|
||
<p>So let's follow the steps:</p>
|
||
<p>Step 1:
|
||
Substitute <b>y = uv</b>, and <span class="center"> <b> <span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> = u <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> + v <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span></b> </span></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">So this:</span><span class="right"><span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> − <span class="intbl">
|
||
<em>3y</em>
|
||
<strong>x</strong>
|
||
</span> = x</span></div>
|
||
<div class="row"><span class="left">Becomes this:</span><span class="right"> u <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> + v <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span> − <span class="intbl">
|
||
<em>3uv</em>
|
||
<strong>x</strong>
|
||
</span> = x</span></div>
|
||
</div>
|
||
<p>Step 2: Factor the parts involving <b>v</b></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Factor <b>v</b>:</span><span class="right">u <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> + v( <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span> − <span class="intbl">
|
||
<em>3u</em>
|
||
<strong>x</strong>
|
||
</span> ) = x</span></div>
|
||
</div>
|
||
<p>Step 3: Put the <b>v</b> term equal to zero</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left"><b>v</b> term = zero:</span><span class="right"> <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span> − <span class="intbl">
|
||
<em>3u</em>
|
||
<strong>x</strong>
|
||
</span> = 0</span></div>
|
||
<div class="row"><span class="left">So:</span><span class="right"> <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span> = <span class="intbl">
|
||
<em>3u</em>
|
||
<strong>x</strong>
|
||
</span> </span></div>
|
||
</div>
|
||
<p>Step 4: Solve using <a href="separation-variables.html">separation of variables</a> to find <b>u</b></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Separate variables:</span><span class="right"> <span class="intbl">
|
||
<em>du</em>
|
||
<strong>u</strong>
|
||
</span> = 3 <span class="intbl">
|
||
<em>dx</em>
|
||
<strong>x</strong>
|
||
</span> </span></div>
|
||
<div class="row"><span class="left">Put integral sign:</span><span class="right"> <span class="integral">∫</span><span class="intbl">
|
||
<em>du</em>
|
||
<strong>u</strong>
|
||
</span> = 3 <span class="integral">∫</span><span class="intbl">
|
||
<em>dx</em>
|
||
<strong>x</strong>
|
||
</span> </span></div>
|
||
<div class="row"><span class="left">Integrate:</span><span class="right">ln(u) = 3 ln(x) + C</span></div>
|
||
<div class="row"><span class="left">Make C = −ln(k):</span><span class="right">ln(u) + ln(k) = 3ln(x)</span></div>
|
||
<div class="row"><span class="left">Then:</span><span class="right">uk = x<sup>3</sup></span></div>
|
||
<div class="row"><span class="left">And so:</span><span class="right">u = <span class="intbl">
|
||
<em>x<sup>3</sup></em>
|
||
<strong>k</strong>
|
||
</span></span></div>
|
||
</div>
|
||
<p>Step 5: Substitute <b>u</b> back into the equation at Step 2</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">(Remember <b>v</b> term equals 0 so can be ignored):</span><span class="right">( <span class="intbl">
|
||
<em>x<sup>3</sup></em>
|
||
<strong>k</strong>
|
||
</span> ) <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> = x </span></div>
|
||
</div>
|
||
<p>Step 6: Solve this to find <b>v</b></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Separate variables:</span><span class="right">dv = k x<sup>-2</sup> dx</span></div>
|
||
<div class="row"><span class="left">Put integral sign:</span><span class="right"> <span class="integral">∫</span>dv = <span class="integral">∫</span>k x<sup>-2</sup> dx</span></div>
|
||
<div class="row"><span class="left">Integrate:</span><span class="right">v = −k x<sup>-1</sup> + D</span></div>
|
||
</div>
|
||
<p>Step 7: Substitute into <b>y = uv</b> to find the solution to the original equation.</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">y = uv:</span><span class="right">y = <span class="intbl">
|
||
<em>x<sup>3</sup></em>
|
||
<strong>k</strong>
|
||
</span> ( −k x<sup>-1</sup> + D )</span></div>
|
||
<div class="row"><span class="left">Simplify:</span><span class="right">y = −x<sup>2</sup> + <span class="intbl">
|
||
<em>D</em>
|
||
<strong>k</strong>
|
||
</span> x<sup>3</sup></span></div>
|
||
<div class="row"><span class="left">Replace <b>D/k</b> with a single constant <b>c</b>: </span><span class="right">y =
|
||
c
|
||
x<sup>3 </sup>− x<sup>2</sup></span></div>
|
||
</div>
|
||
<p>And it produces this nice family of curves:</p>
|
||
<p class="center"><img src="images/diff-eq-lin-a.svg" alt="differential equation at 0.2, 0.4, 0.6 and 0.8" height="211" width="233"> <span class="larger"><br>
|
||
y = c
|
||
x<sup>3 </sup>− x<sup>2</sup></span> for various values of <b>c</b></p>
|
||
</div>
|
||
<p>And one more example, this time even <b> harder</b>:</p>
|
||
<div class="example">
|
||
|
||
<h3>Example 3: Solve this:</h3>
|
||
<p class="center larger"> <span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> + 2xy= −2x<sup>3</sup></p>
|
||
<p>First, is this linear? Yes, as it is in the form</p>
|
||
<p class="center"><span class="larger"><span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> + P(x)y = Q(x)</span><br>
|
||
where <b>P(x) = 2x</b> and <b>Q(x) = −2x<sup>3</sup></b></p>
|
||
<p>So let's follow the steps:</p>
|
||
<p>Step 1:
|
||
Substitute <b>y = uv</b>, and <span class="center"> <b> <span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> = u <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> + v <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span></b> </span></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">So this:</span><span class="right"><span class="intbl">
|
||
<em>dy</em>
|
||
<strong>dx</strong>
|
||
</span> + 2xy= −2x<sup>3</sup></span></div>
|
||
<div class="row"><span class="left">Becomes this:</span><span class="right"> u <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> + v <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span> + 2xuv
|
||
= −2x<sup>3</sup> </span></div>
|
||
</div>
|
||
<p>Step 2: Factor the parts involving <b>v</b></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Factor <b>v</b>:</span><span class="right">u <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> + v( <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span> + 2xu
|
||
) = −2x<sup>3</sup> </span></div>
|
||
</div>
|
||
<p>Step 3: Put the <b>v</b> term equal to zero</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left"><b>v</b> term = zero:</span><span class="right"> <span class="intbl">
|
||
<em>du</em>
|
||
<strong>dx</strong>
|
||
</span> + 2xu = 0</span></div>
|
||
</div>
|
||
<p>Step 4: Solve using <a href="separation-variables.html">separation of variables</a> to find <b>u</b></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Separate variables:</span><span class="right"> <span class="intbl">
|
||
<em>du</em>
|
||
<strong>u</strong>
|
||
</span> = −2x dx</span></div>
|
||
<div class="row"><span class="left">Put integral sign:</span><span class="right"> <span class="integral">∫</span><span class="intbl">
|
||
<em>du</em>
|
||
<strong>u</strong>
|
||
</span> = −2<span class="integral">∫</span>x dx</span></div>
|
||
<div class="row"><span class="left">Integrate:</span><span class="right">ln(u) = −x<sup>2</sup> + C</span></div>
|
||
<div class="row"><span class="left">Make C = −ln(k):</span><span class="right">ln(u) + ln(k) = −x<sup>2</sup></span></div>
|
||
<div class="row"><span class="left">Then:</span><span class="right">uk = e<sup>-x<sup>2</sup></sup></span></div>
|
||
<div class="row"><span class="left">And so:</span><span class="right">u = <span class="intbl">
|
||
<em>e<sup>-x<sup>2</sup></sup></em>
|
||
<strong>k</strong>
|
||
</span></span></div>
|
||
</div>
|
||
<p>Step 5: Substitute <b>u</b> back into the equation at Step 2</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">(Remember <b>v</b> term equals 0 so can be ignored):</span><span class="right">( <span class="intbl">
|
||
<em>e<sup>-x<sup>2</sup></sup></em>
|
||
<strong>k</strong>
|
||
</span> ) <span class="intbl">
|
||
<em>dv</em>
|
||
<strong>dx</strong>
|
||
</span> = −2x<sup>3</sup> </span></div>
|
||
</div>
|
||
<p>Step 6: Solve this to find <b>v</b></p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Separate variables:</span><span class="right">dv = −2k x<sup>3</sup> e<sup>x<sup>2</sup></sup> dx</span></div>
|
||
<div class="row"><span class="left">Put integral sign:</span><span class="right"> <span class="integral">∫</span>dv
|
||
= <span class="integral">∫</span>−2k x<sup>3</sup> e<sup>x<sup>2</sup></sup> dx</span></div>
|
||
<div class="row"><span class="left">Integrate:</span><span class="right">v = oh no! this is hard!</span></div>
|
||
</div>
|
||
<p>Let's see ... we can <a href="integration-by-parts.html">integrate by parts</a>... which says:</p>
|
||
<p class="center larger"><span class="integral">∫</span>RS dx = R<span class="integral">∫</span>S dx − <span class="integral">∫</span>R' ( <span class="integral">∫</span>S dx ) dx</p>
|
||
<p><i>(Side Note: we use R and S here, using u and v could be confusing as they already mean something else.)</i></p>
|
||
<p>Choosing R and S is very important, this is the best choice we found:</p>
|
||
<ul>
|
||
<li><span class="larger">R = −x<sup>2</sup></span> and</li>
|
||
<li><span class="larger">S = 2x e<sup>x<sup>2</sup></sup></span></li>
|
||
</ul>
|
||
<p>So let's go:</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">First pull out k:</span><span class="right">v
|
||
= k<span class="integral">∫</span>−2x<sup>3</sup> e<sup>x<sup>2</sup></sup> dx</span></div>
|
||
<div class="row"><span class="left"><b>R = −x<sup>2</sup></b> and <b>S = 2x e<sup>x<sup>2</sup></sup></b>:</span><span class="right">v
|
||
= k<span class="integral">∫</span>(−x<sup>2</sup>)(2xe<sup>x<sup>2</sup></sup>) dx</span></div>
|
||
<div class="row"><span class="left">Now integrate by parts:</span><span class="right">v
|
||
= kR<span class="integral">∫</span>S dx − k<span class="integral">∫</span>R' ( <span style="font−size:150%;">∫</span> S dx) dx</span></div>
|
||
<div class="row"><span class="left">
|
||
<p>Put in R = −x<sup>2</sup> and S = 2x e<sup>x<sup>2</sup></sup></p>
|
||
<p>And also R' = −2x and <span class="integral">∫</span> S dx = e<sup>x<sup>2</sup></sup></p>
|
||
</span></div>
|
||
<div class="row"><span class="left">So it becomes:</span><span class="right">v
|
||
= −kx<sup>2</sup><span class="integral">∫</span>2x e<sup>x<sup>2</sup></sup> dx − k<span class="integral">∫</span>−2x (e<sup>x<sup>2</sup></sup>) dx</span></div>
|
||
<div class="row"><span class="left">Now Integrate:</span><span class="right">v
|
||
= −kx<sup>2</sup> e<sup>x<sup>2</sup></sup> + k e<sup>x<sup>2</sup></sup> + D</span></div>
|
||
<div class="row"><span class="left">Simplify:</span><span class="right">v
|
||
= ke<sup>x<sup>2</sup></sup> (1−x<sup>2</sup>) + D</span></div>
|
||
</div>
|
||
<p>Step 7: Substitute into <b>y = uv</b> to find the solution to the original equation.</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">y = uv:</span><span class="right">y = <span class="intbl">
|
||
<em>e<sup>-x<sup>2</sup></sup></em>
|
||
<strong>k</strong>
|
||
</span> ( ke<sup>x<sup>2</sup></sup> (1−x<sup>2</sup>) + D )</span></div>
|
||
<div class="row"><span class="left">Simplify:</span><span class="right">y =1 − x<sup>2</sup> + ( <span class="intbl">
|
||
<em>D</em>
|
||
<strong>k</strong>
|
||
</span>)e<sup>-</sup><sup>x<sup>2</sup></sup></span></div>
|
||
<div class="row"><span class="left">Replace <b>D/k</b> with a single constant <b>c</b>: </span><span class="right">y = 1 − x<sup>2</sup> +
|
||
c
|
||
e<sup>-</sup><sup>x<sup>2</sup></sup></span></div>
|
||
</div>
|
||
<p>And we get this nice family of curves:</p>
|
||
<p class="center"><img src="images/diff-eq-lin-d.svg" alt="differential equation" height="340" width="550"> <span class="larger"><br>
|
||
<span class="right">y = 1 − x<sup>2</sup> +
|
||
c
|
||
e<sup>-</sup><sup>x<sup>2</sup></sup></span> </span> for various values of <b>c</b></p>
|
||
</div>
|
||
<p> </p>
|
||
<div class="questions">9429, 9430, 9431, 9432, 9433, 9434, 9435, 9436, 9437, 9438</div>
|
||
|
||
<div class="related">
|
||
<a href="differential-equations.html">Differential Equations</a>
|
||
<a href="separation-variables.html">Separation of Variables</a>
|
||
<a href="index.html">Calculus Index</a>
|
||
</div>
|
||
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