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<h1>Second Order Differential Equations</h1>
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<div class="def">
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<p>Here we learn how to solve equations of this type:</p>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + p<span class="intbl"><em>dy</em><strong>dx</strong></span> + qy = 0</p>
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</div>
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<h2><span class="center">Differential Equation</span></h2>
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<p>A <span class="center">Differential Equation is a</span>n 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="differential equation 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"> <em>dy</em> <strong>dx</strong> </span></b> </p>
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<h2>Order</h2>
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<p>The Order is the <b>highest derivative</b> (is it a first derivative? a <a href="second-derivative.html">second derivative</a>? etc):</p>
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<div class="example">
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<h3>Example:</h3>
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<p class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> + y<sup>2</sup> = 5x</p>
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<p>It has only the first derivative <span class="intbl"> <em>dy</em> <strong>dx</strong> </span> , so is "First Order"</p>
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</div>
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<div class="example">
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<h3>Example:</h3>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + xy = sin(x)</p>
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<p>This has a second derivative <span class="intbl"> <em>d<sup>2</sup>y</em> <strong>dx<sup>2</sup></strong> </span> , so is "Second Order" or "Order 2"</p>
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</div>
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<div class="example">
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<h3>Example:</h3>
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<p class="center large"><span class="intbl"><em>d<sup>3</sup>y</em><strong>dx<sup>3</sup></strong></span> + x<span class="intbl"><em>dy</em><strong>dx</strong></span> + y = e<sup>x</sup></p>
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<p>This has a third derivative <span class="intbl"> <em>d<sup>3</sup>y</em> <strong>dx<sup>3</sup></strong> </span> which outranks the <span class="intbl"> <em>dy</em> <strong>dx</strong> </span> , so is "Third Order" or "Order 3"</p>
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</div>
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<h3>Before tackling second order differential equations, make sure you are familiar with the various methods for <a href="differential-equations-solution-guide.html">solving first order differential equations</a>.</h3>
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<h2>Second Order Differential Equations</h2>
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<div class="def">
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<p>We can solve a second order differential equation of the type:</p>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + P(x)<span class="intbl"><em>dy</em><strong>dx</strong></span> + Q(x)y
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= f(x)</p>
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<p>where P(x), Q(x) and f(x) are functions of x, by using:</p>
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<p class="dotpoint"><a href="differential-equations-undetermined-coefficients.html">Undetermined Coefficients</a> which only works when f(x) is a polynomial, exponential, sine, cosine or a linear combination of those.</p>
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<p class="dotpoint"><a href="differential-equations-variation-parameters.html">Variation of Parameters</a> which is a little messier but works on a wider range of functions.</p>
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</div>
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<p>But here we begin by learning the case where <b>f(x) = 0</b> (this makes it "homogeneous"):</p>
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<p class="center larger"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + P(x)<span class="intbl"><em>dy</em><strong>dx</strong></span> + Q(x)y
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= 0</p>
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<p>and also where the functions P(X) and Q(x) are constants <b>p</b> and <b>q</b>:</p>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + p<span class="intbl"><em>dy</em><strong>dx</strong></span> + qy = 0</p>
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<p>Let's learn to solve them!</p>
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<p> </p>
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<h3><b><i>e</i></b> to the rescue</h3>
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<p>We are going to use a special property of the <a href="derivatives-rules.html">derivative</a> of the <a href="../sets/function-exponential.html">exponential function</a>:</p>
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<p>At any point the slope (derivative) of <i><b>e</b></i><sup>x</sup> equals the value of <i><b>e</b></i><sup>x</sup> :</p>
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<p class="center"><img src="../sets/images/function-exponential-slopes.svg" alt="natural exponential function" height="161" width="263" ></p>
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<p>And when we introduce a value "r" like this:</p>
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<p class="center large">f(x) = e<sup>rx</sup></p>
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<p>We find:</p>
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<ul>
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<li>the first derivative is f'(x) = re<sup>rx</sup></li>
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<li>the second derivative is f''(x) = r<sup>2</sup>e<sup>rx</sup></li>
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</ul>
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<p>In other words, the first and second derivatives of f(x) are both <b>multiples</b> of f(x)</p>
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<p>This is going to help us a lot!</p>
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<div class="example">
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<h3>Example 1: Solve</h3>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + <span class="intbl"><em>dy</em><strong>dx</strong></span> − 6y = 0</p>
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<p>Let y = e<sup>rx</sup> so we get:</p>
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<ul>
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<li><span class="intbl"><em>dy</em><strong>dx</strong></span> = re<sup>rx</sup></li>
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<li><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = r<sup>2</sup>e<sup>rx</sup></li>
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</ul>
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<p>Substitute these into the equation above:</p>
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<p class="center large">r<sup>2</sup>e<sup>rx</sup> + re<sup>rx</sup> − 6e<sup>rx</sup> = 0</p>
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<p>Simplify:</p>
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<p class="center large">e<sup>rx</sup>(r<sup>2</sup> + r − 6) = 0</p>
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<p class="center large">r<sup>2</sup> + r − 6 = 0</p>
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<p>We have reduced the differential equation to an ordinary <a href="../algebra/quadratic-equation.html">quadratic equation</a>!</p>
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<p>This quadratic equation is given the special name of <b>characteristic equation</b>.</p>
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<p>We can factor this one to:</p>
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<p class="center large">(r − 2)(r + 3) = 0</p>
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<p>So <b>r = 2 or −3</b></p>
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<p>And so we have two solutions:</p>
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<p class="center large">y = e<sup>2x</sup></p>
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<p class="center large">y = e<sup>−3x</sup></p>
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<p>But that’s not the final answer because we can combine different <b>multiples</b> of these two answers to get a more general solution:</p>
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<p class="center large">y = Ae<sup>2x</sup> + Be<sup>−3x</sup></p>
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<h3>Check</h3>
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<p>Let us check that answer. First take derivatives:</p>
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<p class="center"><span class="center large">y = Ae<sup>2x</sup> + Be<sup>−3x</sup></span></p>
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<p class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> = 2Ae<sup>2x</sup> − 3Be<sup>−3x</sup></p>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 4Ae<sup>2x</sup> + 9Be<sup>−3x</sup></p>
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<p>Now substitute into the original equation:</p>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + <span class="intbl"><em>dy</em><strong>dx</strong></span> − 6y = 0</p>
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<p class="center large">(4Ae<sup>2x</sup> + 9Be<sup>−3x</sup>) + (2Ae<sup>2x</sup> − 3Be<sup>−3x</sup>) − 6(Ae<sup>2x</sup> + Be<sup>−3x</sup>) = 0</p>
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<p class="center large">4Ae<sup>2x</sup> + 9Be<sup>−3x</sup> + 2Ae<sup>2x</sup> − 3Be<sup>−3x</sup> − 6Ae<sup>2x</sup> − 6Be<sup>−3x</sup> = 0</p>
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<p class="center large">4Ae<sup>2x</sup> + 2Ae<sup>2x</sup> − 6Ae<sup>2x</sup>+ 9Be<sup>−3x</sup>− 3Be<sup>−3x</sup> − 6Be<sup>−3x</sup> = 0</p>
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<p class="center large">0 = 0</p>
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<p>It worked!</p>
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</div>
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<h2>So, does this method work generally?</h2>
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<p>Well, yes and no. The answer to this question depends on the constants <b>p</b> and <b>q</b>.</p>
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<p>With <b>y = e<sup>rx</sup></b> as a solution of the differential equation:</p>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + p<span class="intbl"><em>dy</em><strong>dx</strong></span> + qy = 0</p>
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<p>we get:</p>
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<p class="center large">r<sup>2</sup>e<sup>rx</sup> + pre<sup>rx</sup> + qe<sup>rx</sup> = 0</p>
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<p class="center large">e<sup>rx</sup>(r<sup>2</sup> + pr + q) = 0</p>
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<p class="center large">r<sup>2</sup> + pr + q = 0</p>
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<p>This is a <a href="../algebra/quadratic-equation.html">quadratic equation</a>, and there can be three types of answer:</p>
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<ul>
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<li>two real roots</li>
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<li>one real root (i.e. both real roots are the same)</li>
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<li>two complex roots</li>
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</ul>
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<p>How we solve it depends which type!</p>
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<p>We can easily find which type by calculating the <a href="../algebra/quadratic-equation.html">discriminant</a> <b>p<sup>2</sup> − 4q</b>. When it is</p>
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<ul>
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<li>positive we get two real roots</li>
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<li>zero we get one real root</li>
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<li>negative we get two complex roots</li>
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</ul>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="../algebra/images/quadratic-graph.svg" alt="Quadratic Graph" height="140" width="175" ></p>
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<h2>Two Real Roots</h2>
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<p>When the discriminant <b>p<sup>2</sup> − 4q</b> is <b>positive</b> we can go straight from the differential equation</p>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + p<span class="intbl"><em>dy</em><strong>dx</strong></span> + qy = 0</p>
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<p>through the "characteristic equation":</p>
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<p class="center large">r<sup>2</sup> + pr + q = 0</p>
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<p>to the general solution with two real roots <b>r<sub>1</sub></b> and <b>r<sub>2</sub></b>:</p>
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<p class="center large">y = Ae<sup>r<sub>1</sub>x</sup> + Be<sup>r<sub>2</sub>x</sup></p>
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<div class="example">
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<h3><b>Example 2:</b> Solve</h3>
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<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 9<span class="intbl"><em>dy</em><strong>dx</strong></span> + 20y = 0</p>
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<p>The characteristic equation is:<br></p>
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<p class="center large">r<sup>2</sup> − 9r<strong> </strong>+ 20 = 0</p>
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<p>Factor:</p>
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<p class="center large">(r − 4)(r − 5) = 0</p>
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<p class="center large">r = 4 or 5</p>
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<p>So the general solution of our differential equation is:</p>
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<p class="center large">y = Ae<sup>4x</sup> + Be<sup>5x</sup></p>
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<p>And here are some sample values:</p>
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<p class="center"><img src="images/diff-eq-2nd-order-a.svg" alt="y = Ae^4x + Be^5x" height="340" width="550" ></p>
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</div>
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<div class="example">
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<h3><b>Example 3:</b> Solve</h3>
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<p class="center large">6<span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + 5<span class="intbl"><em>dy</em><strong>dx</strong></span> − 6y = 0</p>
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<p>The characteristic equation is:<br></p>
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<p class="center large">6r<sup>2</sup> + 5r<strong> </strong>− 6 = 0</p>
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<p>Factor:</p>
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<p class="center large">(3r − 2)(2r + 3) = 0</p>
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<p class="center large">r = <span class="intbl"><em>2</em><strong>3</strong></span> or <span class="intbl"><em>−3</em><strong>2</strong></span></p>
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<p>So the general solution of our differential equation is:</p>
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<p class="center large">y = Ae<sup>(<span class="intbl"><em>2</em><strong>3</strong></span>x)</sup> + Be<sup>(<span class="intbl"><em>−3</em><strong>2</strong></span>x)</sup></p>
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</div>
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<div class="example">
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<h3><b>Example 4:</b> Solve</h3>
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<p class="center large">9<span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 6<span class="intbl"><em>dy</em><strong>dx</strong></span> − y = 0</p>
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<p>The characteristic equation is:<br></p>
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<p class="center large">9r<sup>2</sup> − 6r<strong> </strong>− 1 = 0</p>
|
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<p>This does not factor easily, so we use the <a href="../algebra/quadratic-equation.html">quadratic equation formula</a>:</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−b ± √(b<sup>2 </sup>− 4ac)</em> <strong>2a</strong></span></p>
|
||
<p>with a = 9, b = −6 and c = −1</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−(−6) ± √((−6)<sup>2 </sup>− 4×9×(−1))</em> <strong>2×9</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>6 ± √(36+ 36)</em> <strong>18</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>6 ± 6√2</em> <strong>18</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>1 ± √2</em> <strong>3</strong></span></p>
|
||
<p>So the general solution of the differential equation is</p>
|
||
<p class="center large">y = Ae<sup>(<span class="intbl"><em>1 + √2</em> <strong>3</strong></span>)x</sup> + Be<sup>(<span class="intbl"><em>1 − √2</em> <strong>3</strong></span>)x</sup></p>
|
||
</div>
|
||
<p style="float:right; margin: 0 0 5px 10px;"><img src="../algebra/images/quadratic-graph-1root.svg" alt="Quadratic Graph" height="140" width="175" ></p>
|
||
|
||
|
||
<h2>One Real Root</h2>
|
||
|
||
<p>When the discriminant <b>p<sup>2</sup> − 4q</b> is <b>zero</b> we get one real root (i.e. both real roots are equal).</p>
|
||
<p>Here are some examples:</p>
|
||
<div class="example">
|
||
|
||
<h3><b>Example 5:</b> Solve</h3>
|
||
<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 10<span class="intbl"><em>dy</em><strong>dx</strong></span> + 25y = 0</p>
|
||
<p>The characteristic equation is:<br></p>
|
||
<p class="center large">r<sup>2</sup> − 10r<strong> </strong>+ 25 = 0</p>
|
||
<p>Factor:</p>
|
||
<p class="center large">(r − 5)(r − 5) = 0</p>
|
||
<p class="center large">r = 5<span class="intbl"><strong></strong></span></p>
|
||
<p>So we have one solution: <b>y = e<sup>5x</sup></b></p>
|
||
<p> </p>
|
||
<p class="larger"><b>BUT</b> when <b>e<sup>5x</sup></b> is a solution, then <b>xe<sup>5x</sup></b> is <b>also</b> a solution!</p>
|
||
<div class="center80">
|
||
<p>Why? I can show you:</p>
|
||
<p>y = xe<sup>5x</sup><br></p>
|
||
<p><span class="intbl"><em>dy</em><strong>dx</strong></span> = e<sup>5x</sup> + 5xe<sup>5x</sup></p>
|
||
<p><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 5e<sup>5x</sup> + 5e<sup>5x</sup> + 25xe<sup>5x</sup></p>
|
||
<p>So</p>
|
||
<p><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 10<span class="intbl"><em>dy</em><strong>dx</strong></span> + 25y</p>
|
||
<p>= 5e<sup>5x</sup> + 5e<sup>5x</sup> + 25xe<sup>5x</sup> − 10(e<sup>5x</sup> + 5xe<sup>5x</sup>) + 25xe<sup>5x</sup></p>
|
||
<p>= (5e<sup>5x</sup> + 5e<sup>5x</sup> − 10e<sup>5x</sup>) + (25xe<sup>5x</sup> − 50xe<sup>5x</sup> + 25xe<sup>5x</sup>) = 0</p>
|
||
</div>
|
||
<p>So, in this case our solution is:</p>
|
||
<p class="center large">y = Ae<sup>5x</sup> + Bxe<sup>5x</sup></p>
|
||
</div>
|
||
|
||
<h3> </h3>
|
||
|
||
<h3>How does this work in the general case?</h3>
|
||
<p>With <b>y = xe<sup>rx</sup></b> we get the derivatives:</p>
|
||
<ul>
|
||
<li><span class="intbl"><em>dy</em><strong>dx</strong></span> = e<sup>rx</sup> + rxe<sup>rx</sup></li>
|
||
<li><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = re<sup>rx</sup> + re<sup>rx</sup> + r<sup>2</sup>xe<sup>rx</sup></li>
|
||
</ul>
|
||
<p>So</p>
|
||
<p><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + p <span class="intbl"><em>dy</em><strong>dx</strong></span> + qy</p>
|
||
<p>= (re<sup>rx</sup> + re<sup>rx</sup> + r<sup>2</sup>xe<sup>rx</sup>) + p( e<sup>rx</sup> + rxe<sup>rx</sup> ) + q( xe<sup>rx</sup> )</p>
|
||
<p>= e<sup>rx</sup>(r + r + r<sup>2</sup>x + p + prx + qx)</p>
|
||
<p>= e<sup>rx</sup>(2r + p + x(r<sup>2</sup> + pr + q))</p>
|
||
<p>= e<sup>rx</sup>(2r + p) because we already know that r<sup>2</sup> + pr + q = 0</p>
|
||
<p> </p>
|
||
<p>And when <b>r<sup>2</sup> + pr + q</b> has a repeated root, then <b>r = <span class="intbl"><em>−p</em>2</span></b> and <b>2r + p = 0</b></p>
|
||
<p>So if r is a repeated root of the characteristic equation, then the general solution is</p>
|
||
<p class="center large">y = Ae<sup>rx</sup> + Bxe<sup>rx</sup></p>
|
||
<p>Let's try another example to see how quickly we can get a solution:</p>
|
||
<div class="example">
|
||
|
||
<h3><b>Example 6:</b> Solve</h3>
|
||
<p class="center large">4<span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + 4<span class="intbl"><em>dy</em><strong>dx</strong></span> + y = 0</p>
|
||
<p>The characteristic equation is:<br></p>
|
||
<p class="center large">4r<sup>2</sup> + 4r<strong> </strong>+ 1 = 0</p>
|
||
<p>Then:</p>
|
||
<p class="center large">(2r + 1)<sup>2</sup> = 0</p>
|
||
<p class="center large">r = −<span class="intbl"><em>1</em><strong>2</strong></span></p>
|
||
<p>So the solution of the differential equation is:</p>
|
||
<p class="center large">y = Ae<sup>(−½)x</sup> + Bxe<sup>(−½)x</sup></p>
|
||
</div>
|
||
<p style="float:right; margin: 0 0 5px 10px;"><img src="../algebra/images/quadratic-graph-complex.svg" alt="Quadratic Graph with Complex Roots" height="140" width="175" ></p>
|
||
|
||
|
||
<h2>Complex roots</h2>
|
||
|
||
<p>When the discriminant <b>p<sup>2</sup> − 4q</b> is <b>negative</b> we get <a href="../numbers/complex-numbers.html">complex</a> roots.</p>
|
||
<p>Let’s try an example to help us work out how to do this type:</p>
|
||
<div class="example">
|
||
|
||
<h3><b>Example 7:</b> Solve</h3>
|
||
<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 4<span class="intbl"><em>dy</em><strong>dx</strong></span> + 13y = 0</p>
|
||
<p>The characteristic equation is:<br></p>
|
||
<p class="center large">r<sup>2</sup> − 4r<strong> </strong>+ 13 = 0</p>
|
||
<p>This does not factor, so we use the <a href="../algebra/quadratic-equation.html">quadratic equation formula</a>:</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−b ± √(b<sup>2 </sup>− 4ac)</em> <strong>2a</strong></span></p>
|
||
<p>with a = 1, b = −4 and c = 13</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−(−4) ± √((−4)<sup>2 </sup>− 4×1×13)</em> <strong>2×1</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>4 ± √(16− 52)</em> <strong>2</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>4 ± √(−36)</em> <strong>2</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>4 ± 6i</em> <strong>2</strong></span></p>
|
||
<p class="center larger">x = 2 ± 3i</p>
|
||
<p>If we follow the method used for two real roots, then we can try the solution:</p>
|
||
<p class="center large">y = Ae<sup>(2+3i)x</sup> + Be<sup>(2−3i)x</sup><br></p>
|
||
<p>We can simplify this since e<sup>2x</sup> is a common factor:</p>
|
||
<p class="center large">y = e<sup>2x</sup>( Ae<sup>3ix</sup> + Be<sup>−3ix</sup> )</p>
|
||
<p>But we haven't finished yet ... !</p>
|
||
<a href="../algebra/eulers-formula.html">Euler's formula</a> tells us that:
|
||
<p class="center large">e<sup>ix</sup> = cos(x) + i sin(x)</p>
|
||
<p>So now we can follow a whole new avenue to (eventually) make things simpler.</p>
|
||
<p>Looking just at the "A plus B" part:</p>
|
||
<p class="center large">Ae<sup>3ix</sup> + Be<sup>−3ix</sup></p>
|
||
<p class="center large">A(cos(3x) + i sin(3x)) + B(cos(−3x) + i sin(−3x))</p>
|
||
<p class="center large">Acos(3x) + Bcos(−3x) + i(Asin(3x) + Bsin(−3x))</p>
|
||
<p>Now apply the <a href="../algebra/trigonometric-identities.html">Trigonometric Identities</a>: cos(−θ)=cos(θ) and sin(−θ)=−sin(θ):</p>
|
||
<center>
|
||
<p class="center large">Acos(3x) + Bcos(3x) + i(Asin(3x) − Bsin(3x)</p>
|
||
</center>
|
||
<p class="center large">(A+B)cos(3x) + i(A−B)sin(3x)</p>
|
||
<p>Replace A+B by C, and A−B by D:</p>
|
||
<p class="center large">Ccos(3x) + iDsin(3x)</p>
|
||
<p>And we get the solution:<br></p>
|
||
<p class="center large">y = e<sup>2x</sup>( Ccos(3x) + iDsin(3x) )</p>
|
||
<p class="center large"> </p>
|
||
|
||
<h3>Check</h3>
|
||
<p>We have our answer, but maybe we should check that it does indeed satisfy the original equation:</p>
|
||
<p>y = e<sup>2x</sup>( Ccos(3x) + iDsin(3x) )</p>
|
||
<p><span class="intbl"><em>dy</em><strong>dx</strong></span> = e<sup>2x</sup>( −3Csin(3x)+3iDcos(3x) ) + 2e<sup>2x</sup>( Ccos(3x)+iDsin(3x) )</p>
|
||
<p><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = e<sup>2x</sup>( −(6C+9iD)sin(3x) + (−9C+6iD)cos(3x)) + 2e<sup>2x</sup>(2C+3iD)cos(3x) + (−3C+2iD)sin(3x) )</p>
|
||
<p>Substitute:</p>
|
||
<p><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 4<span class="intbl"><em>dy</em><strong>dx</strong></span> + 13y = e<sup>2x</sup>( −(6C+9iD)sin(3x) + (−9C+6iD)cos(3x)) + 2e<sup>2x</sup>(2C+3iD)cos(3x) + (−3C+2iD)sin(3x) ) − 4( e<sup>2x</sup>( −3Csin(3x)+3iDcos(3x) ) + 2e<sup>2x</sup>( Ccos(3x)+iDsin(3x) ) ) + 13( e<sup>2x</sup>(Ccos(3x) + iDsin(3x)) )<br></p>
|
||
<p>... hey, why don't YOU try adding up all the terms to see if they equal zero ... if not please <a href="../contact.html">let me know</a>, OK?</p>
|
||
</div>
|
||
<p><br></p>
|
||
|
||
<h3>How do we generalize this?</h3>
|
||
<p>Generally, when we solve the characteristic equation with complex roots, we will get two solutions <b>r<sub>1</sub> = v + wi</b> and <b>r<sub>2</sub> = v − wi</b></p>
|
||
<p>So the general solution of the differential equation is</p>
|
||
<p class="center large">y = e<sup>vx</sup> ( Ccos(wx) + iDsin(wx) )</p>
|
||
<div class="example">
|
||
|
||
<h3><b>Example 8:</b> Solve</h3>
|
||
<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 6<span class="intbl"><em>dy</em><strong>dx</strong></span> + 25y = 0</p>
|
||
<p>The characteristic equation is:<br></p>
|
||
<p class="center large">r<sup>2</sup> − 6r<strong> </strong>+ 25 = 0</p>
|
||
<p>Use the quadratic equation formula:</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−b ± √(b<sup>2 </sup>− 4ac)</em> <strong>2a</strong></span></p>
|
||
<p>with a = 1, b = −6 and c = 25</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−(−6) ± √((−6)<sup>2 </sup>− 4×1×25)</em> <strong>2×1</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>6 ± √(36− 100)</em> <strong>2</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>6 ± √(−64)</em> <strong>2</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>6 ± 8i</em> <strong>2</strong></span></p>
|
||
<p class="center larger">x = 3 ± 4i</p>
|
||
<p>And we get the solution:<br></p>
|
||
<p class="center large">y = e<sup>3x</sup>(Ccos(4x) + iDsin(4x))</p>
|
||
<p><br></p>
|
||
</div>
|
||
<p><br></p>
|
||
<div class="example">
|
||
|
||
<h3><b>Example 9:</b> Solve</h3>
|
||
<p class="center large">9<span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + 12<span class="intbl"><em>dy</em><strong>dx</strong></span> + 29y = 0</p>
|
||
<p>The characteristic equation is:<br></p>
|
||
<p class="center large">9r<sup>2</sup> + 12r<strong> </strong>+ 29 = 0</p>
|
||
<p>Use the quadratic equation formula:</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−b ± √(b<sup>2 </sup>− 4ac)</em> <strong>2a</strong></span></p>
|
||
<p>with a = 9, b = 12 and c = 29</p>
|
||
<p class="center larger">x = <span class="intbl"> <em>−12 ± √(12<sup>2 </sup>− 4×9×29)</em> <strong>2×9</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>−12 ± √(144− 1044)</em> <strong>18</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>−12 ± √(−900)</em> <strong>18</strong></span></p>
|
||
<p class="center larger">x = <span class="intbl"><em>−12 ± 30i</em> <strong>18</strong></span></p>
|
||
<p class="center larger">x = −<span class="intbl"><em>2</em><strong>3</strong></span>
|
||
± <span class="intbl"><em>5</em><strong>3</strong></span>i</p>
|
||
<p>And we get the solution:<br></p>
|
||
<p class="center large">y = e<sup>(−<span class="intbl"><em>2</em><strong>3</strong></span>)x</sup>(Ccos(<span class="intbl"><em>5</em><strong>3</strong></span>x) + iDsin(<span class="intbl"><em>5</em><strong>3</strong></span>x))</p>
|
||
<p><br></p>
|
||
</div>
|
||
|
||
|
||
<h2>Summary</h2>
|
||
|
||
<p>To solve a linear second order differential equation of the form</p>
|
||
<p class="center large"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> + p<span class="intbl"><em>dy</em><strong>dx</strong></span> + qy = 0</p>
|
||
<p>where <b>p</b> and <b>q</b> are constants, we must find the roots of the characteristic equation</p>
|
||
<p class="center large">r<sup>2</sup> + pr + q = 0</p>
|
||
<p>There are three cases, depending on the discriminant <b>p<sup>2</sup> - 4q</b>. When it is</p>
|
||
<p class="dotpoint"><b>positive</b> we get two real roots, and the solution is</p>
|
||
<p class="center large">y = Ae<sup>r<sub>1</sub>x</sup> + Be<sup>r<sub>2</sub>x</sup></p>
|
||
<p class="dotpoint"><b>zero</b> we get one real root, and the solution is</p>
|
||
<p class="center large">y = Ae<sup>rx</sup> + Bxe<sup>rx</sup></p>
|
||
<p class="dotpoint"><b>negative</b> we get two complex roots <b>r<sub>1</sub> = v + wi</b> and <b>r<sub>2</sub> = v − wi</b>, and the solution is</p>
|
||
<p class="center large">y = e<sup>vx</sup> ( Ccos(wx) + iDsin(wx) )</p>
|
||
<p> </p>
|
||
<div class="questions">9479, 9480, 9481, 9482, 9483, 9484, 9485, 9486, 9487, 9488</div>
|
||
<p> </p>
|
||
|
||
<div class="related">
|
||
<a href="differential-equations-solution-guide.html">Differential Equations Solution Guide</a>
|
||
<a href="index.html">Calculus Index</a>
|
||
</div>
|
||
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