Linear Independence of Functions - Revision history

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If we have an $n^{\mathrm{th}}$ order linear homogenous differential equation $\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac..."$

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Created page with "<p>If we have an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac..."

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<p>If we have an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> where <math>p_1</math>, <math>p_2</math>, …, <math>p_n</math> are continuous on an open interval <math>I</math> and if <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, …, <math>y = y_n(t)</math> are solutions to this differential equation, then provided that <math>W(y_1, y_2, ..., y_n) \neq 0</math> for at least one point <math>t \in I</math>, then <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> form a fundamental set of solutions to this differential equation - that is, for constants <math>C_1</math>, <math>C_2</math>, …, <math>C_n</math>, then every solution to this differential equation can be written in the form:</p>

<div class="math-equation" id="equation-1">\begin{align} \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) \end{align}</div>
<p>We will now look at the connection between the solutions <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.</p>
<table class="wiki-content-table">
<tr>
<td><strong>Definition:</strong> The functions <math>f_1</math>, <math>f_2</math>, …, <math>f_n</math> are said to be <strong>Linearly Independent</strong> on an interval <math>I</math> if for constants <math>k_1</math>, <math>k_2</math>, …, <math>k_n</math> we have that <math>k_1f_1(t) + k_2f_2(t) + ... + k_nf_n(t) = 0</math> implies that <math>k_1 = k_2 = ... = k_n = 0</math> for all <math>t \in I</math>. This set of functions is said to be <strong>Linearly Dependent</strong> if <math>k_1f_1(t) + k_2f_2(t) + ... + k_nf_n(t) = 0</math> where <math>k_1</math>, <math>k_2</math>, …, <math>k_n</math> are not all zero for all <math>t \in I</math>.</td>
</tr>
</table>
<p>Perhaps the simplest linearly independent sets of functions is that set that contains <math>f_1(t) = 1</math>, <math>f_2(t) = t</math>, and <math>f_3(t) = t^2</math>. Let <math>k_1</math>, <math>k_2</math>, and <math>k_3</math> be constants and consider the following equation:</p>

<div class="math-equation" id="equation-2">\begin{align} \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 \end{align}</div>
<p>It's not hard to see that equation above is satisfied if and only if the constants <math>k_1 = k_2 = k_3 = 0</math>.</p>
<p>For another example, consider the functions <math>f_1(t) = \sin t</math> and <math>f_2(t) = \sin (t + \pi)</math> defined on all of <math>\mathbb{R}</math>. This set of functions is not linearly independent. To show this, let <math>k_1</math> and <math>k_2</math> be constants and consider the following equation:</p>

<div class="math-equation" id="equation-3">\begin{align} \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 \end{align}</div>
<p>Now choose <math>t = \pi</math>. Then we have that:</p>

<div class="math-equation" id="equation-4">\begin{align} \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ \end{align}</div>
<p>But the above equation is true for any choice of constants <math>k_1</math> and <math>k_2</math> since <math>\sin \pi = \sin 2\pi = 0</math>, and thus <math>f_1</math> and <math>f_2</math> do not form a linearly independent set on all of <math>\mathbb{R}</math>.</p>
<p>From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for <math>n^{\mathrm{th}}</math> order linear homogenous differential equations.</p>
<table class="wiki-content-table">
<tr>
<td><strong>Theorem 1:</strong> Let <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> be an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation. If <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, …, <math>y = y_n(t)</math> are solutions to this differential equation then <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> form a fundamental set of solutions to this differential equation on the open interval <math>I</math> if and only if <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> are linearly dependent on <math>I</math>.</td>
</tr>
</table>
<ul>
<li><strong>Proof:</strong> Consider the following <math>n^{\mathrm{th}}</math> order linear homogenous differential equation:</li>
</ul>

<div class="math-equation" id="equation-5">\begin{align} \quad \frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0 \end{align}</div>
<ul>
<li><math>\Rightarrow</math> Suppose that <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, …, <math>y = y_n(t)</math> form a fundamental set of solutions to this differential equation on the open interval <math>I</math>. Then this implies that for all <math>t_0 \in I</math> we have that :</li>
</ul>

<div class="math-equation" id="equation-6">\begin{align} \quad W(y_1, y_2, ..., y_n) \biggr \rvert_{t_0} \neq 0 \end{align}</div>
<ul>
<li>Thus this implies that following system of equations have only trivial solution <math>k_1 = k_2 = ... = k_n = 0</math>:</li>
</ul>

<div class="math-equation" id="equation-7">\begin{align} \quad k_1y_1(t) + k_2y_2(t) + ... + k_ny_n(t) = 0 \\ \quad k_1y_1'(t) + k_2y_2'(t) + ... + k_ny_n'(t) = 0 \\ \quad \quad \quad \quad \quad \quad \vdots \quad \quad \quad \quad \quad \quad \\ \quad k_1y_1^{(n-2)}(t) + k_2y_2^{(n-2)}(t) + ... + k_ny_n^{(n-2)}(t) = 0 \\ \quad k_1y_1^{(n-1)}(t) + k_2y_2^{(n-1)}(t) + ... + k_ny_n^{(n-1)}(t) = 0 \end{align}</div>
<ul>
<li>Thus the equation <math>k_1y_1(t) + k_2y_2(t) + ... + k_ny_n(t) = 0</math> implies that <math>k_1 = k_2 = ... = k_n = 0</math>. Thus <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> are linearly independent on <math>I</math>.</li>
</ul>
<ul>
<li><math>\Leftarrow</math> We will prove the converse of Theorem 1 by contradiction. Suppose that <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> are linearly independent on <math>I</math>, and assume that instead <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> do NOT form a fundamental set of solutions on <math>I</math>. Then for some <math>t_0 \in I</math>, the Wronskian <math>W(y_1, y_2, ..., y_n) \biggr \rvert_{t_0} = 0</math>. Thus the system of equations above does not have only the trivial solution. Let the constants <math>k_1^*</math>, <math>k_2^*</math>, …, <math>k_n^*</math> be a nontrivial solution to this system. Define <math>\phi(t)</math> as:</li>
</ul>

<div class="math-equation" id="equation-8">\begin{align} \quad \phi(t) = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}</div>
<ul>
<li>Note that <math>y = \phi(t)</math> satisfies the initial conditions <math>y(t_0) = 0</math>, <math>y'(t_0) = 0</math>, …, <math>y^{(n-1)} (t_0) = 0</math>, and <math>\phi(t)</math> satisfies our <math>n^{\mathrm{th}}</math> order linear homogenous differential equation because <math>\phi(t)</math> is a linear combination of the solutions <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math>.</li>
</ul>
<ul>
<li>Now note that the function <math>y = 0</math> also satisfies the differential equation and the initial conditions. By the existence/uniqueness theorem for <math>n^{\mathrm{th}}</math> order linear homogenous differential equations, this implies that <math>\phi(t) = 0</math> for all <math>t \in I</math>, so:</li>
</ul>

<div class="math-equation" id="equation-9">\begin{align} \quad 0 = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}</div>
<ul>
<li>But <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> are linearly independent which implies that <math>k_1^* = k_2^* = ... = k_n^* = 0</math>. Thus <math>k_1^*</math>, <math>k_2^*</math>, …, <math>k_n^*</math> is a trivial solution to the system above, which is a contradiction. Therefore our assumption that <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> do not form a fundamental set of solutions was false. <math>\blacksquare</math></li>
</ul>

← Older revision		Revision as of 00:04, 29 October 2021
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	<li>But <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> are linearly independent which implies that <math>k_1^* = k_2^* = ... = k_n^* = 0</math>. Thus <math>k_1^</math>, <math>k_2^</math>, …, <math>k_n^*</math> is a trivial solution to the system above, which is a contradiction. Therefore our assumption that <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> do not form a fundamental set of solutions was false. <math>\blacksquare</math></li>		<li>But <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> are linearly independent which implies that <math>k_1^* = k_2^* = ... = k_n^* = 0</math>. Thus <math>k_1^</math>, <math>k_2^</math>, …, <math>k_n^*</math> is a trivial solution to the system above, which is a contradiction. Therefore our assumption that <math>y_1</math>, <math>y_2</math>, …, <math>y_n</math> do not form a fundamental set of solutions was false. <math>\blacksquare</math></li>
	</ul>		</ul>
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/linear-independence-dependence-of-a-set-of-functions Linear Independence Dependence of a Set of Functions, http://mathonline.wikidot.com/] under a CC BY-SA license

@@ Line 30: / Line 30: @@
 </ul>
-<math>\begin{align} \quad W(y_1, y_2, ..., y_n)  \right|_{t_0} \neq 0 \end{align}</math>
+<math>\begin{align} \quad W(y_1, y_2, ..., y_n)  \big|_{t_0} \neq 0 \end{align}</math>
 <ul>
 <li>Thus this implies that following system of equations have only trivial solution <math>k_1 = k_2 = ... = k_n = 0</math>:</li>

@@ Line 1: / Line 1: @@
 <p>If we have an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> where <math>p_1</math>, <math>p_2</math>, &#8230;, <math>p_n</math> are continuous on an open interval <math>I</math> and if <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, &#8230;, <math>y = y_n(t)</math> are solutions to this differential equation, then provided that <math>W(y_1, y_2, ..., y_n) \neq 0</math> for at least one point <math>t \in I</math>, then <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> form a fundamental set of solutions to this differential equation - that is, for constants <math>C_1</math>, <math>C_2</math>, &#8230;, <math>C_n</math>, then every solution to this differential equation can be written in the form:</p>
-<math> \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) </math>
+<math>\begin{align} \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) \end{align}</math>
 <p>We will now look at the connection between the solutions <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.</p>
@@ Line 8: / Line 8: @@
 <p>Perhaps the simplest linearly independent sets of functions is that set that contains <math>f_1(t) = 1</math>, <math>f_2(t) = t</math>, and <math>f_3(t) = t^2</math>. Let <math>k_1</math>, <math>k_2</math>, and <math>k_3</math> be constants and consider the following equation:</p>
-<math> \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 </math>
+<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 \end{align}</math>
 <p>It's not hard to see that equation above is satisfied if and only if the constants <math>k_1 = k_2 = k_3 = 0</math>.</p>
 <p>For another example, consider the functions <math>f_1(t) = \sin t</math> and <math>f_2(t) = \sin (t + \pi)</math> defined on all of <math>\mathbb{R}</math>. This set of functions is not linearly independent. To show this, let <math>k_1</math> and <math>k_2</math> be constants and consider the following equation:</p>
-<math> \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 </math>
+<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 \end{align}</math>
 <p>Now choose <math>t = \pi</math>. Then we have that:</p>
-<math> \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ </math>
+<math>\begin{align} \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ \end{align}</math>
 <p>But the above equation is true for any choice of constants <math>k_1</math> and <math>k_2</math> since <math>\sin \pi = \sin 2\pi = 0</math>, and thus <math>f_1</math> and <math>f_2</math> do not form a linearly independent set on all of <math>\mathbb{R}</math>.</p>
 <p>From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for <math>n^{\mathrm{th}}</math> order linear homogenous differential equations.</p>

@@ Line 1: / Line 1: @@
 <p>If we have an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> where <math>p_1</math>, <math>p_2</math>, &#8230;, <math>p_n</math> are continuous on an open interval <math>I</math> and if <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, &#8230;, <math>y = y_n(t)</math> are solutions to this differential equation, then provided that <math>W(y_1, y_2, ..., y_n) \neq 0</math> for at least one point <math>t \in I</math>, then <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> form a fundamental set of solutions to this differential equation - that is, for constants <math>C_1</math>, <math>C_2</math>, &#8230;, <math>C_n</math>, then every solution to this differential equation can be written in the form:</p>
-<math>\begin{align} \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) \end{align}</math>
+<math> \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) </math>
 <p>We will now look at the connection between the solutions <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.</p>
@@ Line 8: / Line 8: @@
 <p>Perhaps the simplest linearly independent sets of functions is that set that contains <math>f_1(t) = 1</math>, <math>f_2(t) = t</math>, and <math>f_3(t) = t^2</math>. Let <math>k_1</math>, <math>k_2</math>, and <math>k_3</math> be constants and consider the following equation:</p>
-<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 \end{align}</math>
+<math> \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 </math>
 <p>It's not hard to see that equation above is satisfied if and only if the constants <math>k_1 = k_2 = k_3 = 0</math>.</p>
 <p>For another example, consider the functions <math>f_1(t) = \sin t</math> and <math>f_2(t) = \sin (t + \pi)</math> defined on all of <math>\mathbb{R}</math>. This set of functions is not linearly independent. To show this, let <math>k_1</math> and <math>k_2</math> be constants and consider the following equation:</p>
-<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 \end{align}</math>
+<math> \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 </math>
 <p>Now choose <math>t = \pi</math>. Then we have that:</p>
-<math>\begin{align} \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ \end{align}</math>
+<math> \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ </math>
 <p>But the above equation is true for any choice of constants <math>k_1</math> and <math>k_2</math> since <math>\sin \pi = \sin 2\pi = 0</math>, and thus <math>f_1</math> and <math>f_2</math> do not form a linearly independent set on all of <math>\mathbb{R}</math>.</p>
 <p>From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for <math>n^{\mathrm{th}}</math> order linear homogenous differential equations.</p>

@@ Line 30: / Line 30: @@
 </ul>
-<math>\begin{align} \quad W(y_1, y_2, ..., y_n)  \big|_{t_0} \neq 0 \end{align}</math>
+<math>\begin{align} \quad W(y_1, y_2, ..., y_n)  \bigg|_{t_0} \neq 0 \end{align}</math>
 <ul>
 <li>Thus this implies that following system of equations have only trivial solution <math>k_1 = k_2 = ... = k_n = 0</math>:</li>
@@ Line 40: / Line 40: @@
 </ul>
 <ul>
-<li><math>\Leftarrow</math> We will prove the converse of Theorem 1 by contradiction. Suppose that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent on <math>I</math>, and assume that instead <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do NOT form a fundamental set of solutions on <math>I</math>. Then for some <math>t_0 \in I</math>, the Wronskian <math>W(y_1, y_2, ..., y_n)  \right|_{t_0} = 0</math>. Thus the system of equations above does not have only the trivial solution. Let the constants <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> be a nontrivial solution to this system. Define <math>\phi(t)</math> as:</li>
+<li><math>\Leftarrow</math> We will prove the converse of Theorem 1 by contradiction. Suppose that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent on <math>I</math>, and assume that instead <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do NOT form a fundamental set of solutions on <math>I</math>. Then for some <math>t_0 \in I</math>, the Wronskian <math>W(y_1, y_2, ..., y_n)  \bigg|_{t_0} = 0</math>. Thus the system of equations above does not have only the trivial solution. Let the constants <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> be a nontrivial solution to this system. Define <math>\phi(t)</math> as:</li>
 </ul>

@@ Line 1: / Line 1: @@
 <p>If we have an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> where <math>p_1</math>, <math>p_2</math>, &#8230;, <math>p_n</math> are continuous on an open interval <math>I</math> and if <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, &#8230;, <math>y = y_n(t)</math> are solutions to this differential equation, then provided that <math>W(y_1, y_2, ..., y_n) \neq 0</math> for at least one point <math>t \in I</math>, then <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> form a fundamental set of solutions to this differential equation - that is, for constants <math>C_1</math>, <math>C_2</math>, &#8230;, <math>C_n</math>, then every solution to this differential equation can be written in the form:</p>
-<div class="math-equation" id="equation-1">\begin{align} \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) \end{align}</div>
+<math>\begin{align} \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) \end{align}</math>
 <p>We will now look at the connection between the solutions <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.</p>
-<table class="wiki-content-table">
 <td><strong>Definition:</strong> The functions <math>f_1</math>, <math>f_2</math>, &#8230;, <math>f_n</math> are said to be <strong>Linearly Independent</strong> on an interval <math>I</math> if for constants <math>k_1</math>, <math>k_2</math>, &#8230;, <math>k_n</math> we have that <math>k_1f_1(t) + k_2f_2(t) + ... + k_nf_n(t) = 0</math> implies that <math>k_1 = k_2 = ... = k_n = 0</math> for all <math>t \in I</math>. This set of functions is said to be <strong>Linearly Dependent</strong> if <math>k_1f_1(t) + k_2f_2(t) + ... + k_nf_n(t) = 0</math> where <math>k_1</math>, <math>k_2</math>, &#8230;, <math>k_n</math> are not all zero for all <math>t \in I</math>.</td>
-</tr>
 <p>Perhaps the simplest linearly independent sets of functions is that set that contains <math>f_1(t) = 1</math>, <math>f_2(t) = t</math>, and <math>f_3(t) = t^2</math>. Let <math>k_1</math>, <math>k_2</math>, and <math>k_3</math> be constants and consider the following equation:</p>
-<div class="math-equation" id="equation-2">\begin{align} \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 \end{align}</div>
+<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 \end{align}</math>
 <p>It's not hard to see that equation above is satisfied if and only if the constants <math>k_1 = k_2 = k_3 = 0</math>.</p>
 <p>For another example, consider the functions <math>f_1(t) = \sin t</math> and <math>f_2(t) = \sin (t + \pi)</math> defined on all of <math>\mathbb{R}</math>. This set of functions is not linearly independent. To show this, let <math>k_1</math> and <math>k_2</math> be constants and consider the following equation:</p>
-<div class="math-equation" id="equation-3">\begin{align} \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 \end{align}</div>
+<math>\begin{align} \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 \end{align}</math>
 <p>Now choose <math>t = \pi</math>. Then we have that:</p>
-<div class="math-equation" id="equation-4">\begin{align} \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ \end{align}</div>
+<math>\begin{align} \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ \end{align}</math>
 <p>But the above equation is true for any choice of constants <math>k_1</math> and <math>k_2</math> since <math>\sin \pi = \sin 2\pi = 0</math>, and thus <math>f_1</math> and <math>f_2</math> do not form a linearly independent set on all of <math>\mathbb{R}</math>.</p>
 <p>From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for <math>n^{\mathrm{th}}</math> order linear homogenous differential equations.</p>
-<table class="wiki-content-table">
 <td><strong>Theorem 1:</strong> Let <math>\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0</math> be an <math>n^{\mathrm{th}}</math> order linear homogenous differential equation. If <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, &#8230;, <math>y = y_n(t)</math> are solutions to this differential equation then <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> form a fundamental set of solutions to this differential equation on the open interval <math>I</math> if and only if <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly dependent on <math>I</math>.</td>
-</tr>
 <ul>
 <li><strong>Proof:</strong> Consider the following <math>n^{\mathrm{th}}</math> order linear homogenous differential equation:</li>
 </ul>
-<div class="math-equation" id="equation-5">\begin{align} \quad \frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0 \end{align}</div>
+<math>\begin{align} \quad \frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0 \end{align}</math>
 <ul>
 <li><math>\Rightarrow</math> Suppose that <math>y = y_1(t)</math>, <math>y = y_2(t)</math>, &#8230;, <math>y = y_n(t)</math> form a fundamental set of solutions to this differential equation on the open interval <math>I</math>. Then this implies that for all <math>t_0 \in I</math> we have that :</li>
 </ul>
-<div class="math-equation" id="equation-6">\begin{align} \quad W(y_1, y_2, ..., y_n) \biggr \rvert_{t_0} \neq 0 \end{align}</div>
+<math>\begin{align} \quad W(y_1, y_2, ..., y_n)  \right|_{t_0} \neq 0 \end{align}</math>
 <ul>
 <li>Thus this implies that following system of equations have only trivial solution <math>k_1 = k_2 = ... = k_n = 0</math>:</li>
 </ul>
-<div class="math-equation" id="equation-7">\begin{align} \quad k_1y_1(t) + k_2y_2(t) + ... + k_ny_n(t) = 0 \\ \quad k_1y_1'(t) + k_2y_2'(t) + ... + k_ny_n'(t) = 0 \\ \quad \quad \quad \quad \quad \quad \vdots \quad \quad \quad \quad \quad \quad \\ \quad k_1y_1^{(n-2)}(t) + k_2y_2^{(n-2)}(t) + ... + k_ny_n^{(n-2)}(t) = 0 \\ \quad k_1y_1^{(n-1)}(t) + k_2y_2^{(n-1)}(t) + ... + k_ny_n^{(n-1)}(t) = 0 \end{align}</div>
+<math>\begin{align} \quad k_1y_1(t) + k_2y_2(t) + ... + k_ny_n(t) = 0 \\ \quad k_1y_1'(t) + k_2y_2'(t) + ... + k_ny_n'(t) = 0 \\ \quad \quad \quad \quad \quad \quad \vdots \quad \quad \quad \quad \quad \quad \\ \quad k_1y_1^{(n-2)}(t) + k_2y_2^{(n-2)}(t) + ... + k_ny_n^{(n-2)}(t) = 0 \\ \quad k_1y_1^{(n-1)}(t) + k_2y_2^{(n-1)}(t) + ... + k_ny_n^{(n-1)}(t) = 0 \end{align}</math>
 <ul>
 <li>Thus the equation <math>k_1y_1(t) + k_2y_2(t) + ... + k_ny_n(t) = 0</math> implies that <math>k_1 = k_2 = ... = k_n = 0</math>. Thus <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent on <math>I</math>.</li>
 </ul>
 <ul>
-<li><math>\Leftarrow</math> We will prove the converse of Theorem 1 by contradiction. Suppose that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent on <math>I</math>, and assume that instead <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do NOT form a fundamental set of solutions on <math>I</math>. Then for some <math>t_0 \in I</math>, the Wronskian <math>W(y_1, y_2, ..., y_n) \biggr \rvert_{t_0} = 0</math>. Thus the system of equations above does not have only the trivial solution. Let the constants <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> be a nontrivial solution to this system. Define <math>\phi(t)</math> as:</li>
+<li><math>\Leftarrow</math> We will prove the converse of Theorem 1 by contradiction. Suppose that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent on <math>I</math>, and assume that instead <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do NOT form a fundamental set of solutions on <math>I</math>. Then for some <math>t_0 \in I</math>, the Wronskian <math>W(y_1, y_2, ..., y_n)  \right|_{t_0} = 0</math>. Thus the system of equations above does not have only the trivial solution. Let the constants <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> be a nontrivial solution to this system. Define <math>\phi(t)</math> as:</li>
 </ul>
-<div class="math-equation" id="equation-8">\begin{align} \quad \phi(t) = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}</div>
+<math>\begin{align} \quad \phi(t) = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}</math>
 <ul>
 <li>Note that <math>y = \phi(t)</math> satisfies the initial conditions <math>y(t_0) = 0</math>, <math>y'(t_0) = 0</math>, &#8230;, <math>y^{(n-1)} (t_0) = 0</math>, and <math>\phi(t)</math> satisfies our <math>n^{\mathrm{th}}</math> order linear homogenous differential equation because <math>\phi(t)</math> is a linear combination of the solutions <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math>.</li>
@@ Line 55: / Line 51: @@
 </ul>
-<div class="math-equation" id="equation-9">\begin{align} \quad 0 = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}</div>
+<math>\begin{align} \quad 0 = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}</math>
 <ul>
 <li>But <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> are linearly independent which implies that <math>k_1^* = k_2^* = ... = k_n^* = 0</math>. Thus <math>k_1^*</math>, <math>k_2^*</math>, &#8230;, <math>k_n^*</math> is a trivial solution to the system above, which is a contradiction. Therefore our assumption that <math>y_1</math>, <math>y_2</math>, &#8230;, <math>y_n</math> do not form a fundamental set of solutions was false. <math>\blacksquare</math></li>
 </ul>