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<h1>Method of Undetermined Coefficients</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><p><br></p>
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<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 these equations:</p>
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<p class="dotpoint"><b>Undetermined Coefficients</b> (that we learn here) 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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<h2>Undetermined Coefficients</h2>
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<p>To keep things simple, we only 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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<p>where <b>p</b> and <b>q</b> are constants.</p>
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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>The <b>general solution</b> of the
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homogeneous equation</li>
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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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<li><b>Particular solutions</b> of the
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non-homogeneous equation</li>
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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 = f(x)</p>
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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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<div class="example">
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<h3>Example 1: <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>
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<p>(For the moment trust me regarding these solutions)</p>
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<p>The homogeneous equation <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − y = 0 has a general solution</p>
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<p class="center">y = Ae<sup>x</sup> + Be<sup>-x</sup></p>
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<p>The non-homogeneous 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 has a particular solution</p>
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<p class="center">y = −2x<sup>2 </sup>+ x − 1</p>
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<p>So the complete solution of the differential equation is</p>
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<p class="center large">y = Ae<sup>x</sup> + Be<sup>-x</sup> − 2x<sup>2 </sup>+ x − 1</p>
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<p>Let’s check if the answer is correct:</p>
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<p class="so">y = Ae<sup>x</sup> + Be<sup>-x</sup> − 2x<sup>2 </sup>+ x − 1</p>
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<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = Ae<sup>x</sup> − Be<sup>-x</sup> − 4x + 1</p>
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<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = Ae<sup>x</sup> + Be<sup>-x</sup> − 4</p>
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<p>Putting it together:</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> − y = Ae<sup>x</sup> + Be<sup>-x</sup> − 4 − (Ae<sup>x</sup> + Be<sup>-x</sup> − 2x<sup>2 </sup>+ x − 1)</p>
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<p class="center">= Ae<sup>x</sup> + Be<sup>-x</sup> − 4 − Ae<sup>x</sup> − Be<sup>-x</sup> + 2x<sup>2 </sup>− x + 1</p>
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<p class="center">= 2x<sup>2</sup> − x − 3</p>
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</div>
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<p>So in this case we have shown that the answer is correct, but how do we
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find the particular solutions?</p>
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<p>We can try <em>guessing</em> ... !</p>
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<p>This method is only easy to apply if f(x) is one of the following:</p>
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<div class="tbl">
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<div class="row"><span class="left">Either:</span><span class="right">f(x)
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is a polynomial function.</span></div>
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<div class="row"><span class="left">Or:</span><span class="right">f(x)
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is a linear combination of sine and cosine functions.</span></div>
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<div class="row"><span class="left">Or:</span><span class="right">f(x)
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is an exponential function.</span></div>
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</div>
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<p>And here is a guide to help us with a guess:</p>
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<div class="simple">
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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<th align="center">f(x)</th>
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<th align="center">y(x) guess</th>
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</tr>
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<tr>
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<td style="text-align:center;">ae<sup>bx</sup></td>
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<td style="text-align:center;">Ae<sup>bx</sup></td>
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</tr>
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<tr>
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<td style="text-align:center;">a cos(cx) + b sin(cx)</td>
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<td style="text-align:center;">A cos(cx) + B sin(cx)</td>
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</tr>
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<tr>
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<td style="text-align:center;">kx<sup>n</sup> <i>(n=0, 1, 2,...)</i></td>
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<td style="text-align:center;">A<sub>n</sub>x<sup>n</sup> + A<sub>n−1</sub>x<sup>n−1</sup> + … + A<sub>0</sub></td>
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</tr>
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</tbody></table>
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</div>
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<p>But there is one important rule that must be applied:</p>
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<p class="center"><b>You must first find the general solution to the
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homogeneous equation.</b></p>
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<p>You will see why as we continue on.</p>
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<div class="example">
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<h3><strong>Example 1 (again)</strong>: 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>
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<p>1. Find the general solution of</p>
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<p><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − y = 0</p>
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<p>The characteristic equation is: r<sup>2</sup> − 1 = 0</p>
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<p>Factor: (r − 1)(r + 1) = 0</p>
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<p class="so">r = 1 or −1</p>
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<p>So the general solution of the differential equation is</p>
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<p>y = Ae<sup>x</sup> + Be<sup>-x</sup></p>
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<p>2. Find the particular solution of</p>
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<p><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − y = 2x<sup>2</sup> − x − 3</p>
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<p>We make a guess:</p>
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<p>Let y = ax<sup>2</sup> + bx + c</p>
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<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = 2ax + b</p>
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<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 2a</p>
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<p>Substitute these values into <span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − y = 2x<sup>2</sup> − x − 3</p>
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<p class="so">2a − (ax<sup>2</sup> + bx + c) =
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2x<sup>2</sup> − x − 3</p>
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<p class="so">2a − ax<sup>2</sup> − bx − c =
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2x<sup>2</sup> − x − 3</p>
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<p class="so">− ax<sup>2</sup> − bx + (2a − c) = 2x<sup>2</sup> − x − 3</p>
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<p>Equate coefficients:</p>
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<div class="beach">
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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<td style="text-align:center; width:150px;">x<sup>2</sup> coefficients: </td>
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<td style="text-align: right;">−a = 2 <span style="font-size:140%;">⇒ </span>a = −2 ... (1) </td>
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</tr>
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<tr>
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<td style="text-align:center; width:150px;">x coefficients: </td>
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<td style="text-align: right;">−b = −1 <span style="font-size:140%;">⇒ </span>b = 1 ... (2)</td>
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</tr>
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<tr>
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<td style="text-align:center; width:150px;">Constant coefficients:</td>
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<td style="text-align: right;">2a − c = −3
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... (3)</td>
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</tr>
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</tbody>
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</table>
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</div><br>
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<p>Substitute a = −2 from (1) into (3)</p>
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<p class="so">−4 − c = −3</p>
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<p class="so">c = −1</p>
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<p>a = −2, b = 1 and c = −1, so the particular solution of the
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differential equation is</p>
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<p style="text-align: center;"><strong>y = − 2x<sup>2</sup> + x − 1</strong></p>
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<p>Finally, we combine our two answers to get the complete solution:</p>
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<p style="text-align: center;"><strong>y = Ae<sup>x</sup> + Be<sup>-x</sup> − 2x<sup>2 </sup>+ x − 1</strong></p>
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<p>Why did we guess y = ax<sup>2</sup> + bx + c (a quadratic function)
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and not include a cubic term (or higher)?</p>
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<p>The answer is simple. The function f(x) on the right side of the
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differential equation has no cubic term (or higher); so, if y did have
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a cubic term, its coefficient would have to be zero.</p>
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<p><b>Hence, for a differential equation of the type<strong> </strong></b><strong><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) </strong><b><strong></strong>where
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f(x) is a polynomial of degree n, our guess for y will also be a
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polynomial of degree n.</b></p>
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</div><br>
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<div class="example">
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<h3><strong>Example 2:</strong> 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> − 13<span class="intbl"><em>dy</em><strong>dx</strong></span> − 5y = 5x<sup>3</sup> +
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39x<sup>2</sup> − 36x − 10</p>
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1. Find the general solution of 6<span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 13<span class="intbl"><em>dy</em><strong>dx</strong></span> − 5y = 0
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<p>The characteristic equation is: 6r<sup>2</sup> − 13r − 5 = 0</p>
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<p>Factor: (2r − 5)(3r + 1) = 0</p>
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<p class="so">r = <span class="intbl"><em>5</em><strong>2</strong></span> or −<span class="intbl"><em>1</em><strong>3</strong></span></p>
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<p>So the general solution of the differential equation is</p>
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<p style="text-align: center;"><strong>y = Ae<sup>(5/2)x</sup> + Be<sup>(−1/3)x </sup></strong></p>
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<p>2. Find the particular solution of 6<span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 13<span class="intbl"><em>dy</em><strong>dx</strong></span> − 5y = 5x<sup>3</sup> +
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39x<sup>2</sup> − 36x − 10</p>
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<p>Guess a cubic polynomial because 5x<sup>3</sup> + 39x<sup>2</sup> − 36x − 10 is cubic.</p>
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<p>Let y = ax<sup>3</sup> + bx<sup>2</sup> + cx + d</p>
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<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = 3ax<sup>2</sup> + 2bx + c</p>
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<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 6ax + 2b</p>
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<p>Substitute these values into 6<span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> − 13<span class="intbl"><em>dy</em><strong>dx</strong></span> −5y = 5x<sup>3</sup> +
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39x<sup>2</sup> −36x −10</p>
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<p class="so">6(6ax + 2b) − 13(3ax<sup>2</sup> + 2bx + c) − 5(ax<sup>3</sup> + bx<sup>2</sup> + cx + d) = 5x<sup>3</sup> + 39x<sup>2</sup> − 36x − 10</p>
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<p class="so">36ax + 12b − 39ax<sup>2 </sup>− 26bx − 13c − 5ax<sup>3</sup> − 5bx<sup>2</sup> − 5cx − 5d = 5x<sup>3</sup> + 39x<sup>2</sup> − 36x − 10</p>
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<p class="so">−5ax<sup>3</sup> + (−39a − 5b)x<sup>2</sup> + (36a − 26b −
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5c)x + (12b − 13c − 5d) = 5x<sup>3</sup> + 39x<sup>2</sup> − 36x − 10</p>
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<p>Equate coefficients:</p>
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<div class="beach">
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|
||
<table style="border: 0; margin:auto;">
|
||
<tbody>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">x<sup>3</sup> coefficients: </td>
|
||
<td style="text-align: right;">−5a = 5 <span style="font-size:140%;">⇒ </span><strong>a = −1</strong></td>
|
||
</tr>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">x<sup>2</sup> coefficients: </td>
|
||
<td style="text-align: right;">−39a −5b = 39 <span style="font-size:140%;">⇒ </span><strong>b = 0</strong></td>
|
||
</tr>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">x coefficients: </td>
|
||
<td style="text-align: right;">36a −26b −5c = −36 <span style="font-size:140%;">⇒ </span><strong>c = 0</strong></td>
|
||
</tr>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">Constant coefficients:</td>
|
||
<td style="text-align: right;">12b − 13c −5d = −10 <span style="font-size:140%;">⇒ </span><strong>d = 2</strong></td>
|
||
</tr>
|
||
</tbody>
|
||
</table>
|
||
</div>
|
||
<p>So the particular solution is:</p>
|
||
<p style="text-align: center;"><strong>y = −x<sup>3</sup> + 2</strong></p>
|
||
<p>Finally, we combine our two answers to get the complete solution:</p>
|
||
<p style="text-align: center;"><strong>y = Ae<sup>(5/2)x</sup> + Be<sup>(−1/3)x</sup> − x<sup>3</sup> + 2</strong></p>
|
||
<p>And here are some sample curves:</p>
|
||
<p class="center"><img src="images/diff-eq-undetermined-ex2.svg" height="340" width="550" ></p>
|
||
</div><br>
|
||
<div class="example">
|
||
|
||
<h3><strong>Example 3:</strong> 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> − 10y = −130cos(x) + 16e<sup>3x</sup></h3><br>
|
||
In this case we need to solve three differential equations:
|
||
<p>1. Find the general solution to <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> − 10y = 0</p>
|
||
<p>2. Find the particular solution to <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> − 10y = −130cos(x)</p>
|
||
<p>3. Find the particular solution to <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> − 10y = 16e<sup>3x</sup></p>
|
||
<p> </p>
|
||
<p>So, here’s how we do it:</p>
|
||
<p>1. Find the general solution to <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> − 10y = 0</p>
|
||
<p>The characteristic equation is: r<sup>2</sup> + 3r − 10 = 0</p>
|
||
<p>Factor: (r − 2)(r + 5) = 0</p>
|
||
<p class="so">r = 2 or −5</p>
|
||
<p>So the general solution of the differential equation is:</p>
|
||
<p style="text-align: center;"><strong>y = Ae<sup>2x</sup>+Be<sup>-5x</sup><br>
|
||
</strong></p>
|
||
<div style="text-align: center;"><strong> </strong></div>
|
||
<p>2. Find the particular solution to <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> − 10y = −130cos(x)</p>
|
||
<p>Guess. Since f(x) is a cosine function, we guess that <em>y</em> is
|
||
a linear combination of sine and cosine functions:</p>
|
||
<p>Try y = acos(x) + bsin(x)</p>
|
||
<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = − asin(x) + bcos(x)</p>
|
||
<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = − acos(x) − bsin(x)</p>
|
||
<p>Substitute these values into <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> − 10y = −130cos(x)</p>
|
||
<p class="so">−acos(x) − bsin(x) +
|
||
3[−asin(x) + bcos(x)] − 10[acos(x)+bsin(x)] = −130cos(x)</p>
|
||
<p class="so">cos(x)[−a + 3b − 10a] +
|
||
sin(x)[−b − 3a − 10b] = −130cos(x)</p>
|
||
<p class="so">cos(x)[−11a + 3b] +
|
||
sin(x)[−11b − 3a] = −130cos(x)</p>
|
||
<p>Equate coefficients:</p>
|
||
<div class="beach">
|
||
|
||
<table style="border: 0; margin:auto;">
|
||
<tbody>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">Coefficients of cos(x): </td>
|
||
<td style="text-align: right;">−11a + 3b = −130
|
||
... (1)<strong></strong></td>
|
||
</tr>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">Coefficients of sin(x):</td>
|
||
<td style="text-align: right;">−11b − 3a = 0
|
||
... (2)<strong></strong></td>
|
||
</tr>
|
||
</tbody>
|
||
</table>
|
||
</div>
|
||
<p>From equation (2), a = −<span class="intbl"><em>11b</em><strong>3</strong></span></p>
|
||
<p>Substitute into equation (1)</p>
|
||
<p class="so"><span class="intbl"><em>121b</em><strong>3</strong></span> + 3b = −130</p>
|
||
<p class="so"><span class="intbl"><em>130b</em><strong>3</strong></span> = −130</p>
|
||
<p class="so">b = −3</p>
|
||
<p class="so">a = −<span class="intbl"><em>11(−3)</em><strong>3</strong></span> =
|
||
11</p>
|
||
<p>So the particular solution is:</p>
|
||
<p style="text-align: center;"><strong>y = 11cos(x) − 3sin(x)</strong></p>
|
||
<p>3. Find the particular solution to <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> − 10y = 16e<sup>3x</sup></p>
|
||
<p>Guess.</p>
|
||
<p>Try y = ce<sup>3x</sup></p>
|
||
<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = 3ce<sup>3x</sup></p>
|
||
<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 9ce<sup>3x</sup></p>
|
||
<p>Substitute these values into <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> − 10y = 16e<sup>3x</sup></p>
|
||
<p class="so">9ce<sup>3x</sup> + 9ce<sup>3x</sup> − 10ce<sup>3x</sup> = 16e<sup>3x</sup></p>
|
||
<p class="so">8ce<sup>3x</sup> = 16e<sup>3x</sup></p>
|
||
<p class="so">c = 2</p>
|
||
So the particular solution is:
|
||
<p class="center"><b>y = 2e<sup>3x</sup></b></p>
|
||
<p>Finally, we combine our three answers to get the complete solution:</p>
|
||
<p style="text-align: center;"><strong>y = Ae<sup>2x</sup> + Be<sup>-5x</sup> + 11cos(x) − 3sin(x) + 2e<sup>3x</sup></strong></p>
|
||
</div><br>
|
||
<div class="example">
|
||
|
||
<h3><strong>Example 4:</strong> 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> − 10y = −130cos(x) + 16e<sup>2x</sup></h3>
|
||
<p>This is exactly the same as Example 3 except for the final term,
|
||
which has been replaced by 16e<sup>2x</sup>.</p>
|
||
<p>So Steps 1 and 2 are exactly the same. On to step 3:</p>
|
||
<p>3. Find the particular solution to <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> − 10y = 16e<sup>2x</sup></p>
|
||
<p>Guess.</p>
|
||
<p>Try y = ce<sup>2x</sup></p>
|
||
<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = 2ce<sup>2x</sup></p>
|
||
<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 4ce<sup>2x</sup></p>
|
||
<p>Substitute these values into <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> − 10y = 16e<sup>2x</sup></p>
|
||
<p class="so">4ce<sup>2x</sup> + 6ce<sup>2x</sup> − 10ce<sup>2x</sup> = 16e<sup>2x</sup></p>
|
||
<p class="so">0 = 16e<sup>2x</sup></p>
|
||
<p>Oh dear! Something seems to have gone wrong. How can 16e<sup>2x</sup> = 0?</p>
|
||
<p>Well, it can’t, and there is nothing wrong here except that there is
|
||
no particular solution to 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> − 10y = 16e<sup>2x</sup></p>
|
||
...Wait a minute!<br>
|
||
The general solution to the homogeneous 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> − 10y = 0<b>,</b> which is
|
||
y = Ae<sup>2x</sup> + Be<sup>-5x</sup>, already has a term Ae<sup>2x</sup>,
|
||
so our guess y = ce<sup>2x</sup> already satisfies 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> − 10y = 0 (it was just a
|
||
different constant.)<br>
|
||
<p>So we must guess y = cxe<sup>2x</sup><br>
|
||
<br>
|
||
Let's see what happens:</p>
|
||
<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = ce<sup>2x</sup> + 2cxe<sup>2x</sup></p>
|
||
<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 2ce<sup>2x </sup>+ 4cxe<sup>2x</sup> + 2ce<sup>2x</sup> = 4ce<sup>2x</sup> + 4cxe<sup>2x</sup></p>
|
||
<p>Substitute these values into <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> − 10y = 16e<sup>2x</sup></p>
|
||
<p class="so">4ce<sup>2x</sup> + 4cxe<sup>2x </sup>+ 3ce<sup>2x</sup> + 6cxe<sup>2x </sup>− 10cxe<sup>2x</sup> =
|
||
16e<sup>2x</sup></p>
|
||
<p class="so">7ce<sup>2x</sup> = 16e<sup>2x</sup></p>
|
||
<p class="so">c = <span class="intbl"><em>16</em>7</span></p>
|
||
<p>So in the present case our particular solution is</p>
|
||
<p class="center large">y = <span class="intbl"><em>16</em><strong>7</strong></span>xe<sup>2x</sup></p>
|
||
Thus, our final complete solution in this case is:
|
||
<p class="center large">y = Ae<sup>2x</sup> + Be<sup>-5x</sup> +
|
||
11cos(x) − 3sin(x) + <span class="intbl"><em>16</em><strong>7</strong></span>xe<sup>2x</sup><br>
|
||
<strong></strong></p>
|
||
</div><br>
|
||
<div class="example">
|
||
|
||
<h3><strong>Example 5:</strong> 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 = 5e<sup>-2x</sup></h3>
|
||
<p>1. Find the general solution to <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>
|
||
<p>The characteristic equation is: r<sup>2</sup> − 6r + 9 = 0</p>
|
||
<p class="so">(r − 3)<sup>2</sup> = 0</p>
|
||
<p class="so">r = 3,
|
||
which is a repeated root.</p>
|
||
<p>Then the general solution of the differential equation is <strong>y = Ae<sup>3x</sup> + Bxe<sup>3x</sup></strong></p>
|
||
<p>2. Find the particular solution to <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 = 5e<sup>-2x</sup></p>
|
||
<p>Guess.</p>
|
||
<p>Try y = ce<sup>-2x</sup></p>
|
||
<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = −2ce<sup>-2x</sup></p>
|
||
<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = 4ce<sup>-2x</sup></p>
|
||
<p>Substitute these values into <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 = 5e<sup>-2x</sup></p>
|
||
<p class="so">4ce<sup>-2x</sup> + 12ce<sup>-2x</sup> + 9ce<sup>-2x</sup> = 5e<sup>-2x</sup></p>
|
||
<p class="so">25e<sup>-2x</sup> = 5e<sup>-2x</sup></p>
|
||
<p class="so">c = <span class="intbl"><em>1</em><strong>5</strong></span></p>
|
||
<p>So the particular solution is:</p>
|
||
<p class="center large">y= <span class="intbl"><em>1</em><strong>5</strong></span>e<sup>-2x </sup></p>
|
||
<p>Finally, we combine our two answers to get
|
||
the complete solution:</p>
|
||
<p class="center large">y= Ae<sup>3x</sup> + Bxe<sup>3x </sup>+ <span class="intbl"><em>1</em><strong>5</strong></span>e<sup>-2x </sup></p>
|
||
</div><br>
|
||
<div class="example">
|
||
|
||
<h3><strong>Example 6: </strong>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> + 34y = 109cos(5x)</h3>
|
||
<p>1. Find the general solution to <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> + 34y = 0</p>
|
||
<p>The characteristic equation is: r<sup>2</sup> + 6r + 34 = 0</p>
|
||
<p>Use the <u><a href="../algebra/quadratic-equation.html">quadratic
|
||
equation formula</a></u></p>
|
||
<p>r = <span class="intbl"> <em>−b ± √(b<sup>2 </sup>− 4ac)</em><strong>2a</strong></span></p>
|
||
<p>with a = 1, b = 6 and c = 34</p>
|
||
<p>So</p>
|
||
<p>r = <span class="intbl"> <em>−6 ± √[6<sup>2 </sup>− 4(1)(34)]</em><strong>2(1)</strong></span></p>
|
||
<p>r = <span class="intbl"> <em>−6 ± √(36−136)</em><strong>2</strong></span></p>
|
||
<p>r = <span class="intbl"> <em>−6 ± √(−100)</em><strong>2</strong></span></p>
|
||
<p>r = −3 ± 5i</p>
|
||
<p>And we get:</p>
|
||
<p class="so">y =e<sup>-3x</sup>(Acos(5x) +
|
||
iBsin(5x))</p>
|
||
2. Find the particular solution to <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> + 34y = 109sin(5x)
|
||
<p>Since f(x) is a sine function, we assume that y is a linear
|
||
combination of sine and cosine functions:</p>
|
||
<p>Guess.</p>
|
||
<p>Try y = acos(5x) + bsin(5x)</p>
|
||
<p>Note: since we do not have sin(5x) or cos(5x) in the solution to the
|
||
homogeneous equation (we have e<sup>-3x</sup>cos(5x) and e<sup>-3x</sup>sin(5x),
|
||
which are different functions), our guess should work.</p>
|
||
<p>Let’s continue and see what happens:</p>
|
||
<p class="so"><span class="intbl"><em>dy</em><strong>dx</strong></span> = −5asin(5x) + 5bcos(5x)</p>
|
||
<p class="so"><span class="intbl"><em>d<sup>2</sup>y</em><strong>dx<sup>2</sup></strong></span> = −25acos(5x) − 25bsin(5x)</p>
|
||
<p>Substitute these values into <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> + 34y = 109sin(5x)</p>
|
||
<p class="so">−25acos(5x) − 25bsin(5x) +
|
||
6[−5asin(5x) + 5bcos(5x)] + 34[acos(5x) + bsin(5x)] = 109sin(5x)</p>
|
||
<p class="so">cos(5x)[−25a + 30b + 34a] +
|
||
sin(5x)[−25b − 30a + 34b] = 109sin(5x)</p>
|
||
<p class="so">cos(5x)[9a + 30b] + sin(5x)[9b −
|
||
30a] = 109sin(5x)<br>
|
||
<span class="intbl"></span></p>
|
||
<p>Equate coefficients of cos(5x) and sin(5x):</p>
|
||
<div class="beach">
|
||
|
||
<table style="border: 0; margin:auto;">
|
||
<tbody>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">Coefficients of cos(5x): </td>
|
||
<td style="text-align: right;">9a + 30b = 0
|
||
... (1)<strong></strong></td>
|
||
</tr>
|
||
<tr>
|
||
<td style="text-align:center; width:150px;">Coefficients of sin(5x):</td>
|
||
<td style="text-align: right;">9b − 30a = 109
|
||
... (2)<strong></strong></td>
|
||
</tr>
|
||
</tbody>
|
||
</table>
|
||
</div>
|
||
<p>From equation (1), b = <span class="intbl"><em>−3a</em><strong>10</strong></span></p>
|
||
<p>Substitute into equation (2)</p>
|
||
<p class="so">9(<span class="intbl"><em>−3a</em><strong>10</strong></span>) − 30a = 109</p>
|
||
<p class="so">−27a − 300a = 1090</p>
|
||
<p class="so">−327a = 1090</p>
|
||
<p class="so">a = <span class="intbl"><em>−10</em><strong>3</strong></span></p>
|
||
<p class="so">b = 1</p>
|
||
So the particular solution is:
|
||
<p class="center large">y = <span class="intbl"><strong></strong></span><span class="intbl"><em>−10</em><strong>3</strong></span>cos(5x) + sin(5x)</p>
|
||
<p>Finally, we combine our answers to get the complete solution:</p>
|
||
<p class="center large">y = e<sup>-3x</sup>(Acos(5x) +
|
||
iBsin(5x)) − <span class="intbl"><em>10</em><strong>3</strong></span>cos(5x) + sin(5x) </p>
|
||
</div>
|
||
<p> </p>
|
||
<div class="questions">9509, 9510, 9511, 9512, 9513, 9514, 9515, 9516, 9517, 9518</div>
|
||
|
||
<div class="related">
|
||
<a href="homogeneous-function.html">Homogeneous Functions</a>
|
||
<a href="differential-equations.html">Differential Equation</a>
|
||
<a href="differential-equations-solution-guide.html">Differential Equations Solution Guide</a>
|
||
<a href="index.html">Calculus Index</a>
|
||
</div>
|
||
<!-- #EndEditable -->
|
||
|
||
</article>
|
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