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<h1 class="center">The Method of Variation of Parameters</h1>
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<div class="def">
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<p>This page is about second order differential 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(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.</p>
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</div><div class="center80">
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<p>Please read <a href="differential-equations-second-order.html">Introduction to Second Order Differential Equations</a> first, it shows how to solve the simpler "homogeneous" case where f(x)=0</p></div>
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<h2>Two Methods</h2>
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<p>There are two main methods to solve equations like</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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= f(x)</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"><span class="center"><b>Variation of Parameters</b></span> (that we will learn here) which works on a wide range of functions but is a little messy to use.</p>
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<h2>Variation of Parameters</h2>
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<p>To keep things simple, we are only going to look at the case:</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 = f(x)</p>
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where p and q are constants and f(x) is a non-zero function of x.
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<p>The <strong>complete solution</strong> to such an equation can be found
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by combining two types of solution:</p>
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<ol>
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<li style="text-align: justify;">The <b>general solution</b> of the
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homogeneous equation <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</li>
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<li style="text-align: justify;"><b>Particular solutions</b> of the
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non-homogeneous equation <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 =
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f(x)</li>
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</ol>
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<p>Note that f(x) could be a single function or a sum of two or more
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functions.</p>
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<p>Once we have found the general solution and all the particular
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solutions, then the final complete solution is found by adding all the
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solutions together.</p>
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<p>This method relies on <a href="integration-introduction.html">integration.</a></p>
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<p>The problem with this method is that, although it may yield a solution,
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in some cases the solution has to be left as an integral.</p>
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<h2>Start with the General Solution</h2>
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<p>On <a href="differential-equations-second-order.html">Introduction to Second Order Differential Equations</a> we learn how to find the general solution.</p>
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<p>Basically we take the 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>and reduce it to 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>Which is a quadratic equation that has three possible solution types depending on the discriminant <b>p<sup>2</sup> − 4q</b>. When <b>p<sup>2</sup> − 4q</b> is</p>
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<p class="dotpoint"><b>positive</b> we get two real roots, and the solution is</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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<p class="dotpoint"><b>zero</b> we get one real root, and the solution is</p>
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<p class="center large">y = Ae<sup>rx</sup> + Bxe<sup>rx</sup></p>
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<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>
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<p class="center large">y = e<sup>vx</sup> ( Ccos(wx) + iDsin(wx) )</p>
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<h2>The Fundamental Solutions of The Equation</h2>
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<p>In all three cases above "y" is made of two parts:</p>
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<ul>
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<li><b>y = Ae<sup>r<sub>1</sub>x</sup> + Be<sup>r<sub>2</sub>x</sup></b> is made of <b>y<sub>1</sub> = Ae<sup>r<sub>1</sub>x</sup></b> and <b>y<sub>2</sub> = Be<sup>r<sub>2</sub>x</sup></b></li>
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<li><b>y = Ae<sup>rx</sup> + Bxe<sup>rx</sup></b> is made of <b>y<sub>1</sub> = Ae<sup>rx</sup></b> and <b>y<sub>2</sub> = Bxe<sup>rx</sup></b></li>
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<li><b>y = e<sup>vx</sup> ( Ccos(wx) + iDsin(wx) )</b> is made of <b>y<sub>1</sub> = e<sup>vx</sup>Ccos(wx)</b> and <b>y<sub>2</sub> = e<sup>vx</sup>iDsin(wx)</b></li>
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</ul>
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<p>y<sub>1</sub> and y<sub>2</sub> are known as the fundamental solutions of the equation</p>
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<p>And y<sub>1</sub> and y<sub>2</sub> are said to be <b>linearly
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independent</b> because neither function is a constant multiple of the
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other.</p>
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<h2>The Wronskian</h2>
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<p>When y<sub>1</sub> and y<sub>2</sub> are the two fundamental solutions
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of the homogeneous equation</p>
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<p class="center"><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>then the Wronskian W(y<sub>1</sub>, y<sub>2</sub>) is the <a href="../algebra/matrix-determinant.html">determinant
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of the matrix</a></p>
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<p class="center"> <img src="images/wronskian-matrix.svg" alt="matrix for the Wronskian" height="81" width="119"> </p>
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<p>So</p>
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<p class="center larger">W(y<sub>1</sub>, y<sub>2</sub>) = y<sub>1</sub>y<sub>2</sub>'
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− y<sub>2</sub>y<sub>1</sub>'</p>
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<p class="words">The <b>Wronskian</b> is named after the Polish mathematician and
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philosopher Józef Hoene-Wronski (1776−1853).</p>
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<p>Since y<sub>1</sub> and y<sub>2</sub> are linearly independent, the value of the Wronskian cannot equal zero.</p>
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<h2>The Particular Solution</h2>
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<p>Using the Wronskian we can now find the particular solution of the differential equation</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<span class="intbl"><em>dy</em><strong>dx</strong></span> + qy = f(x)</p>
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<p>using the formula:</p>
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<p class="center large">y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
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+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
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<p class="center large"><br></p>
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<div class="example">
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<h3>Example 1: Solve <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 3<span class="intbl"><em>dy</em><strong>dx</strong></span> + 2y =
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e<sup>3x</sup></h3>
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<p><b>1. Find the general solution of</b> <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 3<span class="intbl"><em>dy</em><strong>dx</strong></span> + 2y = 0</p>
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<p>The characteristic equation is: r<sup>2</sup> − 3r + 2 = 0</p>
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<p>Factor: (r − 1)(r − 2) = 0</p>
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<p class="so">r = 1 or 2</p>
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<p>So the general solution of the differential equation is <span class="larger">y = Ae<sup>x</sup>+Be<sup>2x</sup></span></p>
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<p>So in this case the fundamental solutions and their derivatives are:</p>
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<p class="so">y<sub>1</sub>(x) = e<sup>x</sup></p>
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<p class="so">y<sub>1</sub>'(x) = e<sup>x</sup></p>
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<p class="so">y<sub>2</sub>(x) = e<sup>2x</sup></p>
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<p class="so">y<sub>2</sub>'(x) = 2e<sup>2x</sup></p>
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<p><b>2. Find the Wronskian:</b></p>
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<p class="center">W(y<sub>1</sub>, y<sub>2</sub>) = y<sub>1</sub>y<sub>2</sub>'
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− y<sub>2</sub>y<sub>1</sub>' = 2e<sup>3x</sup> − e<sup>3x</sup> = e<sup>3x</sup></p>
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<p><b>3. Find the particular solution using the formula:</b></p>
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<div style="text-align: center;"> y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
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+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx </div>
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<p><b>4. First we solve the integrals:</b></p>
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<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p><br>
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= <span class="integral">∫</span><span class="intbl"><em>e<sup>2x</sup>e<sup>3x</sup></em><strong>e<sup>3x</sup></strong></span>dx
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<p>= <span class="integral">∫</span>e<sup>2x</sup>dx</p>
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<p>= <span class="intbl"><em>1</em>2</span>e<sup>2x</sup></p>
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<p>So:</p>
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<p class="center">−y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
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= −(e<sup>x</sup>)(<span class="intbl"><em>1</em>2</span>e<sup>2x</sup>)
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= −<span class="intbl"><em>1</em>2</span>e<sup>3x</sup></p>
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<p><b>And also:</b></p>
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<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p><br>
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= <span class="integral">∫</span><span class="intbl"><em>e<sup>x</sup>e<sup>3x</sup></em><strong>e<sup>3x</sup></strong></span>dx
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<p>= <span class="integral">∫</span>e<sup>x</sup>dx</p>
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<p>= e<sup>x</sup></p>
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<p>So:</p>
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<p class="center">y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
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= (e<sup>2x</sup>)(<span class="intbl"></span>e<sup>x</sup>) = e<sup>3x</sup></p>
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<p><b>Finally:</b></p>
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<p>y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
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+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
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<p>= −<span class="intbl"><em>1</em>2</span>e<sup>3x</sup> + e<sup>3x</sup></p>
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<p>= <span class="intbl"><em>1</em>2</span>e<sup>3x</sup></p>
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<p>and the complete solution of the differential equation <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 3<span class="intbl"><em>dy</em><strong>dx</strong></span> + 2y = e<sup>3x</sup> is</p>
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<p class="center large">y = Ae<sup>x</sup> + Be<sup>2x</sup> + <span class="intbl"><em>1</em>2</span>e<sup>3x</sup></p>
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<p>Which looks like this (example values of A and B):</p>
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<p class="center"><img src="images/graph-aex-be2x-2-e3x.svg" alt="Aex + Be2x + 12e3x " height="340" width="550"></p>
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</div>
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<div><br>
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</div>
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<div class="example">
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<h3>Example 2: Solve <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − y = 2x<sup>2</sup> − x − 3</h3><br>
|
||
<b>1. Find the general solution of</b> <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − y = 0
|
||
<p>The characteristic equation is: r<sup>2</sup> − 1 = 0</p>
|
||
<p>Factor: (r − 1)(r + 1) = 0</p>
|
||
<p class="so">r = 1 or −1</p>
|
||
<p>So the general solution of the differential equation is y = Ae<sup>x</sup>+Be<sup>−x</sup></p>
|
||
<p>So in this case the fundamental solutions and their derivatives are:</p>
|
||
<p class="so">y<sub>1</sub>(x) = e<sup>x</sup></p>
|
||
<p class="so">y<sub>1</sub>'(x) = e<sup>x</sup></p>
|
||
<p class="so">y<sub>2</sub>(x) = e<sup>−x</sup></p>
|
||
<p class="so">y<sub>2</sub>'(x) = −e<sup>−x</sup></p>
|
||
<p><b>2. Find the Wronskian:</b></p>
|
||
<p class="center">W(y<sub>1</sub>, y<sub>2</sub>) = y<sub>1</sub>y<sub>2</sub>'
|
||
− y<sub>2</sub>y<sub>1</sub>' = −e<sup>x</sup>e<sup>−x</sup> − e<sup>x</sup>e<sup>−x</sup> = −2</p>
|
||
<p><b>3. Find the particular solution using the formula:</b></p>
|
||
<p class="center large">y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<p><b>4. Solve the integrals:</b></p>
|
||
<div style="text-align: left;">Each of the integrals can be obtained
|
||
by using <a href="integration-by-parts.html">Integration
|
||
by Parts</a> twice:</div>
|
||
<div style="text-align: left;"><br>
|
||
</div>
|
||
<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p><br>
|
||
= <span class="integral">∫</span><span class="intbl"><em>e<sup>−x </sup>(2x<sup>2</sup>−x−3)</em><strong>−2</strong></span>dx
|
||
|
||
<p>= −<span class="intbl"><em>1</em>2</span> <span class="integral">∫</span>(2x<sup>2</sup>−x−3)<span class="integral"></span>e<sup>−x</sup>dx</p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>[ −(2x<sup>2</sup>−x−3)e<sup>−x</sup> + <span class="integral">∫</span>(4x−1)<span class="integral"></span>e<sup>−x </sup>dx ]</p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>[ −(2x<sup>2</sup>−x−3)e<sup>−x</sup> − (4x − 1)e<sup>−x</sup> + <span class="integral">∫</span>4e<sup>−x</sup>dx
|
||
]</p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>[ −(2x<sup>2</sup>−x−3)e<sup>−x</sup> − (4x − 1)e<sup>−x</sup> − 4e<sup>−x</sup> ]</p>
|
||
<p>= <span class="intbl"><em>e<sup>−x</sup></em>2</span>[ 2x<sup>2</sup> − x − 3 + 4x −1 + 4 ]</p>
|
||
<p>= <span class="intbl"><em>e<sup>−x</sup></em>2</span>[ 2x<sup>2</sup> + 3x ]</p>
|
||
<p>So:</p>
|
||
<p class="center">−y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
= (−e<sup>x</sup>)[<span class="intbl"><em>e<sup>−x</sup></em>2</span>(
|
||
2x<sup>2</sup> + 3x )] = −<span class="intbl"><em>1</em>2</span>(2x<sup>2</sup> + 3x)</p>
|
||
<p><b>And this one:</b></p>
|
||
<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p><br>
|
||
= <span class="integral">∫</span><span class="intbl"><em>e<sup>x </sup>(2x<sup>2</sup>−x−3)</em><strong>−2</strong></span>dx
|
||
|
||
<p>= −<span class="intbl"><em>1</em>2</span> <span class="integral">∫</span>(2x<sup>2</sup>−x−3)<span class="integral"></span>e<sup>x</sup>dx</p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>[ (2x<sup>2</sup>−x−3)e<sup>x</sup> − <span class="integral">∫</span>(4x−1)<span class="integral"></span>e<sup>x </sup>dx ]</p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>[ (2x<sup>2</sup>−x−3)e<sup>x</sup> − (4x − 1)e<sup>x</sup> + <span class="integral">∫</span>4e<sup>x</sup>dx
|
||
]</p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>[ (2x<sup>2</sup>−x−3)e<sup>x</sup> − (4x − 1)e<sup>x</sup> + 4e<sup>x</sup> ]</p>
|
||
<p>= <span class="intbl"><em>−e<sup>x</sup></em>2</span>[ 2x<sup>2</sup> − x − 3 − 4x + 1 + 4 ]</p>
|
||
<p>= <span class="intbl"><em>−e<sup>x</sup></em>2</span>[ 2x<sup>2</sup> − 5x + 2 ]<br>
|
||
</p>
|
||
<p>So:</p>
|
||
<p class="center">y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
= (e<sup>−x</sup>)[<span class="intbl"></span><span class="intbl"><em>−e<sup>x</sup></em>2</span>(
|
||
2x<sup>2</sup> − 5x + 2 ) ] = −<span class="intbl"><em>1</em>2</span>(
|
||
2x<sup>2</sup> − 5x + 2 )<sup> </sup></p>
|
||
<p><b>Finally:</b></p>
|
||
<p>y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>( 2x<sup>2</sup> + 3x ) − <span class="intbl"><em>1</em>2</span>( 2x<sup>2</sup> − 5x + 2 ) </p>
|
||
<p>= −<span class="intbl"><em>1</em>2</span>( 4x<sup>2</sup> − 2x + 2 )</p>
|
||
<p>= −2x<sup>2</sup> + x − 1</p>
|
||
<p>and the complete solution of the differential equation <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − y = 2x<sup>2</sup> − x − 3 is</p>
|
||
<p class="center"><strong>y = Ae<sup>x</sup> + Be<sup>−x</sup> − 2x<sup>2</sup> + x − 1 </strong></p>
|
||
<p>(This is the same answer that we got in Example 1 on the page Method of undetermined coefficients.)</p>
|
||
</div>
|
||
<div class="example">
|
||
|
||
<h3>Example 3: Solve <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> + 9y =<span class="intbl"><em>1</em><strong>x</strong></span></h3><br>
|
||
<b>1. Find the general solution of</b> <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> + 9y = 0
|
||
<p>The characteristic equation is: r<sup>2</sup> − 6r + 9 = 0</p>
|
||
<p>Factor: (r − 3)(r − 3) = 0</p>
|
||
<p class="so">r = 3</p>
|
||
<p>So the general solution of the differential equation is y = Ae<sup>3x</sup> + Bxe<sup>3x</sup></p>
|
||
<p>And so in this case the fundamental solutions and their derivatives are:</p>
|
||
<p class="so">y<sub>1</sub>(x) = e<sup>3x</sup></p>
|
||
<p class="so">y<sub>1</sub>'(x) = 3e<sup>3x</sup></p>
|
||
<p class="so">y<sub>2</sub>(x) = xe<sup>3x</sup></p>
|
||
<p class="so">y<sub>2</sub>'(x) = (3x + 1)e<sup>3x</sup></p>
|
||
<p><b>2. Find the Wronskian:</b></p>
|
||
<p class="center">W(y<sub>1</sub>, y<sub>2</sub>) = y<sub>1</sub>y<sub>2</sub>'
|
||
− y<sub>2</sub>y<sub>1</sub>' = (3x + 1)e<sup>3x</sup>e<sup>3x</sup> −
|
||
3xe<sup>3x</sup>e<sup>3x</sup> = e<sup>6x</sup></p>
|
||
<p><b>3. Find the particular solution using the formula:</b></p>
|
||
<div style="text-align: center;"> y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx </div>
|
||
<div style="text-align: left;"><br>
|
||
</div>
|
||
<div style="text-align: left;"><b>4. Solve the integrals:</b></div>
|
||
<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p><br>
|
||
= <span class="integral">∫</span><span class="intbl"><em>(xe<sup>3x</sup>)x<sup>−1</sup> </em><strong>e<sup>6x</sup></strong></span>dx (Note: <span class="intbl"><em>1</em><strong>x</strong></span> = x<sup>−1</sup>)
|
||
|
||
<p>= <span class="integral">∫</span>e<sup>−3x</sup>dx</p>
|
||
<p>= −<span class="intbl"><em>1</em>3</span>e<sup>−3x</sup></p>
|
||
<p>So:</p>
|
||
<p class="center">−y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
= −(e<sup>3x</sup>)(−<span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>)
|
||
= <span class="intbl"><em>1</em>3</span></p>
|
||
<p><b>And this one:</b></p>
|
||
<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p><br>
|
||
= <span class="integral">∫</span><span class="intbl"><em>e<sup>3x</sup>x<sup>−1</sup></em><strong>e<sup>6x</sup></strong></span>dx
|
||
|
||
<p>= <span class="integral">∫</span>e<sup>−3x</sup>x<sup>−1</sup>dx</p>
|
||
<p>This cannot be integrated, so this is an example where the answer has
|
||
to be left as an integral.</p>
|
||
<p>So:</p>
|
||
<p>y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
= ( xe<sup>3x </sup>)( <span class="integral">∫</span>e<sup>−3x</sup>x<sup>−1</sup>dx
|
||
) = xe<sup>3x</sup><span class="integral">∫</span>e<sup>−3x</sup>x<sup>−1</sup>dx</p>
|
||
<p><b>Finally:</b></p>
|
||
<p>y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<p>= <span class="intbl"><em>1</em>3</span> + xe<sup>3x</sup><span class="integral">∫</span>e<sup>−3x</sup>x<sup>−1</sup>dx</p>
|
||
<p>So the complete solution of the differential equation <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> + 9y = <span class="intbl"><em>1</em>x</span> is</p>
|
||
<p class="center"><strong>y = Ae<sup>3x</sup> + Bxe<sup>3x</sup> + <span class="intbl"><em>1</em>3</span> + xe<sup>3x</sup><span class="integral">∫</span>e<sup>−3x</sup>x<sup>−1</sup>dx<span class="intbl"></span></strong></p>
|
||
</div>
|
||
<div class="example">
|
||
|
||
<h3>Example 4 (Harder example): Solve <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> + 13y =
|
||
195cos(4x)<span class="intbl"><strong></strong></span></h3><br>
|
||
<div class="fun">
|
||
<p><strong>This example uses the following <a href="../algebra/trigonometric-identities.html">trigonometric
|
||
identities</a></strong></p>
|
||
<p class="center">sin<sup>2</sup>(θ) + cos<sup>2</sup>(θ) = 1</p>
|
||
<p class="center">sin(θ ± φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)</p>
|
||
<p class="center">cos(θ ± φ) = cos(θ)cos(φ) <img src="../images/symbols/minus-plus.svg" alt="minus/plus" height="19" width="13"> sin(θ)sin(φ)</p>
|
||
<p class="center">sin(θ)cos(φ) = <span class="intbl"><em>1</em><strong>2</strong></span>[sin(θ
|
||
+ φ) + sin(θ − φ)]<br>
|
||
cos(θ)cos(φ) = <span class="intbl"><em>1</em><strong>2</strong></span>[cos(θ
|
||
− φ) + cos(θ + φ)]</p>
|
||
</div><br>
|
||
<b>1. Find the general solution of</b> <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> + 13y = 0
|
||
<p>The characteristic equation is: r<sup>2</sup> − 6r + 13 = 0</p>
|
||
<p>Use the <a href="../algebra/quadratic-equation.html">quadratic equation
|
||
formula</a></p>
|
||
<p class="center">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 = 13</p>
|
||
<p>So:</p>
|
||
<p>r = <span class="intbl"><em>−(−6) ± √[(−6)<sup>2 </sup>− 4(1)(13)]</em> <strong>2(1)</strong></span></p>
|
||
<p>= <span class="intbl"><em>6 ± √[36−52]</em> <strong>2</strong></span></p>
|
||
<p>= <span class="intbl"><em>6 ± √[−16]</em> <strong>2</strong></span></p>
|
||
<p>= <span class="intbl"><em>6 ± 4i</em> <strong>2</strong></span></p>
|
||
<p>= 3 ± 2i</p>
|
||
<p>So α = 3 and β = 2</p>
|
||
<p><span style="font-size:140%;">⇒</span> y = e<sup>3x</sup>[Acos(2x) +
|
||
iBsin(2x)]</p>
|
||
<p>So in this case we have:</p>
|
||
<p class="so">y<sub>1</sub>(x) = e<sup>3x</sup>cos(2x)</p>
|
||
<p class="so">y<sub>1</sub>'(x) = e<sup>3x</sup>[3cos(2x) − 2sin(2x)]</p>
|
||
<p class="so">y<sub>2</sub>(x) = e<sup>3x</sup>sin(2x)</p>
|
||
<p class="so">y<sub>2</sub>'(x) = e<sup>3x</sup>[3sin(2x) + 2cos(2x)]</p>
|
||
<p><b>2. Find the Wronskian:</b></p>
|
||
<p>W(y<sub>1</sub>, y<sub>2</sub>) = y<sub>1</sub>y<sub>2</sub>' − y<sub>2</sub>y<sub>1</sub>'</p>
|
||
<p>= e<sup>6x</sup>cos(2x)[3sin(2x) + 2cos(2x)] − e<sup>6x</sup>sin(2x)[3cos(2x)
|
||
− 2sin(2x)]</p>
|
||
<p>= e<sup>6x</sup>[3cos(2x)sin(2x) +2cos<sup>2</sup>(2x) − 3sin(2x)cos(2x) + 2sin<sup>2</sup>(2x)]</p>
|
||
<p>=2e<sup>6x</sup></p>
|
||
<p><br>
|
||
<b>3. Find the particular solution using the formula:</b></p>
|
||
<p class="center">y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<div style="text-align: left;"><br>
|
||
</div>
|
||
<div style="text-align: left;"><b>4. Solve the integrals:</b></div>
|
||
<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p><br>
|
||
= <span class="integral">∫</span><span class="intbl"><em>e<sup>3x</sup>sin(2x)[195cos(4x)] </em><strong>2e<sup>6x</sup></strong></span>dx
|
||
|
||
<p>= <span class="intbl"><em>195</em><strong>2</strong></span><span class="integral">∫</span>e<sup>−3x</sup>sin(2x)cos(4x)dx</p>
|
||
<p>= <span class="intbl"><em>195</em><strong>4</strong></span><span class="integral">∫</span>e<sup>−3x</sup>[sin(6x)
|
||
− sin(2x)]dx ... (1)</p>
|
||
<p>In this case, we won’t do the integration yet, for reasons that will
|
||
become clear in a moment.</p>
|
||
<p>The other integral is:</p>
|
||
<p class="center large"><span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<p>= <span class="integral">∫</span><span class="intbl"><em>e<sup>3x</sup>cos(2x)[195cos(4x)]</em><strong>2e<sup>6x</sup></strong></span>dx</p>
|
||
<p>= <span class="intbl"><em>195</em><strong>2</strong></span><span class="integral">∫</span>e<sup>−3x</sup>cos(2x)cos(4x)dx</p>
|
||
<p>= <span class="intbl"><em>195</em><strong>4</strong></span><span class="integral">∫</span>e<sup>−3x</sup>[cos(6x)
|
||
+ cos(2x)]dx ... (2)</p>
|
||
<p><br>
|
||
<br>
|
||
From equations (1) and (2) we see that there are four very similar
|
||
integrations that we need to perform:</p>
|
||
<p><em>I</em><sub>1</sub> = <span class="integral">∫</span>e<sup>−3x</sup>sin(6x)dx<br>
|
||
<em>I</em><sub>2</sub> = <span class="integral">∫</span>e<sup>−3x</sup>sin(2x)dx<br>
|
||
<em>I</em><sub>3</sub> = <span class="integral">∫</span>e<sup>−3x</sup>cos(6x)dx<br>
|
||
<em>I</em><sub>4</sub> = <span class="integral">∫</span>e<sup>−3x</sup>cos(2x)dx</p>
|
||
<p>Each of these could be obtained by using Integration by Parts twice,
|
||
but there’s an easier method:</p>
|
||
<p><em>I</em><sub>1</sub> = <span class="integral">∫</span>e<sup>−3x</sup>sin(6x)dx
|
||
= −<span class="intbl"><em>1</em><strong>6</strong></span>e<sup>−3x</sup>cos(6x)
|
||
− <span class="intbl"><em>3</em><strong>6</strong></span><span class="integral"></span><span class="integral"></span><span class="integral">∫</span>e<sup>−3x</sup>cos(6x)dx
|
||
= − <span class="intbl"><em>1</em><strong>6</strong></span>e<sup>−3x</sup>cos(6x)
|
||
− <span class="intbl"><em>1</em>2</span><em>I</em><sub>3</sub></p>
|
||
<p><span style="font-size:140%;">⇒ </span><strong>2<em>I</em><sub>1 </sub>+ <em>I</em><sub>3</sub> = − <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>cos(6x)
|
||
... (3)</strong></p>
|
||
<p><em>I</em><sub>2</sub> = <span class="integral">∫</span>e<sup>−3x</sup>sin(2x)dx
|
||
= −<span class="intbl"><em>1</em><strong>2</strong></span>e<sup>−3x</sup>cos(2x)
|
||
− <span class="intbl"><em>3</em>2</span><span class="integral"></span><span class="integral"></span><span class="integral">∫</span>e<sup>−3x</sup>cos(2x)dx
|
||
= − <span class="intbl"><em>1</em>2</span>e<sup>−3x</sup>cos(2x) − <span class="intbl"><em>3</em>2</span><em>I</em><sub>4</sub></p>
|
||
<p><span style="font-size:140%;">⇒ </span><strong>2<em>I</em><sub>2 </sub>+
|
||
3<em>I</em><sub>4</sub> = − e<sup>−3x</sup>cos(2x)
|
||
... (4)</strong></p>
|
||
<p><em>I</em><sub>3</sub> = <span class="integral">∫</span>e<sup>−3x</sup>cos(6x)dx
|
||
= <span class="intbl"><em>1</em><strong>6</strong></span>e<sup>−3x</sup>sin(6x)
|
||
+ <span class="intbl"><em>3</em><strong>6</strong></span><span class="integral"></span><span class="integral"></span><span class="integral">∫</span>e<sup>−3x</sup>sin(6x)dx
|
||
= <span class="intbl"><em>1</em><strong>6</strong></span>e<sup>−3x</sup>sin(6x)
|
||
+ <span class="intbl"><em>1</em>2</span><em>I</em><sub>1</sub><br>
|
||
<span style="font-size:140%;">⇒ </span><strong>2<em>I</em><sub>3 </sub>− <em>I</em><sub>1</sub> = <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>sin(6x)
|
||
... (5)</strong><br>
|
||
<em>I</em><sub>4</sub> = <span class="integral">∫</span>e<sup>−3x</sup>cos(2x)dx
|
||
= <span class="intbl"><em>1</em><strong>2</strong></span>e<sup>−3x</sup>sin(2x)
|
||
+ <span class="intbl"><em>3</em>2</span><span class="integral"></span><span class="integral"></span><span class="integral">∫</span>e<sup>−3x</sup>sin(2x)dx
|
||
= <span class="intbl"><em>1</em>2</span>e<sup>−3x</sup>sin(2x) + <span class="intbl"><em>3</em>2</span><em>I</em><sub>2</sub></p>
|
||
<p><span style="font-size:140%;">⇒ </span><strong>2<em>I</em><sub>4 </sub>−
|
||
3<em>I</em><sub>2</sub> = e<sup>−3x</sup>sin(2x)
|
||
... (6)</strong></p>
|
||
<p>Solve equations (3) and (5) simultaneously:</p>
|
||
<p><span style="font-size:140%;"></span>2<em>I</em><sub>1 </sub>+ <em>I</em><sub>3</sub> = − <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>cos(6x)
|
||
... (3)</p>
|
||
<p>2<em>I</em><sub>3 </sub>− <em>I</em><sub>1</sub> = <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>sin(6x)
|
||
... (5)</p>
|
||
<p>Multiply equation (5) by 2 and add them together (term <em>I</em><sub>1</sub> will neutralize):</p>
|
||
<p><span style="font-size:140%;">⇒ </span>5<em>I</em><sub>3 </sub>=
|
||
− <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>cos(6x) + <span class="intbl"><em>2</em>3</span>e<sup>−3x</sup>sin(6x)</p>
|
||
<p> = <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>[2sin(6x)
|
||
− cos(6x)]</p>
|
||
<p><span style="font-size:140%;">⇒ </span><strong><em>I</em><sub>3 </sub>= <span class="intbl"><em>1</em>15</span>e<sup>−3x</sup>[2sin(6x)
|
||
− cos(6x)]</strong></p>
|
||
<p>Multiply equation (3) by 2 and subtract (term <em>I</em><sub>3</sub> will neutralize):</p>
|
||
<p><span style="font-size:140%;">⇒ </span>5<em>I</em><sub>1 </sub>=
|
||
− <span class="intbl"><em>2</em>3</span>e<sup>−3x</sup>cos(6x) − <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>sin(6x)</p>
|
||
<p> = − <span class="intbl"><em>1</em>3</span>e<sup>−3x</sup>[2cos(6x)
|
||
+ sin(6x)]</p>
|
||
<p><span style="font-size:140%;">⇒ </span><strong><em>I</em><sub>1 </sub>= − <span class="intbl"><em>1</em>15</span>e<sup>−3x</sup>[2cos(6x)
|
||
+ sin(6x)]</strong></p>
|
||
<p>Solve equations (4) and (6) simultaneously:</p>
|
||
<p><span style="font-size:140%;"></span>2<em>I</em><sub>2 </sub>+ 3<em>I</em><sub>4</sub> = − e<sup>−3x</sup>cos(2x) ... (4)</p>
|
||
<p><span style="font-size:140%;"></span>2<em>I</em><sub>4 </sub>− 3<em>I</em><sub>2</sub> = e<sup>−3x</sup>sin(2x) ... (6)</p>
|
||
<p>Multiply equation (4) by 3 and equation (6) by 2 and add (term <em>I</em><sub>2</sub> will neutralize):</p>
|
||
<p><span style="font-size:140%;">⇒ </span>13<em>I</em><sub>4 </sub>=
|
||
− 3e<sup>−3x</sup>cos(2x) + 2e<sup>−3x</sup>sin(2x)</p>
|
||
<p> =<span class="intbl"></span>e<sup>−3x</sup>[2sin(2x)
|
||
− 3 cos(2x)]</p>
|
||
<p><span style="font-size:140%;">⇒ </span><strong><em>I</em><sub>4 </sub>= <span class="intbl"><em>1</em>13</span>e<sup>−3x</sup>[2sin(2x)
|
||
− 3cos(2x)]</strong></p>
|
||
<p>Multiply equation (4) by 2 and equation (6) by 3 and subtract (term <em>I</em><sub>4</sub> will neutralize):</p>
|
||
<p><span style="font-size:140%;">⇒ </span>13<em>I</em><sub>2 </sub>=
|
||
− 2e<sup>−3x</sup>cos(2x) − 3e<sup>−3x</sup>sin(2x)</p>
|
||
<p> =<span class="intbl"></span>−
|
||
e<sup>−3x</sup>[2cos(2x) + 3 sin(2x)]</p>
|
||
<p><span style="font-size:140%;">⇒ </span><strong><em>I</em><sub>2 </sub>= − <span class="intbl"><em>1</em>13</span>e<sup>−3x</sup>[2cos(2x)
|
||
+ 3sin(2x)]</strong></p>
|
||
<p>Substitute into (1) and (2):</p>
|
||
<p><span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<p>= <span class="intbl"><em>195</em>4</span><span class="integral">∫</span>e<sup>−3x</sup>[sin(6x)
|
||
− sin(2x)]dx ... (1) </p>
|
||
<p>= <span class="intbl"><em>195</em>4</span>[<strong>−</strong> <span class="intbl"><em>1</em>15</span>e<sup>−3x</sup>[2cos(6x) + sin(6x)]
|
||
− [−<span class="intbl"><em>1</em>13</span>e<sup>−3x</sup>[2cos(2x) +
|
||
3sin(2x)]]]</p>
|
||
<p>= <span class="intbl"><em>e<sup>−3x</sup></em>4</span>[−13(2cos(6x)+sin(6x))+15(2
|
||
cos(2x)+3sin(2x))] <strong></strong><strong></strong></p>
|
||
<p><span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<p>= <span class="intbl"><em>195</em><strong>4</strong></span><span class="integral">∫</span>e<sup>−3x</sup>[cos(6x)
|
||
+ cos(2x)]dx ... (2)</p>
|
||
<p>= <span class="intbl"><em>195</em>4</span>[<strong></strong><span class="intbl"><em>1</em>15</span>e<sup>−3x</sup>[2sin(6x)
|
||
− cos(6x)] + <span class="intbl"><em>1</em>13</span>e<sup>−3x</sup>[2sin(2x)
|
||
− 3cos(2x)]]</p>
|
||
<p>= <span class="intbl"><em>e<sup>−3x</sup></em>4</span>[13(2sin(6x) −
|
||
cos(6x)) + 15(2sin(2x) − 3cos(2x))]</p>
|
||
<p>So y<sub>p</sub>(x) = −y<sub>1</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>2</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx
|
||
+ y<sub>2</sub>(x)<span class="integral">∫</span><span class="intbl"><em>y<sub>1</sub>(x)f(x)</em><strong>W(y<sub>1</sub>, y<sub>2</sub>)</strong></span>dx</p>
|
||
<p>= − e<sup>3x</sup>cos(2x)<span class="intbl"><em>e<sup>−3x</sup></em>4</span>[−13(2cos(6x)+sin(6x))
|
||
+ 15(2 cos(2x)+3sin(2x))] + e<sup>3x</sup>sin(2x)<span class="intbl"><em>e<sup>−3x</sup></em>4</span>[13(2sin(6x)
|
||
− cos(6x)) + 15(2sin(2x) − 3cos(2x))]</p>
|
||
<p>= − <span class="intbl"><em>1</em>4</span>cos(2x) [−13(2cos(6x) −
|
||
sin(6x)) + 15(2 cos(2x) + 3sin(2x))] +<span class="intbl"><em>1</em>4</span> sin(2x)[13(2sin(6x) − cos(6x)) + 15(2 sin(2x) −
|
||
3cos(2x))] </p>
|
||
<p>= <span class="intbl"><em>1</em>4</span>[26cos(2x)cos(6x) +
|
||
13cos(2x)sin(6x) − 30cos<sup>2</sup>(2x) − 45cos(2x)sin(2x) +
|
||
26sin(2x)sin(6x) − 13sin(2x)cos(6x) + 30sin<sup>2</sup>(2x) −
|
||
45sin(2x)cos(2x)]</p>
|
||
<p>= <span class="intbl"><em>1</em>4</span>[26[cos(2x)cos(6x) +
|
||
sin(2x)sin(6x)] + 13[cos(2x)sin(6x) − sin(2x)cos(6x)] − 30[cos<sup>2</sup>(2x)
|
||
− sin<sup>2</sup>(2x)] − 45[cos(2x)sin(2x) + sin(2x)cos(2x)]]</p>
|
||
<p>= <span class="intbl"><em>1</em>4</span>[26cos(4x) + 13sin(4x) −
|
||
30cos(4x) − 45sin(4x)]</p>
|
||
<p>= <span class="intbl"><em>1</em>4</span>[−4cos(4x) − 32sin(4x)]</p>
|
||
<p>= −cos(4x) − 8 sin(4x)</p>
|
||
<p>So the complete solution of the differential equation <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> + 13y =
|
||
195cos(4x) is</p>
|
||
<p class="center"><strong> y = e<sup>3x</sup>(Acos(2x) + iBsin(2x)) −
|
||
cos(4x) − 8sin(4x)<br>
|
||
<span class="intbl"></span></strong></p>
|
||
</div>
|
||
<p> </p>
|
||
<div class="questions">9529, 9530, 9531, 9532, 9533, 9534, 9535, 9536, 9537, 9538</div>
|
||
|
||
<div class="related">
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<a href="differential-equations-solution-guide.html">Differential Equations Solution Guide</a>
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