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<h1>Exact Equations and Integrating Factors</h1>
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<p>Hi! You might like to learn about <a href="differential-equations.html">differential equations</a> and <a href="derivatives-partial.html">partial derivatives</a> first!</p>
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<h2>Exact Equation</h2>
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<p>An "exact" equation is where a first-order differential equation like this:</p>
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<p class="center larger">M(x, y)dx + N(x, y)dy = 0</p>
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<p>has some special function <span class="larger">I(x, y)</span> whose <a href="derivatives-partial.html">partial derivatives</a> can be put in place of M and N like this:</p>
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<p class="center larger"><span class="intbl"><em>∂I</em><strong>∂x</strong></span>dx + <span class="intbl"><em>∂I</em><strong>∂y</strong></span>dy = 0</p>
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<p>and our job is to find that magical function <span class="larger">I(x, y)</span> if it exists.</p>
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<div class="def">
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<h3>We can know at the start if it is an exact equation or not!</h3>
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<p>Imagine we do these further partial derivatives:</p>
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<p class="center large"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = <span class="intbl"><em>∂<sup>2</sup>I</em><strong>∂y ∂x</strong></span></p>
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<p class="center large"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = <span class="intbl"><em>∂<sup>2</sup>I</em><strong>∂y ∂x</strong></span></p>
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<p>they end up <b>the same</b>! And so this will be true:</p>
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<p class="center large"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = <span class="intbl"><em>∂N</em><strong>∂x</strong></span></p>
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<p>When it is true we have an an "exact equation" and we can proceed.</p>
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<p>And to discover <b>I(x, y)</b> we do <b>EITHER</b>:</p>
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<ul>
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<li>I(x, y) = <span class="integral">∫</span>M(x, y) dx (with <b>x</b> as an independent variable), <b>OR</b></li>
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<li>I(x, y) = <span class="integral">∫</span>N(x, y) dy (with <b>y</b> as an independent variable)</li>
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</ul>
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<p>And then there is some extra work (we will show you) to arrive at the <b>general solution</b></p>
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<p class="center large">I(x, y) = C</p>
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</div>
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<p> </p>
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<p>Let's see it in action.</p>
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<div class="example">
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<h3><b>Example 1:</b> Solve</h3>
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<p class="center large">(3x<sup>2</sup>y<sup>3</sup> − 5x<sup>4</sup>) dx + (y + 3x<sup>3</sup>y<sup>2</sup>) dy = 0</p>
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<p>In this case we have:</p>
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<ul>
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<li>M(x, y) = 3x<sup>2</sup>y<sup>3</sup> −
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5x<sup>4</sup></li>
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<li>N(x, y) = y + 3x<sup>3</sup>y<sup>2</sup></li>
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</ul>
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<p>We evaluate the partial derivatives to check for exactness.</p>
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<ul>
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<li><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = 9x<sup>2</sup>y<sup>2</sup></li>
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<li><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = 9x<sup>2</sup>y<sup>2</sup></li>
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</ul><br>
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<p class="so">They are the same! So our equation is exact.</p>
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<p>We can proceed.</p>
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<p>Now we want to discover I(x, y)</p>
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<p>Let's do the integration with <b>x</b> as an independent variable:</p>
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<p class="center large">I(x, y) = <span class="integral">∫</span>M(x, y) dx</p>
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<p class="center large">= <span class="integral">∫</span>(3x<sup>2</sup>y<sup>3</sup> − 5x<sup>4</sup>) dx</p>
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<p class="center large">= x<sup>3</sup>y<sup>3</sup> − x<sup>5</sup> +<em> f(y)</em></p>
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<p class="def">Note: <em> f(y)</em> is our version of the constant of integration "C" because (due to the partial derivative) we had <strong>y</strong> as a fixed parameter that we know is really a variable.</p>
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<p>So now we need to discover f(y)</p>
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<p>At the very start of this page we said that N(x, y) can be replaced by <span class="intbl"><em>∂I</em><strong>∂y</strong></span>, so:</p>
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<p class="center large"><span class="intbl"><em>∂I</em><strong>∂y</strong></span> = N(x, y)</p>
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<p>Which gets us:</p>
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<p class="center large">3x<sup>3</sup>y<sup>2</sup> + <span class="intbl"><em>df</em><strong>dy</strong></span> = y + 3x<sup>3</sup>y<sup>2</sup></p>
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<p>Cancelling terms:</p>
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<p class="center large"><span class="intbl"><em>df</em><strong>dy</strong></span> = y</p>
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<p>Integrating both sides:</p>
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<p class="center large">f(y) = <span class="intbl"><em>y<sup>2</sup></em><strong>2</strong></span> + C</p>
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<p><b>We have f(y).</b> Now just put it in place:</p>
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<p class="center large">I(x, y) = x<sup>3</sup>y<sup>3</sup> − x<sup>5</sup> + <span class="intbl"><em>y<sup>2</sup></em><strong>2</strong></span> + C</p>
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<p>and the <b>general solution</b> (as mentioned before this example) is:</p>
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<p class="center large">I(x, y) = C</p>
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<p class="beach dotpoint"><b>Ooops! That "C" can be a different value to the "C" just before.</b> But they both mean "any constant", so let's call them C<sub>1</sub> and C<sub>2</sub> and then roll them into a new C below by saying C=C<sub>1</sub>+C<sub>2</sub></p>
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<p>So we get:</p>
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<p class="center larger">x<sup>3</sup>y<sup>3</sup> − x<sup>5</sup> + <span class="intbl"><em>y<sup>2</sup></em><strong>2</strong></span> = C</p>
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<p>And that's how this method works!</p> </div>
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<p>Since that was our first example, let's go further and make sure our solution is correct.</p>
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<p>Let's derive I(x, y) with respect
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to x, that is:</p>
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<div class="center80">
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<p class="center">Evaluate <span class="intbl"><em>∂I</em><strong>∂x</strong></span></p>
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</div>
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<p>Start with:</p>
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<p class="center large">I(x, y) = x<sup>3</sup>y<sup>3</sup> − x<sup>5</sup> + <span class="intbl"><em>y<sup>2</sup></em><strong>2</strong></span></p>
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<p>Using <a href="implicit-differentiation.html">implicit
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differentiation</a> we get</p>
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<p class="center large"><span class="intbl"><em>∂I</em><strong>∂x</strong></span> = x<sup>3</sup>3y<sup>2</sup>y'
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+ 3x<sup>2</sup>y<sup>3</sup> − 5x<sup>4</sup> + yy'</p>
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<p>Simplify</p>
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<p class="center large"><span class="intbl"><em>∂I</em><strong>∂x</strong></span> = 3x<sup>2</sup>y<sup>3</sup> − 5x<sup>4</sup> + y'(y + 3x<sup>3</sup>y<sup>2</sup>)</p>
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<p>We use the facts that <span class="larger">y' = <span class="intbl"><em>dy</em><strong>dx</strong></span></span> and <span class="larger"><span class="intbl"><em>∂I</em><strong>∂x</strong></span> = 0</span>, then multiply everything by <span class="larger">dx</span> to finally get:</p>
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<p class="center large">(y + 3x<sup>3</sup>y<sup>2</sup>)dy
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+ (3x<sup>2</sup>y<sup>3</sup> − 5x<sup>4</sup>)dx =
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0</p>
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<p>which is our original differential equation.</p>
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<p>And so we know our solution is correct.</p>
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<p><br></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">(3x<sup>2</sup> − 2xy + 2)dx + (6y<sup>2</sup> − x<sup>2</sup> + 3)dy = 0</p>
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<ul>
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<li>M = 3x<sup>2</sup> − 2xy + 2</li>
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<li>N = 6y<sup>2</sup> − x<sup>2</sup> + 3</li>
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</ul>
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<p>So:</p>
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<ul>
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<li><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = −2x</li>
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<li><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = −2x</li>
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</ul><br>
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<p class="so">The equation is exact!</p>
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<p>Now we are going to find the function I(x, y)</p>
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<p>This time let's try I(x, y) = <span class="integral">∫</span>N(x, y)dy</p>
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<p>So I(x, y) = <span class="integral">∫</span>(6y<sup>2</sup> − x<sup>2</sup> + 3)dy</p>
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<p class="center large">I(x, y) = 2y<sup>3</sup> − x<sup>2</sup>y
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+ 3y + g(x) <span class="center hilite">(equation 1)</span></p>
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<p>Now we differentiate I(x, y) with respect to x and set that equal to M:</p>
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<p class="center large"><span class="intbl"><em>∂I</em><strong>∂x</strong></span> = M(x, y)</p>
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<p class="center large">0 − 2xy + 0 + g'(x) = 3x<sup>2</sup> − 2xy + 2</p>
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<p class="center large">−2xy + g'(x) = 3x<sup>2</sup> − 2xy + 2</p>
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<p class="center large">g'(x) = 3x<sup>2</sup> + 2</p>
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<p>And integration yields:</p>
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<p class="center large">g(x) = x<sup>3</sup> +
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2x + C <span class="center hilite">(equation 2)</span></p>
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<p>Now we can replace the g(x) in equation 2 in equation 1:</p>
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<p class="center large">I(x, y) = 2y<sup>3</sup> − x<sup>2</sup>y + 3y + x<sup>3</sup> + 2x + C</p>
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<p>And the general solution is of the form</p>
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<p class="center large">I(x, y) = C</p>
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<p>and so (remembering that the previous two "C"s are different constants that can be rolled into one by using C=C<sub>1</sub>+C<sub>2</sub>) we get:</p>
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<p class="center larger">2y<sup>3</sup> − x<sup>2</sup>y + 3y + x<sup>3</sup> + 2x = C</p>
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<p><strong>Solved!</strong></p>
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</div><br>
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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">(xcos(y) − y)dx + (xsin(y) + x)dy = 0</p>
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<p>We have:</p>
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<p>M = (xcos(y) − y)dx</p>
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<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = −xsin(y) − 1</p>
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<p>N = (xsin(y) + x)dy</p>
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<p class="so"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = sin(y) +1</p><br>
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Thus
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<p class="center large"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> ≠ <span class="intbl"><em>∂N</em><strong>∂x</strong></span></p><br>
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<div class="center80"><strong>So this equation
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is not exact!</strong><br>
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</div><br>
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</div><br>
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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">[y<sup>2</sup> − x<sup>2</sup>sin(xy)]dy +
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[cos(xy) − xy sin(xy) + e<sup>2x</sup>]dx = 0</p>
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<p>M = cos(xy) − xy sin(xy) + e<sup>2x</sup></p>
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<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = −x<sup>2</sup>y cos(xy) − 2x sin(xy)</p>
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<p>N = y<sup>2</sup> − x<sup>2</sup>sin(xy)</p>
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<p class="so"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = −x<sup>2</sup>y cos(xy) − 2x sin(xy)</p>
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<p class="so">They are the same! So our equation is exact.</p>
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<p>This time we will evaluate I(x, y) = <span class="integral">∫</span>M(x, y)dx</p>
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<p class="center large">I(x, y) = <span class="integral">∫</span>(cos(xy) − xy sin(xy) + e<sup>2x</sup>)dx</p>
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<p> Using Integration by Parts we get:</p>
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<p class="center large">I(x, y) = <span class="intbl"><em>1</em><strong>y</strong></span>sin(xy) +
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x cos(xy) − <span class="intbl"><em>1</em><strong>y</strong></span>sin(xy)
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+ <span class="intbl"><em>1</em><strong>2</strong></span>e<sup>2x</sup> + f(y)</p>
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<p class="center large">I(x, y) = x cos(xy) + <span class="intbl"><em>1</em><strong>2</strong></span>e<sup>2x</sup> + f(y)</p>
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<p>Now we evaluate the derivative with respect to y</p>
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<p class="center large"><span class="intbl"><em>∂I</em><strong>∂y</strong></span> =
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−x<sup>2</sup>sin(xy) + f'(y)</p>
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<p>And that is equal to N, that equal to M: </p>
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<p class="center large"><span class="intbl"><em>∂I</em><strong>∂y</strong></span> = N(x, y)</p>
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<p class="center large">−x<sup>2</sup>sin(xy) +
|
||
f'(y) = y<sup>2</sup> − x<sup>2</sup>sin(xy)</p>
|
||
<p class="center large">f'(y) = y<sup>2</sup> − x<sup>2</sup>sin(xy) + x<sup>2</sup>sin(xy)</p>
|
||
<p class="center large">f'(y) = y<sup>2</sup> </p>
|
||
<p class="center large">f(y) = <span class="intbl"><em>1</em><strong>3</strong></span>y<sup>3</sup></p>
|
||
<p>So our general solution of I(x, y) = C becomes:</p>
|
||
<p class="center larger">xcos(xy) + <span class="intbl"><em>1</em><strong>2</strong></span>e<sup>2x</sup> + <span class="intbl"><em>1</em><strong>3</strong></span>y<sup>3</sup> = C</p>
|
||
<p>Done!</p>
|
||
</div>
|
||
|
||
|
||
<h2>Integrating Factors</h2>
|
||
|
||
<p>Some equations that are not exact may be multiplied by some factor, a
|
||
function <b>u(x, y)</b>, to make them exact.</p>
|
||
<p>When this function u(x, y) exists it is called an <b>integrating factor</b>.
|
||
It will make valid the following expression:</p>
|
||
<p class="center"><strong><span class="intbl"><em>∂(u·N(x, y))</em>∂x</span> = <span class="intbl"><em>∂(u·M(x, y))</em>∂y</span></strong></p>
|
||
<div class="def"> There are some special cases:<br>
|
||
<br>
|
||
<ul>
|
||
<li><strong>u(x, y) = x<sup>m</sup>y<sup>n</sup> </strong></li>
|
||
<li><strong> u(x, y) = u(x) </strong> (that is, u is a function only of x)</li>
|
||
<li><strong>u(x, y) = u(y) </strong> (that
|
||
is, u is a function only of y)</li>
|
||
</ul>
|
||
</div>
|
||
<p>Let's look at those cases ...</p>
|
||
<p> </p>
|
||
|
||
<h3>Integrating Factors using u(x, y) = x<sup>m</sup>y<sup>n</sup></h3>
|
||
<div class="example">
|
||
|
||
<h3><b>Example 5:</b> <span class="center large">(y<sup>2</sup> + 3xy<sup>3</sup>)dx + (1 −
|
||
xy)dy = 0 </span></h3>
|
||
<p><br>
|
||
M = y<sup>2</sup> + 3xy<sup>3</sup></p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = 2y + 9xy<sup>2</sup></p>
|
||
<p>N = 1 − xy</p>
|
||
<p class="so"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = −y</p>
|
||
<p>So it's clear that <span class="intbl"><em>∂M</em><strong>∂y</strong></span> ≠ <span class="intbl"><em>∂N</em><strong>∂x</strong></span></p>
|
||
<p>But we can try to <em>make it exact</em> by multiplying each part of the
|
||
equation by<b> x<sup>m</sup>y<sup>n</sup></b>:</p>
|
||
<div class="center large">(x<sup>m</sup>y<sup>n</sup>y<sup>2</sup> + x<sup>m</sup>y<sup>n</sup>3xy<sup>3</sup>) dx + (x<sup>m</sup>y<sup>n</sup> − x<sup>m</sup>y<sup>n</sup>xy) dy = 0</div>
|
||
<!---− (x^m~y^n~y^2 + x^m~y^n~3xy^3~) dx + (x^m~y^n − x^m~y^n~xy) dy = 0---->
|
||
<p>Which "simplifies" to:</p>
|
||
<p class="center large">(x<sup>m</sup>y<sup>n+2</sup> + 3x<sup>m+1</sup>y<sup>n+3</sup>)dx
|
||
+ (x<sup>m</sup>y<sup>n</sup> − x<sup>m+1</sup>y<sup>n+1</sup>)dy =
|
||
0</p>
|
||
<p>And now we have:</p>
|
||
<p>M = x<sup>m</sup>y<sup>n+2</sup> + 3x<sup>m+1</sup>y<sup>n+3</sup></p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = (n + 2)x<sup>m</sup>y<sup>n+1</sup> + 3(n + 3)x<sup>m+1</sup>y<sup>n+2</sup></p>
|
||
<p>N = x<sup>m</sup>y<sup>n</sup> − x<sup>m+1</sup>y<sup>n+1</sup></p>
|
||
<p class="so"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = mx<sup>m−1</sup>y<sup>n</sup> − (m + 1)x<sup>m</sup>y<sup>n+1</sup> </p>
|
||
<p>And we <b>want</b> <span class="intbl"><em>∂M</em><strong>∂y</strong></span> = <span class="intbl"><em>∂N</em><strong>∂x</strong></span></p>
|
||
<p class="large center">So let's choose the right values of <strong>m</strong><em> </em>and <strong>n</strong> to make the equation exact.</p>
|
||
<p>Set them equal:</p>
|
||
<p class="center large">(n + 2)x<sup>m</sup>y<sup>n+1</sup> + 3(n + 3)x<sup>m+1</sup>y<sup>n+2</sup> = mx<sup>m−1</sup>y<sup>n</sup> − (m + 1)x<sup>m</sup>y<sup>n+1</sup> </p>
|
||
<p>Re-order and simplify:</p>
|
||
<p class="center large">[(m + 1) + (n + 2)]x<sup>m</sup>y<sup>n+1</sup> + 3(n + 3)x<sup>m+1</sup>y<sup>n+2</sup> − mx<sup>m−1</sup>y<sup>n</sup> = 0 </p>
|
||
<p><br>
|
||
For it to be equal to zero, <b>every</b> coefficient must be equal to zero, so:</p>
|
||
<ol>
|
||
<li>(m + 1) + (n + 2) = 0</li>
|
||
<li>3(n + 3) = 0</li>
|
||
<li>m = 0</li>
|
||
</ol>
|
||
<p>That last one, <strong>m = 0</strong>, is a big help! With m=0 we can figure that <strong>n = −3</strong></p>
|
||
<p>And the result is:</p>
|
||
<p class="center large">x<sup>m</sup>y<sup>n</sup> =
|
||
y<sup>−3</sup></p>
|
||
<p>We now know to multiply our original differential equation by <b>y<sup>−3</sup></b>:</p>
|
||
<div class="center large">(y<sup>−3</sup>y<sup>2</sup> + y<sup>−3</sup>3xy<sup>3</sup>) dx + (y<sup>−3</sup> − y<sup>−3</sup>xy) dy</div>
|
||
<!---− (y^−3~y^2 + y^−3~3xy^3~) dx + (y^−3 - y^−3~xy) dy ---->
|
||
<p>Which becomes:</p>
|
||
<p class="center large">(y<sup>−1</sup> + 3x)dx + (y<sup>−3</sup> − xy<sup>−2</sup>)dy = 0</p><br>
|
||
<p>And this new equation <i>should</i> be exact, but let's check again:<br>
|
||
<br>
|
||
M = y<sup>−1</sup> + 3x</p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = −y<sup>−2</sup></p>
|
||
<p>N = y<sup>−3</sup> − xy<sup>−2</sup></p>
|
||
<p class="so"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = −y<sup>−2</sup></p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = <span class="intbl"><em>∂N</em><strong>∂x</strong></span></p>
|
||
<p><br>
|
||
They are the same! <strong>Our equation is now exact</strong>!<br>
|
||
<br>
|
||
So let's continue:</p>
|
||
<p class="center large">I(x, y) = <span class="integral">∫</span>N(x, y)dy</p>
|
||
<p class="center large">I(x, y) = <span class="integral">∫</span>(y<sup>−3</sup> − xy<sup>−2</sup>)dy</p>
|
||
<p class="center large">I(x, y) = <span class="intbl"><em>−1</em><strong>2</strong></span>y<sup>−2</sup> + xy<sup>−1</sup> + g(x)</p>
|
||
<p>Now, to determine the function g(x) we evaluate</p>
|
||
<p class="center large"><span class="intbl"><em>∂I</em><strong>∂x</strong></span> = y<sup>−1</sup> + g'(x)</p>
|
||
<p>And that equals M = y<sup>−1</sup> + 3x, so:</p>
|
||
<p class="center large">y<sup>−1</sup> + g'(x) = y<sup>−1</sup> + 3x</p>
|
||
<p>And so:</p>
|
||
<p class="center large">g'(x) = 3x</p>
|
||
<p class="center large">g(x) = <span class="intbl"><em>3</em><strong>2</strong></span>x<sup>2</sup></p>
|
||
<p>So our general solution of I(x, y) = C is:</p>
|
||
<p class="center larger"><span class="intbl"><em>−1</em><strong>2</strong></span>y<sup>−2</sup> + xy<sup>−1</sup> + <span class="intbl"><em>3</em><strong>2</strong></span>x<sup>2</sup> = C</p>
|
||
</div>
|
||
<p> </p>
|
||
|
||
<h3>Integrating Factors using u(x, y) = u(x)</h3>
|
||
<p>For <b>u(x, y) = u(x)</b> we must check for this important condition:</p>
|
||
<div class="def">
|
||
<p>The expression:</p>
|
||
<div class="center large">Z(x) = <span class="intbl"><em>1</em><strong>N</strong></span> [<span class="intbl"><em>∂M</em><strong>∂y</strong></span> − <span class="intbl"><em>∂N</em><strong>∂x</strong></span>]</div>
|
||
<!-- Z(x) = 1/N [DAYM/DAYy - DAYN/DAYx ] -->
|
||
<p>must <b>not</b> have the <b>y</b> term, so that the integrating factor is only a function of <b>x</b></p>
|
||
</div>
|
||
<p><br>
|
||
If the above condition is true then our integrating factor is:</p>
|
||
<p class="center large">u(x) = e<sup><span class="integral">∫</span>Z(x)dx</sup></p>
|
||
<p>Let's try an example:</p>
|
||
<div class="example">
|
||
|
||
<h3><b>Example 6:</b> (3xy − y<sup>2</sup>)dx + x(x − y)dy = 0</h3>
|
||
<div></div>
|
||
<p>M = 3xy − y<sup>2</sup></p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = 3x − 2y</p>
|
||
<p>N = x(x − y)</p>
|
||
<p class="so"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = 2x − y</p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> ≠ <span class="intbl"><em>∂N</em><strong>∂x</strong></span></p>
|
||
So, our equation is <b>not</b> exact.<br>
|
||
<br>
|
||
Let us work out Z(x):
|
||
<div class="center large">Z(x) = <span class="intbl"><em>1</em><strong>N</strong></span> [<span class="intbl"><em>∂M</em><strong>∂y</strong></span> − <span class="intbl"><em>∂N</em><strong>∂x</strong></span> ]</div>
|
||
<!-- Z(x) = 1/N [DAYM/DAYy - DAYN/DAYx ] -->
|
||
<div class="center large">= <span class="intbl"><em>1</em><strong>N</strong></span> [ 3x−2y − (2x−y) ]</div>
|
||
<!-- = 1/N [ 3x−2y − (2x−y) ] -->
|
||
<div class="center large">= <span class="intbl"><em>x−y</em><strong>x(x−y)</strong></span></div>
|
||
<!-- = x-y/x(x-y) -->
|
||
<div class="center large">= <span class="intbl"><em>1</em><strong>x</strong></span></div>
|
||
<!-- = 1/x -->
|
||
<p>So Z(x) is a function only of x, yay!</p>
|
||
<p><br>
|
||
So our <strong>integrating factor</strong> is<br>
|
||
<br>
|
||
u(x) = e<sup><span class="integral">∫</span>Z(x)dx</sup></p>
|
||
<p>= e<sup><span class="integral">∫</span>(1/x)dx</sup></p>
|
||
<p>= e<sup>ln(x)</sup></p>
|
||
<p>= <strong>x</strong></p>
|
||
<p>Now that we found the integrating factor, let's multiply the
|
||
differential equation by it.</p>
|
||
<p class="center large">x[(3xy − y<sup>2</sup>)dx +
|
||
x(x − y)dy = 0]</p>
|
||
<p>and we get</p>
|
||
<p class="center large">(3x<sup>2</sup>y − xy<sup>2</sup>)dx
|
||
+ (x<sup>3</sup> − x<sup>2</sup>y)dy = 0</p>
|
||
<p>It should now be exact. Let's test it:</p>
|
||
<p>M = 3x<sup>2</sup>y − xy<sup>2</sup></p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = 3x<sup>2</sup> − 2xy</p>
|
||
<p>N = x<sup>3</sup> − x<sup>2</sup>y</p>
|
||
<p class="so"><span class="intbl"><em>∂N</em><strong>∂x</strong></span> = 3x<sup>2</sup> − 2xy</p>
|
||
<p class="so"><span class="intbl"><em>∂M</em><strong>∂y</strong></span> = <span class="intbl"><em>∂N</em><strong>∂x</strong></span></p>
|
||
<p>So our equation is exact!</p>
|
||
<p>Now we solve in the same way as the previous examples.</p>
|
||
<p class="center large">I(x, y) = <span class="integral">∫</span>M(x, y)dx</p>
|
||
<p class="center large">= <span class="integral">∫</span>(3x<sup>2</sup>y − xy<sup>2</sup>)dx</p>
|
||
<p class="center large">= x<sup>3</sup>y − <span class="intbl"><em>1</em><strong>2</strong></span>x<sup>2</sup>y<sup>2</sup> + c<sub>1</sub></p>And we get the general solution I(x, y) = c :
|
||
<p class="center larger">x<sup>3</sup>y − <span class="intbl"><em>1</em><strong>2</strong></span>x<sup>2</sup>y<sup>2</sup> + c<sub>1</sub> = c</p>
|
||
<p>Combine the constants:</p>
|
||
<p class="center larger">x<sup>3</sup>y − <span class="intbl"><em>1</em><strong>2</strong></span>x<sup>2</sup>y<sup>2</sup> = c</p>
|
||
<p>Solved!</p>
|
||
</div>
|
||
<p> </p>
|
||
|
||
<h3>Integrating Factors using u(x, y) = u(y)</h3>
|
||
<p><b>u(x, y) = u(y)</b> is very
|
||
similar to the previous case <strong>u(x, y)</strong> <strong>=
|
||
u(x)</strong></p>
|
||
<p>So, in a similar way, we have:</p>
|
||
<div class="def">
|
||
<p>The expression</p>
|
||
<p class="center large"><span class="intbl"><em>1</em><strong>M</strong></span>[<span class="intbl"><em>∂N</em><strong>∂x</strong></span>−<span class="intbl"><em>∂M</em><strong>∂y</strong></span>]</p>
|
||
<p>must <b>not</b> have the <strong>x</strong> term in order for the
|
||
integrating factor to be a function of only <strong>y</strong>.</p>
|
||
</div>
|
||
<p>And if that condition is true, we call that expression <strong>Z(y)</strong> and our integrating factor is</p>
|
||
<p class="center large">u(y) = e<sup><span class="integral">∫</span>Z(y)dy</sup></p>
|
||
<p>And we can continue just like the previous example</p>
|
||
<p> </p>
|
||
<p>And there you have it!</p>
|
||
|
||
<div class="related">
|
||
<a href="index.html">Calculus Index</a>
|
||
</div>
|
||
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