<p>I'm posting some additional clarification to the original post here here because I don't want to confuse the original. Attached here is one the original question sheet. The context is that It's from a 2020 University Exam in Mathematical Economics, 3rd year module.</p>
<ol>
<li>Original Question sheet.</li>
<li>Answer to part c) where I believe having a different general solution causes problems.</li>
</ol>
<p>Specifically we note that if we use my answer from diagonalisation:</p>
<ul>
<li><span class="math-container">$X(t) = -\alpha e^{ct} + \beta e^{-ct}$</span></li>
<li><span class="math-container">$Y(t) = \alpha e^{ct} + \beta e^{-ct}$</span></li>
</ul>
<p>Instead of the official answer</p>
<ul>
<li><span class="math-container">$X(t) = \alpha e^{ct} + \beta e^{-ct}$</span></li>
<li><span class="math-container">$Y(t) = -\alpha e^{ct} + \beta e^{-ct}$</span></li>
</ul>
<p>Then using what's given in part c) we would get:
<span class="math-container">$\alpha + \beta &gt; -\alpha + \beta \implies \alpha &gt; -\alpha \implies \alpha &gt; 0$</span></p>
<p><strong>Solution</strong></p>
<p>While writing this and re-reading Baker Streets answer I think I understand what I should have done now to avoid confusion.</p>
<p>– I should have solved my Matrix in terms of different arbitrary constants, say <span class="math-container">$c$</span> and <span class="math-container">$d$</span> and then just converted them i.e. saying <span class="math-container">$-c = \alpha$</span> and <span class="math-container">$d = \beta$</span>.
– The reason I would have known that I had to do this, is that they put a specific condition on <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> interns of <span class="math-container">$X(0)$</span> and <span class="math-container">$Y(0)$</span>, which would indicate that the answer needs to be a specific general solutions i.e. <span class="math-container">$\alpha &lt; 0$</span> so that the rest of the analysis plays out.</p>
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CC BY-SA 4.0
2023-05-05T10:36:36.293Z
Sorry for reading your answer so late. As for the first part, you wrote "I think I understand what I should have done now to avoid confusion.

– I should have solved my Matrix in terms of different arbitrary constants, say 
and  and then just converted them i.e. saying −= and =. ". Yes, perfect!
BakerStreet
CC BY-SA 4.0
2023-04-22T18:13:31.710Z
2023-04-22T18:13:31.710Z
CormJack
<p>But <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> in the solutions are <strong>arbitrary constants</strong>, true?  I suppose they are, as they are solutions of differential equations.</p>
<p>Therefore,  there is no difference between your solution  and their (but <em>they</em> who?)  solution, the solutions simply say that the first terms of <span class="math-container">$X(t)$</span> and <span class="math-container">$Y(t)$</span> have opposite coefficients.</p>
<p><span class="math-container">$\;$</span></p>
<p>The point is <strong>what means that letters <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> are <em>arbitrary constants</em>.</strong></p>
<p>What does it mean? That they vary, say in <span class="math-container">$\mathbb{R}$</span>, <strong>taking all possible values in <span class="math-container">$\mathbb{R}$</span>.</strong></p>
<p>Therefore, if you write a letter, say <span class="math-container">$C$</span>,  to mean an arbitrary value, any number you can put before it, multiplying it, doesn't change the result. That is, what you obtain is always the same, all numbers in <span class="math-container">$\mathbb{R}$</span>.</p>
<p><strong>Example:</strong></p>
<ul>
<li><p>The coefficient  is <span class="math-container">$C$</span>.  Varying <span class="math-container">$C$</span> in <span class="math-container">$\mathbb{R}$</span> you obtain all <span class="math-container">$\mathbb{R}$</span>, obviuosly .</p>
</li>
<li><p>the coefficient is <span class="math-container">$2C$</span>. What do you obtain varying <span class="math-container">$C$</span> in <span class="math-container">$\mathbb{R}$</span>? All <span class="math-container">$\mathbb{R}$</span>.</p>
</li>
<li><p>the coefficient is <span class="math-container">$- C$</span>.  What do you obtain varying <span class="math-container">$C$</span> in <span class="math-container">$\mathbb{R}$</span>? All <span class="math-container">$\mathbb{R}$</span>.</p>
</li>
</ul>
<p>Because, given any number c, any number in  <span class="math-container">$\mathbb{R}$</span>, say <span class="math-container">$K$</span>, can be obtained as solution  of <span class="math-container">$cC=K$</span>:  you will always find a <span class="math-container">$C$</span> such that, given <span class="math-container">$c$</span>, gives you <span class="math-container">$K$</span>.</p>
<p><strong>Conclusion:</strong> the coefficients in the solutions of the differential equations give you all real numbers, irrespective of how you write it with specific numbers.</p>
<p>If you write a solution of a differential equation as</p>
<p><span class="math-container">$Y(t) = C e^{at} + B e^{bt}$</span></p>
<p>or</p>
<p><span class="math-container">$Y(t) = 2C e^{at} - 4B e^{bt}$</span></p>
<p>it is exactly the same. The coefficients in both cases <strong>are just all real numbers.</strong></p>
<p>So the differences are only formal, they are only 'names' for the arbitrary constants.</p>
<p><span class="math-container">$\;$</span></p>
<p><strong>If you want the appearence of the two solutions of your problem in the question to be the same,</strong> with the same sign in the coefficients,  just take your solution</p>
<p><span class="math-container">$X(t) = -\alpha e^{ct} + \beta e^{-ct}$</span></p>
<p><span class="math-container">$Y(t) = \alpha e^{ct} + \beta e^{-ct}$</span></p>
<p>then set <span class="math-container">$-\alpha=\gamma$</span> and you obtain</p>
<p><span class="math-container">$X(t) = \gamma  e^{ct} + \beta e^{-ct}$</span></p>
<p><span class="math-container">$Y(t) = -\gamma e^{ct} + \beta e^{-ct}$</span>,</p>
<p>which is also formally the same as the other solution.</p>
<p><span class="math-container">$\;$</span></p>
<blockquote>
<p>This then goes on to affect my analysis, for instance it turns out
that <span class="math-container">$\alpha &lt; 0$</span>  therefore <span class="math-container">$-\alpha = |\alpha|$</span> which matters which
equation it's in.</p>
</blockquote>
<p>This is not clear, how can your analysis be affected?
As <span class="math-container">$\alpha$</span> is an arbitary constant, taking all value in <span class="math-container">$\mathbb{R}$</span>,  what could mean that 'it turns out that <span class="math-container">$\alpha &lt; 0$</span>'?
It is meaningless.</p>
<p>The arbitrary constants can take a specific value if you addres a <strong>further problem</strong>, an <em>initial value problem</em> (or <em>Cauchy's Problem</em>), but this is another matter.</p>

CC BY-SA 4.0
2023-04-22T18:14:54.493Z
Thanks as always for the timely response.  So while responding to this in the post below, I think I worked out how I should think about it. The original answer sheet is attached, as well as the answer that required $-\alpha = |\alpha|$. Let me know if I got it!
CormJack
CC BY-SA 4.0
2023-04-22T11:51:02.923Z
2023-04-22T15:01:26.597Z
2023-04-22T15:01:26.597Z
BakerStreet
<p>We have the following system of ordinary differential equation. To solve this using diagonlsation we do the following. My concern is that my general solution is a subtly different result, I've checked with a matrix calculator and mine seems to be correct? But the subtle difference really goes on to affect the analysis I think.</p>
<ol>
<li><span class="math-container">$\frac{dX}{dt} = -cY$</span></li>
<li><span class="math-container">$\frac{dY}{dt} = -cX$</span></li>
</ol>
<p><span class="math-container">$\begin{bmatrix} \frac{dX}{dt} \\ \frac{dY}{dt} \end{bmatrix} = \begin{bmatrix} 0 &amp; -c \\ -c &amp; 0 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}$</span> Let's denote the matrix <span class="math-container">$A =\begin{bmatrix} 0 &amp; -c \\ -c &amp; 0 \end{bmatrix}$</span></p>
<p><span class="math-container">$\det \begin{bmatrix} -\lambda &amp; -c \\ -c &amp; -\lambda \end{bmatrix} = \lambda^2 - c^2 = 0$</span></p>
<p>The eigenvalues are: <span class="math-container">$\lambda_1 = c, \quad \lambda_2 = -c$</span></p>
<ol>
<li><span class="math-container">$\lambda_1 = c$</span>: <span class="math-container">$(A - cI)v_1 = 0$</span></li>
</ol>
<p><span class="math-container">$\begin{bmatrix} -c &amp; -c \\ -c &amp; -c \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = 0$</span></p>
<p><span class="math-container">$v_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$</span></p>
<ol start="2">
<li><span class="math-container">$\lambda_2 = -c$</span>: <span class="math-container">$(A + cI)v_2 = 0$</span></li>
</ol>
<p><span class="math-container">$\begin{bmatrix} c &amp; -c \\ -c &amp; c \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = 0$</span></p>
<p><span class="math-container">$v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$</span></p>
<p>This gives me the general solution:</p>
<p><span class="math-container">$X(t) = -\alpha e^{ct} + \beta e^{-ct}$</span></p>
<p><span class="math-container">$Y(t) = \alpha e^{ct} + \beta e^{-ct}$</span></p>
<p>But what they get is</p>
<p><span class="math-container">$X(t) = \alpha e^{ct} + \beta e^{-ct}$</span></p>
<p><span class="math-container">$Y(t) = -\alpha e^{ct} + \beta e^{-ct}$</span></p>
<p>This then goes on to affect my analysis, for instance it turns out that <span class="math-container">$\alpha &lt; 0$</span>  therefore <span class="math-container">$-\alpha = |\alpha|$</span> which matters which equation it's in.</p>
<p><strong>Note:</strong> They used substitution, and i wonder if that's how they got the different answer? But this would highlight exactly the concerns i had in a <a href="https://economics.stackexchange.com/questions/55072/substitution-vs-diagonalization-for-system-of-differential-equations">previous post</a>, regarding the different general results for the two methods, particularly in say an exam setting.</p>
<p><strong>Note:</strong> Technically these were &quot;Lanchester Equations&quot; - but this seems irrelevant to the  analysis.</p>

<macroeconomics><mathematical-economics><finance>
General Solution Differential Equation
CormJack
# General Solution Differential Equation

(Asked by: username_0 on 2023-04-22)

We have the following system of ordinary differential equation. To solve this using diagonlsation we do the following. My concern is that my general solution is a subtly different result, I've checked with a matrix calculator and mine seems to be correct? But the subtle difference really goes on to affect the analysis I think.

1. $\frac{dX}{dt} = -cY$
2. $\frac{dY}{dt} = -cX$

$\begin{bmatrix} \frac{dX}{dt} \\ \frac{dY}{dt} \end{bmatrix} = \begin{bmatrix} 0 & -c \\ -c & 0 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}$ Let's denote the matrix $A =\begin{bmatrix} 0 & -c \\ -c & 0 \end{bmatrix}$

$\det \begin{bmatrix} -\lambda & -c \\ -c & -\lambda \end{bmatrix} = \lambda^2 - c^2 = 0$

The eigenvalues are: $\lambda\_1 = c, \quad \lambda\_2 = -c$

1. $\lambda\_1 = c$: $(A - cI)v\_1 = 0$

$\begin{bmatrix} -c & -c \\ -c & -c \end{bmatrix} \begin{bmatrix} v\_{11} \\ v\_{12} \end{bmatrix} = 0$

$v\_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$

2. $\lambda\_2 = -c$: $(A + cI)v\_2 = 0$

$\begin{bmatrix} c & -c \\ -c & c \end{bmatrix} \begin{bmatrix} v\_{21} \\ v\_{22} \end{bmatrix} = 0$

$v\_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

This gives me the general solution:

$X(t) = -\alpha e^{ct} + \beta e^{-ct}$

$Y(t) = \alpha e^{ct} + \beta e^{-ct}$

But what they get is

$X(t) = \alpha e^{ct} + \beta e^{-ct}$

$Y(t) = -\alpha e^{ct} + \beta e^{-ct}$

This then goes on to affect my analysis, for instance it turns out that $\alpha < 0$ therefore $-\alpha = |\alpha|$ which matters which equation it's in.

**Note:** They used substitution, and i wonder if that's how they got the different answer? But this would highlight exactly the concerns i had in a [previous post](https://economics.stackexchange.com/questions/55072/substitution-vs-diagonalization-for-system-of-differential-equations), regarding the different general results for the two methods, particularly in say an exam setting.

**Note:** Technically these were "Lanchester Equations" - but this seems irrelevant to the analysis.

**Tags:** macroeconomics, mathematical-economics, finance


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## 2 answers

### Answer by username_1 on 2023-04-22. Score: 2 | Accepted answer

But $\alpha$ and $\beta$ in the solutions are **arbitrary constants**, true? I suppose they are, as they are solutions of differential equations.

Therefore, there is no difference between your solution and their (but *they* who?) solution, the solutions simply say that the first terms of $X(t)$ and $Y(t)$ have opposite coefficients.

$\;$

The point is **what means that letters $\alpha$ and $\beta$ are *arbitrary constants*.**

What does it mean? That they vary, say in $\mathbb{R}$, **taking all possible values in $\mathbb{R}$.**

Therefore, if you write a letter, say $C$, to mean an arbitrary value, any number you can put before it, multiplying it, doesn't change the result. That is, what you obtain is always the same, all numbers in $\mathbb{R}$.

**Example:**

* The coefficient is $C$. Varying $C$ in $\mathbb{R}$ you obtain all $\mathbb{R}$, obviuosly .
* the coefficient is $2C$. What do you obtain varying $C$ in $\mathbb{R}$? All $\mathbb{R}$.
* the coefficient is $- C$. What do you obtain varying $C$ in $\mathbb{R}$? All $\mathbb{R}$.

Because, given any number c, any number in $\mathbb{R}$, say $K$, can be obtained as solution of $cC=K$: you will always find a $C$ such that, given $c$, gives you $K$.

**Conclusion:** the coefficients in the solutions of the differential equations give you all real numbers, irrespective of how you write it with specific numbers.

If you write a solution of a differential equation as

$Y(t) = C e^{at} + B e^{bt}$

or

$Y(t) = 2C e^{at} - 4B e^{bt}$

it is exactly the same. The coefficients in both cases **are just all real numbers.**

So the differences are only formal, they are only 'names' for the arbitrary constants.

$\;$

**If you want the appearence of the two solutions of your problem in the question to be the same,** with the same sign in the coefficients, just take your solution

$X(t) = -\alpha e^{ct} + \beta e^{-ct}$

$Y(t) = \alpha e^{ct} + \beta e^{-ct}$

then set $-\alpha=\gamma$ and you obtain

$X(t) = \gamma e^{ct} + \beta e^{-ct}$

$Y(t) = -\gamma e^{ct} + \beta e^{-ct}$,

which is also formally the same as the other solution.

$\;$

> This then goes on to affect my analysis, for instance it turns out
> that $\alpha < 0$ therefore $-\alpha = |\alpha|$ which matters which
> equation it's in.

This is not clear, how can your analysis be affected?
As $\alpha$ is an arbitary constant, taking all value in $\mathbb{R}$, what could mean that 'it turns out that $\alpha < 0$'?
It is meaningless.

The arbitrary constants can take a specific value if you addres a **further problem**, an *initial value problem* (or *Cauchy's Problem*), but this is another matter.

> **Comments:**
> **username_0**: Thanks as always for the timely response.  So while responding to this in the post below, I think I worked out how I should think about it. The original answer sheet is attached, as well as the answer that required $-\alpha = |\alpha|$. Let me know if I got it!  

---

### Answer by username_0 on 2023-04-22. Score: 1

I'm posting some additional clarification to the original post here here because I don't want to confuse the original. Attached here is one the original question sheet. The context is that It's from a 2020 University Exam in Mathematical Economics, 3rd year module.

1. Original Question sheet.
2. Answer to part c) where I believe having a different general solution causes problems.

Specifically we note that if we use my answer from diagonalisation:

* $X(t) = -\alpha e^{ct} + \beta e^{-ct}$
* $Y(t) = \alpha e^{ct} + \beta e^{-ct}$

Instead of the official answer

* $X(t) = \alpha e^{ct} + \beta e^{-ct}$
* $Y(t) = -\alpha e^{ct} + \beta e^{-ct}$

Then using what's given in part c) we would get:
$\alpha + \beta > -\alpha + \beta \implies \alpha > -\alpha \implies \alpha > 0$

**Solution**

While writing this and re-reading Baker Streets answer I think I understand what I should have done now to avoid confusion.

– I should have solved my Matrix in terms of different arbitrary constants, say $c$ and $d$ and then just converted them i.e. saying $-c = \alpha$ and $d = \beta$.
– The reason I would have known that I had to do this, is that they put a specific condition on $\alpha$ and $\beta$ interns of $X(0)$ and $Y(0)$, which would indicate that the answer needs to be a specific general solutions i.e. $\alpha < 0$ so that the rest of the analysis plays out.

[![enter image description here](https://i.sstatic.net/lndii.jpg)](https://i.sstatic.net/lndii.jpg)
[![enter image description here](https://i.sstatic.net/2O946.jpg)](https://i.sstatic.net/2O946.jpg)

> **Comments:**
> **username_1**: Sorry for reading your answer so late. As for the first part, you wrote "I think I understand what I should have done now to avoid confusion.  – I should have solved my Matrix in terms of different arbitrary constants, say  and  and then just converted them i.e. saying −= and =. ". Yes, perfect!  

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