Mathematics - Ordinary first-order Differential equations (part 2) by drifter1

mathematics · @drifter1 · Mar 6 '18 (edited)

$14.04

Mathematics - Ordinary first-order Differential equations (part 2)

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<p>&nbsp;<img src="https://s14.postimg.org/d890sqd7l/part2.jpg" width="800" height="800"/></p>
<p>&nbsp;&nbsp;&nbsp;&nbsp;Hey it's me again drifter1! Today we continue with the 2nd part of <strong>first-order ODE's</strong> from my <strong>Mathematics </strong>series. There are 6 basic types in total and here we will discuss the 3 remaining. Exercises will come in the next post and will be about all the 1st-order types.</p>
<p>So, let's get straight into it!</p>
<p><br></p>
<h3>Linear n-order</h3>
<p>&nbsp;&nbsp;&nbsp;&nbsp;To understand what a 1st-order linear ODE is, we might first want to get into how n-order ones look and which constraints they have.</p>
<p>A <strong>n-order ODE is linear</strong> when:</p>
<ul>
  <li>The function and it's derivatives (y, y', y'', ..., y^(n)) don't come with exponentials/powers</li>
  <li>The coefficients of the function and it's derivatives are a function of x and so f(x) or a number c</li>
</ul>
<p><br></p>
<p>You might remember from last time that Homogeneous are converted into Separable ones.</p>
<p>&nbsp;&nbsp;&nbsp;&nbsp;In the same way, the 1st-order linear ODE's are very useful, cause Bernoulli ones get converted into a Linear ODE. Also, Riccati ones get converted into a Bernoulli and then into a Linear.</p>
<p>So, let's get into how we solve 1st-order linear ODE's.</p>
<h3>Solving Linear 1st-order ODE's</h3>
<p>A linear 1st-order ODE is of the form:</p>
<p><em><strong>y' + P(x)*y = Q(x)</strong></em></p>
<p><br></p>
<p>There are two main solving methods.</p>
<p><strong>Solution 1:</strong></p>
<p>Multiply both sides of the equation with a new function u = u(x).</p>
<p>We want a u so that: <em>u'y = Puy</em></p>
<p>Doing some calculations we end up with:</p>
<p><em>u = e ^ integral[P(x)dx]</em></p>
<p>and</p>
<p><em>y = (1/u) * integral[u*Q(x)dx]&nbsp;</em></p>
<p>That way we can find our solution y.</p>
<p><br></p>
<p><strong>Solution 2:</strong></p>
<p>We substitute y by setting <em>y(x) = g(x)*Y(x)</em></p>
<p>We want a g so that: <em>g'/g + P = 0</em> =&gt; <em>Y' = Q/g</em></p>
<p>By doing some calculations we end up with:</p>
<p><em>g = e ^[-integral(Pdx)]</em></p>
<p>and</p>
<p><em>Y = integral[(Q/g)dx] + c</em></p>
<p>That way with substitution again we get our y from g and Y.</p>
<p><br></p>
<h3>Solving Bernoulli ODE's</h3>
<p>A Bernoulli ODE is of the form:</p>
<p><em><strong>y' + P(x)y +Q(x)y^a = 0</strong></em>, where<em><strong> </strong></em><strong>a != 0,1</strong><em><strong> </strong></em>and real.</p>
<p>We solve<strong> u(x) = y(x) ^ (1-a)</strong> that will give us a linear ODE.</p>
<p>Solving the linear ODE with the previous method we get u.</p>
<p>From u we can then get our solution y.</p>
<p><br></p>
<h3>Solving Riccati ODE's</h3>
<p>A Riccati ODE is of the form:<br>
<em><strong>y' + P(x)*y^2 + Q(x)*y + R(x) = 0</strong></em></p>
<p><br></p>
<p>By substituting y with <em>y = Y + h</em> we end up with a <strong>sum of two ODE's</strong>:</p>
<ol>
  <li>A Bernoulli ODE with Y</li>
  <li>A Riccati with h</li>
</ol>
<p>&nbsp;&nbsp;&nbsp;&nbsp;We <strong>solve 1 to find Y from the Bernoulli ODE</strong> and <strong>h is given to us from the beginning as a solution of our ODE</strong>, or we can find it easily.</p>
<p>That way by substituting we get our final solution y.</p>
<p><br></p>
<blockquote>You can see that in general <strong>we try to do</strong> <strong>changes of the form</strong>: <em>y =gY + h</em></blockquote>
<p><br></p>
<h3>Image sources:</h3>
<p><a href="http://www.ivyzen.net/wp-content/uploads/2016/09/maths-problems-marilyn-ftr.jpg">http://www.ivyzen.net/wp-content/uploads/2016/09/maths-problems-marilyn-ftr.jpg</a><br>
</p>
<h3>Previous posts of the series:</h3>
<p><a href="https://steemit.com/mathematics/@drifter1/mathematics-differential-equations-introduction">Introduction </a>-&gt; Definition and Applications</p>
<p><a href="https://steemit.com/mathematics/@drifter1/mathematics-ordinary-first-order-differential-equations-part-1">First-order part(1) </a>-&gt; &nbsp;Separable, homogeneous and exact 1st-order ODE's</p>
<p><br>
And this is it for today and I hope that you enjoyed it!</p>
<p>Next time we will get into exercises for 1st-order ODE's.</p>
<p>Ciao!&nbsp;</p>
</html>

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