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Mathematics - Linear second-order Differential equations with constant coefficients by drifter1

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Mathematics - Linear second-order Differential equations with constant coefficients
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<p>&nbsp;&nbsp;&nbsp;&nbsp;Hello it's a me again drifter1! Today we continue with <strong>Mathematics </strong>to talk about <strong>Differential equations</strong> that are <strong>second-order</strong>, <strong>linear </strong>and have <strong>constant coefficients</strong>! Next time we will get into other simple types of 2nd-order ODE's.</p>
<p>So, without further do, let's get straight into it!</p>
<h2>Second-order ODE's</h2>
<p>&nbsp;&nbsp;&nbsp;&nbsp;Before getting into the main topic of this post, we might first want to talk about 2nd-order ODE's in general.</p>
<p>An <strong>2nd-order ODE</strong> is of the form:</p>
<p><em><strong>y" + P(x, y)y' + Q(x, y)y = R(x)</strong></em></p>
<blockquote>The right side is a function of x, cause if it had any y then it would be a part of Q(x, y) in the left side.</blockquote>
<h3><br></h3>
<h3>Actually 1st-order</h3>
<p>If y=y(x) is not a part of an 2nd-order ODE, then we can solve it as an 1st-order ODE.</p>
<p>By <strong>substituting y' = u</strong> we get a <strong>1-st order ODE with variable u</strong>.</p>
<p>This "new" ODE will be one of the 6 types we covered in my previous posts.</p>
<p>By solving it and finding u we can then find y by finding the integral of u:</p>
<p><strong>y(x) = integral(u(x)dx) + c</strong></p>
<p><br></p>
<p><strong>For example:</strong></p>
<p>y" + 5xy' = x^2 + 3</p>
<p>By setting y' = u we get:</p>
<p>u' + 5xu = x^2 + 3 [linear 1st-order ODE]</p>
<p><br></p>
<h3>Linear second-order ODE's</h3>
<p>A linear ODE of any order contains P(x), Q(x) etc. "coefficient functions" of x.</p>
<p>That way a <strong>second-order linear ODE</strong> is of the <strong>form</strong>:</p>
<p><em><strong>y" + P(x)y' + Q(x)y = R(x)</strong></em></p>
<p>If R(x) = 0 then we get the <strong>homogeneous form</strong>:</p>
<p><em><strong>y" + P(x)y' + Q(x)y = 0</strong></em></p>
<p><br></p>
<p>This last one is pretty <strong>important </strong>cause the following <strong>statement </strong>is true:</p>
<p>If <strong>y1(x) and y2(x) are two solutions</strong> of this homogeneous OE then:</p>
<p><em><strong>y0 = c1*y1(x) + c2*y2(x)</strong></em> is the <strong>general solution</strong> of the homogeneous ODE,</p>
<p>where c1, c2 are real numbers</p>
<p><br></p>
<p>&nbsp;&nbsp;&nbsp;&nbsp;If the homogeneous is the <strong>corresponding homogeneous</strong> of an 2nd-order linear ODE (by setting R(x) = 0) then we can find the solution of the given ODE by using:</p>
<p><em><strong>y = y0 + yp</strong></em>, where y is the <strong>general solution</strong> of our 2nd-order linear ODE.</p>
<p>As told before y0 is the general solution of our homogeneous.</p>
<p>The other part (yp) is any solution of the given (non-homogeneous) ODE.</p>
<p><br></p>
<p>After this introduction let's now get into the main topic of this post.</p>
<h2>Linear second-order ODE's with const coeffs</h2>
<p>&nbsp;&nbsp;&nbsp;&nbsp;When <strong>P(x) and Q(x) don't contain x</strong> as an variable <strong>and are real numbers</strong> then we have an linear second-order ODE with constant coefficients.</p>
<p>Note that <strong>R(x) is still a function of x</strong> (it can be 0 or any real number too tho).</p>
<p>So, the <strong>basic form</strong> of such an ODE is:</p>
<p><em><strong>ay" + by' + cy = R(x)</strong></em></p>
<p><br></p>
<p><strong>For example:</strong></p>
<p>5y" + 3y' -2y = x^2 + e^x</p>
<p><br></p>
<h3>Solution</h3>
<p>To solve such a ODE we first get the<strong> corresponding homogeneous</strong>:</p>
<p><em><strong>ay" + by" + cy = 0</strong></em></p>
<p>We suppose that <strong>y = e^px is a solution of the homogeneous </strong>(can be proven).</p>
<p>Doing that we get the <strong>characteristic equation</strong>:</p>
<p><em><strong>ap^2 + bp + c = 0</strong></em></p>
<p>This is a 2nd-order polynomial equation that can be solved very easily.</p>
<p>There are <strong>3 possible outcomes</strong>:</p>
<ol>
  <li>2 real solutions: p1, p2 =&gt; y1 = e^(p1*x) and y2 = e^(p2*x) solutions of the homogeneous.</li>
  <li>1 double-solution: p =&gt; y1 = e^px and y2 = x*e^px (can be proven) solutions of the homogeneous</li>
  <li>2 complex solutions: p1, p2 =&gt; We use the Euler equation (e^iφ = cosφ + i*sinφ) and get y1= e^(i*x) = cosx, y2 = e^(-i*x) = sinx.</li>
</ol>
<p><br></p>
<p><strong>For example:</strong></p>
<p>y" - 2y' - 3y = 2cos(x) &nbsp;&nbsp;&nbsp;[given ODE]</p>
<p>y" - 2y' - 3y = 0 &nbsp;&nbsp;&nbsp;&nbsp;[corresponding homogeneous ODE]</p>
<p>p^2 - 2p - 3 = 0 &nbsp;&nbsp;&nbsp;[characteristic equation]</p>
<p>p1 = 3 and p2 = -1 are the 2 real solutions and so:</p>
<p>y1 = e^3x and y2 = e^-x are the solutions of the homogeneous.</p>
<p>The general solutions will be: y0 = c1e^3x + c2e^-x</p>
<p><br></p>
<p>After that we have to <strong>find one solution of the given ODE</strong>.</p>
<p>There are some <strong>Cases </strong>that depend on the form of R(x):</p>
<p><strong>1. If R(x) = 0</strong> then yp = 0 and so:</p>
<p>y = y0 = c1*y1 + c2*y2, where c1, c2 reals</p>
<p><br></p>
<p><strong>2. If R(x) = e^(k*x) </strong>[<strong>Exponential</strong>], where k a real number</p>
<p>Then we have some <strong>cases that depend on the solutions p1, p2</strong>:</p>
<ul>
  <li><strong>k != p1, p2 </strong>=&gt;<strong> yp = λ*e^(k*x) </strong>and we find λ with substituting in the given ODE.</li>
  <li><strong>k = p1 or k = p2</strong> =&gt; <strong>yp = λ*x*e^(k*x)</strong> and we again find λ with substitution.</li>
  <li><strong>p is double-solution</strong> =&gt; <strong>yp = λ*x^2*e^(k*x)</strong> and we again find λ.</li>
</ul>
<p><br></p>
<p><strong>3. If R(x) = an*x^n + ... + a1*x + a0</strong> [<strong>Polynomial </strong>of x], where ai reals</p>
<p>Then we have <strong>cases that depend on c</strong>:</p>
<ul>
  <li><strong>c!=0</strong> =&gt; <strong>yp = bn*x^n + ... + b1*x + b0</strong>, bi reals and we substitute to find those coefficients.</li>
  <li><strong>c==0 </strong>=&gt; <strong>yp = x*(bn*x^n + ... + b1*x + b0)</strong>, bi reals and we again substitute.</li>
</ul>
<p><br></p>
<p><strong>4. If R(x) = k*cos(n*x) + λ*sin(n*x)</strong> [Trigonometric], where k, l, n are reals</p>
<p>Then we have <strong>cases that depend on the solutions</strong>:</p>
<ul>
  <li><strong>cos(nx) and sin(nx) ARE NOT solutions of the homogeneous </strong>=&gt;<strong> yp = s*cos(n*x) + t*sin(n*x)</strong>, where s, t are reals and we have to find s, t with substitution.</li>
  <li><strong>cos(nx) and sin(nx) ARE solutions of the homogeneous =&gt; yp = x*[s*cos(n*x) + t*sin(n*x)]</strong>, where we find s, t with substitution.</li>
</ul>
<p><br></p>
<p><strong>5. If R(x): Combination</strong> of the previous cases</p>
<p>R(x) is of the form:</p>
<p><em><strong>R(x) = R1(x) + R2(x) + ... + Rn(x)</strong></em>, where Ri(x) one of the cases 2-4.</p>
<p>We find the <strong>solution</strong> for each <strong>independently</strong>.</p>
<p>ay" + by' + cy = R1(x) =&gt; yp1</p>
<p>ay" + by' + cy = R2(x) =&gt; yp2</p>
<p>...</p>
<p>ay" + by' + cy = Rn(x) =&gt; ypn</p>
<p>Then we <strong>sum </strong>all of these together:</p>
<p><em><strong>yp = yp1 + yp2 + ... + ypn</strong></em></p>
<p><br></p>
<p><strong>For example:</strong></p>
<p>R(x) = e^2x - 3x^2 + 4x - 5 + 7cos(3x) - 8sin(3x) &nbsp;&nbsp;&nbsp;[Combination]</p>
<p>We can split it into:</p>
<p>R1(x) = e^2x &nbsp;&nbsp;&nbsp;[Exponential]</p>
<p>R2(x) = -3x^2 + 4x - 5 &nbsp;&nbsp;&nbsp;[Polynomial]</p>
<p>R3(x) = 7cos(3x) - 8sin(3x) &nbsp;&nbsp;&nbsp;[Trigonometric]</p>
<p><br></p>
<p>For all those cases the <strong>general solution</strong> of our given ODE is then:</p>
<p><em><strong>y = y0 + yp</strong></em></p>
<p><br></p>
<p>&nbsp;&nbsp;&nbsp;&nbsp;I will get into practical examples of 2nd-order linear ODE's when we cover the other special forms too (next post).</p>
<p><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><a href="https://steemit.com/mathematics/@drifter1/mathematics-ordinary-first-order-differential-equations-part-2">First-order part(2)</a> -&gt; Linear, Bernoulli and Riccati first-order ODE's</p>
<p><a href="https://steemit.com/mathematics/@drifter1/mathematics-first-order-differential-equation-exercises">First-order exercises</a> -&gt; Exercises for all the 1st-order ODE types</p>
<p><br>
And this is actually it!</p>
<p>Next time we will get into linear 2nd-order ODE's that are of other special forms!</p>
<p>C ya!</p>
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@daradina · 
if seen, @drifter1 a mathematician?
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@drifter1 · 
Haha, hell no!
I'm studying Computer science and we have lots of Mathematics.
Designing algorithms and calculating and comparing the performance/complexity of them with other's needs a lot of maths.
That's why I started posting a little bit about Mathematics...
Physics that is used in games, also uses a lot of those more complex mathematics.
I try to explain everything in a way that is understandable and simple!
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@daradina · 
Thank you, may the science you give useful to me and others in a way that easily understand like you say.
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@paaddor · 
Great! It's refreshing to read posts like this that stand out from most others.

What is actually, if R (x) is a power series? If R (x) does not fall into any of the cases you deal with, it would have to be represented as a power series.
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@drifter1 · (edited)
There are of course more R(x) cases, but these are the "simplest".
If R(x) doesn't fall into any of the cases then we find yp (solution of the given ODE) with the Lagrange method.
Using a specific equation we can get yp using y1, y2 and the Wronsky determinant.
More about that in a sec, I am writing the post for today at the moment! :)
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