<p>You are dealing with <strong>a linear system of two differential equations.</strong></p>
<p>This system has solutions that you found using two different methods, that is solving directly the system of two equations with the eigenvalue-eigen vector method, or reducing it to a second order differential equation.</p>
<p><strong>The solutions are infinite</strong>, as they vary as the coefficients of the solution, <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, <span class="math-container">$D$</span> vary.</p>
<p><strong>Point 1</strong></p>
<blockquote>
<p>Does the coefficients ABCD in the general equations tell me anything
meaningful about the system?</p>
</blockquote>
<p>No, the coefficients are just <strong>arbitrary constant</strong>, giving rise to different solutions of the system as they vary.</p>
<p>You could have used the same letters in both cases, for example <span class="math-container">$C$</span> and <span class="math-container">$D$</span> in the first method of solution, it is the same, they are just 'names' of arbitrary constants.</p>
<p><strong>Point 2</strong></p>
<p>You wrote:</p>
<blockquote>
<p>I then tested what they gave in the first methods and then checked the
second gave the same particular solution. It did, but would this
always work for picking random initial conditions?</p>
</blockquote>
<p>The answer is <strong>yes</strong>.</p>
<p>No possible '<em>break down</em>' of the system, as you wrote in the comments.</p>
<p>Why? This is linked to a very important theorem, the <em><strong>Theorem of existence and unicity</strong></em> for the solutions of differential equations.</p>
<p>This is a <strong>fundamental theorem</strong> of the theory of differential equations.</p>
<p>What you made, imposing two initial conditions, is finding a solution of a so-called <em><strong>Cauchy's problem</strong></em>, which for a first order differential equation is defined as follow:</p>
<p><span class="math-container">$$y'(x)= f(x, y(x)) \;\;\;\;       x\in I \qquad (1.1)$$</span></p>
<p><span class="math-container">$$ y(x_0)=0 \qquad \qquad \qquad \qquad \;(1.2)$$</span></p>
<p>where the equation in <span class="math-container">$(1.1)$</span> is a first order differential equation in <em>normal form</em><span class="math-container">$^1$</span>, <span class="math-container">$I \subseteq \mathbb{R}$</span>, <span class="math-container">$x_0 \in \mathbb{R}$</span>, <span class="math-container">$f$</span> is a known function.</p>
<p>Equation <span class="math-container">$(1.1)$</span> together with the initial condition <span class="math-container">$(1.2)$</span> is called the <strong><em>Cauchy's Problem</em> for the differential equation <span class="math-container">$(1.1)$</span>: that is, a differential equation of first order together with an initial condition.</strong></p>
<p>Analogously, <strong>for a differential equation of order <span class="math-container">$n$</span> (in normal form), the Cauchy's problem</strong> is:</p>
<p><span class="math-container">$$ y^{(n)}= f(x, y(x), y'(x),..., y^{(n-1)}) \qquad x\in I \qquad (2)$$</span>
<span class="math-container">$$y(x_0)= y_0$$</span>
<span class="math-container">$$y'(x_0)= y_1$$</span>
<span class="math-container">$$...$$</span>
<span class="math-container">$$y^{(n-1)}= y_{n-1}$$</span></p>
<p>where <span class="math-container">$x_0, y_0, y_1, y_{n-1}$</span> are given real numbers.</p>
<p>As you can notice, in this case there are <strong><span class="math-container">$n$</span> initial conditions,</strong> they are about the function <span class="math-container">$y(x)$</span> and its derivatives, until order <span class="math-container">$n-1$</span>.</p>
<p><strong>The Theorem of existence and unicity says that, under appropriate assumptions, the Cauchy's problems above have one and only one solution</strong> <span class="math-container">$^2$</span>.</p>
<p>Therefore, you are assured that the solution exists, it is unique, given any arbitrary initial condition,  no 'break down' occurs, and obviuosly it is indipendent of the method of solution.</p>
<p><strong>REMARK</strong>: It can be proved that any differental equation of order <span class="math-container">$n$</span> in normal form <strong>can be transformed (is equivalent to) a system of linear <span class="math-container">$n$</span> equations of the first order, and vice versa.</strong></p>
<p>That is the reason why you can solve the system through the substitution method.</p>
<p>Therefore, the theorem of existence and unicity <strong>immediately holds also for systems of first order differential equations</strong> as your system.</p>
<hr />
<p><span class="math-container">$^1$</span> <sup>A differential equation is said in <em>normal form</em> if  a function <span class="math-container">$f$</span> exists such that the equation can be written as <span class="math-container">$ y^{(n)}= f(x, y(x), y'(x),..., y^{(n-1)})$</span>. </sup></p>
<p><span class="math-container">$^2$</span> <sup> One must be careful, because the assumptions of the theorem are not trivial (and also the proofs of the theorem, there are two standard proofs in the literature and both are not trivial).  But those assumptions <strong>are always satisfied in the case of linear differential equations</strong> as in your case. Moreover, this theorem is a <strong>local result</strong>, holding only in an appropriate subset of <span class="math-container">$\mathbb{R^n}$</span>. </sup></p>

CC BY-SA 4.0
2023-04-14T20:32:57.223Z
Another great answer @BakerStreet, thank you! Just to slightly rephrase my first question. Is it meaningful that I have produced two seemingly different general solutions? For example I find it weird that in an sam let's say, two completely different correct answers could be produce.

Another way of thinking about this is. the matrix that Giskard produced in the comments. Is there a theory determine what that matrix would be, that allows us to correspond between the general solution from diagonalisation vs, the general solution from substitution?
CormJack
CC BY-SA 4.0
2023-04-14T20:35:03.990Z
Also your remark/reminder that any normal form differential equation corresponds to a system is helpful. Does it work in reverse order always. The way I started with a system here and turned it into a single second order equation. I assume for a linear system of differential equations it must?
CormJack
CC BY-SA 4.0
2023-04-14T20:35:35.653Z
Also apologies I must remember to add reference numbers to my key equations, as you mentioned before!!
CormJack
CC BY-SA 4.0
2023-04-14T20:49:45.010Z
@ CormJack Don’t mind of the numbers beside the equations!
BakerStreet
CC BY-SA 4.0
2023-04-14T20:50:42.877Z
As for your question above, your general solutions, obtained with the two methods,  aren’t different! Don’t mind the numbers before the $e$! As there are arbitrary constants, the solutions obtained are the same. You can obtain whichever number you want before the $e$ just varying the arbitrary constants. 
Thanks for your kindness.
BakerStreet
CC BY-SA 4.0
2023-04-14T14:23:56.160Z
2023-04-14T14:23:56.160Z
BakerStreet
<p>Take the following system of differential equations.</p>
<ul>
<li><span class="math-container">$x'(t) = x + 2y$</span></li>
<li><span class="math-container">$y'(t) = 3x + 2y$</span></li>
</ul>
<p>I have used two different methods (substitution and diagonalisation) to solve the system. The working is just fine, and is included for context. I want to ask the following questions about the two different approaches.</p>
<ol>
<li><p>Does the coefficients ABCD in the <strong>general equations</strong> tell me anything meaningful about the system? I ask because two different methods give two seemingly different general equations, that only converge once a general solution is provided. Is there a relationship between them? Is one more &quot;right&quot; than the other?</p>
</li>
<li><p>This question didn't actually come with a particular solutions. The solutions I use,<span class="math-container">$x(0) = 2$</span> , <span class="math-container">$y(0) = 1$</span>, I invented. I then tested what they gave in the first methods and then checked the second gave the same particular solution. It did, but would this always work for picking random initial conditions?</p>
</li>
<li><p>The second method doesn't directly give you the Eigen Vectors which are useful for drawing the phase portraits? Would i need to take the roots from the characteristic equation and use them to find Eigen Vectors in the normal way. Or they implicit somewhere else in the second approach!.</p>
</li>
</ol>
<ul>
<li>Thanks. Working below for context!</li>
</ul>
<p><strong>Method 1: Diagonalisation</strong></p>
<p><span class="math-container">$λ_1 = 4$</span> &amp;  <span class="math-container">$λ_2 = -1$</span></p>
<p>Eigen Vectors <span class="math-container">$V_1 = (2,3)^T$</span> &amp; <span class="math-container">$V_2 = (-1, 1)^T$</span></p>
<p><span class="math-container">$P = \begin{pmatrix} 2&amp; -1 \\3&amp; \;1 \end{pmatrix}$</span></p>
<ul>
<li><strong>General System 1:</strong></li>
<li><span class="math-container">$x(t) = 2Ae^{4t} - Be^{-t}$</span></li>
<li><span class="math-container">$y(t) = 3Ae^{4t} + Be^{-t}$</span></li>
</ul>
<p>Working with the particular solutions.
<span class="math-container">$x(0) = 2$</span> , <span class="math-container">$y(0) = 1$</span></p>
<p>Solving the System: <span class="math-container">$\begin{pmatrix} 2&amp; -1 \\3&amp; \;1 \end{pmatrix} \begin{pmatrix} A \\B\end{pmatrix} = \begin{pmatrix} 2 \\1\end{pmatrix}$</span>  You get: <span class="math-container">$A = \frac{3}{5}$</span> , <span class="math-container">$B = -\frac{4}{5}$</span></p>
<p>Which then plugging back  into the system gives you</p>
<ul>
<li><span class="math-container">$x(t) = \frac{6}{5}e^{4t} + \frac{4}{5}e^{-t}$</span></li>
<li><span class="math-container">$y(t) = \frac{9}{5}e^{4t} - \frac{4}{5}e^{-t}$</span></li>
</ul>
<p><strong>Method 2: Substitution</strong></p>
<p><span class="math-container">$x'(t) = x + 2y$</span>
<span class="math-container">$y'(t) = 3x + 2y$</span></p>
<p>Through substitution we get the Linear Second Order Homogenous ODE in <span class="math-container">$x(t)$</span></p>
<p><span class="math-container">$x''(t) - 3x'(t) - 4x(t) = 0$</span></p>
<p>The characteristic equation <span class="math-container">$(λ - 4)(λ + 1)$</span> gives the expected: <span class="math-container">$λ_1 = 4 \;, λ_2 = -1$</span></p>
<ul>
<li><strong>General System 2</strong></li>
<li><span class="math-container">$x(t) = Ce^{4t} + De^{-t}$</span></li>
</ul>
<p>Then using: <span class="math-container">$y(t) = \frac{1}{2}[x'(t) - x(t)]  =  \frac{1}{2}[x'(t) - x(t)] = \frac{1}{2}[4Ce^{4t} - De^{-t} - (Ce^{4t} + De^{-t})]$</span></p>
<ul>
<li><span class="math-container">$y(t) =  \frac{3}{2}Ce^{4t} - De^{-t}$</span></li>
</ul>
<p>So then solving the system: <span class="math-container">$\begin{pmatrix} 1&amp; \;1 \\\frac{3}{2}&amp; -1 \end{pmatrix} \begin{pmatrix} C \\D\end{pmatrix} = \begin{pmatrix} 2 \\1\end{pmatrix}$</span> <span class="math-container">$C = \frac{6}{5}$</span> , <span class="math-container">$D = \frac{4}{5}$</span></p>
<p>Which then plugging back into the system gives the desired result:</p>
<ul>
<li><span class="math-container">$x(t) = \frac{6}{5}e^{4t} + \frac{4}{5}e^{-t}$</span></li>
<li><span class="math-container">$y(t) = \frac{18}{10}e^{4t} - \frac{4}{5}e^{-t} = \frac{9}{5}e^{4t} - \frac{4}{5}e^{-t}$</span></li>
</ul>

CC BY-SA 4.0
2023-04-14T07:25:31.840Z
Hi! Could you please elaborate on your question 2., I am not sure I understand. If the two solution methods both work, they should come to the same conclusion. Are you asking if the people who invented them made a mistake?
Giskard
CC BY-SA 4.0
2023-04-14T08:18:45.397Z
hey Giskard, thanks for your response! No, the second question what I'm asking is. The original question came with no particular solutions. So I just invented those particular solutions that you see. Like I randomly just said okay x(0) = 2, y(0) = 1. And then I worked through what that would give me with the first method, and then I worked through what that would give me with the second method, and they both gave the same (I assume) correct answer.
CormJack
CC BY-SA 4.0
2023-04-14T08:20:10.090Z
My Question is. Will that always work. Can I always just make up a random particular x(o) and y(0), and that will always give me some meaningful particular solution. Or could I have generated a random x(0) and y(0) where perhaps the system would have broke down. A symptom of that maybe being that the two different approaches give a different particular solution or something?
CormJack
CC BY-SA 4.0
2023-04-14T08:21:28.910Z
The reason it was interesting to me, and why these questions are totally related. UP until the very last second where I solved for the particular solution of the second method. I had no idea if any of my working was "correct" because I had no answer sheet, and also. The general solutions for each method look very different, which I found strange and have hence asked a question on that as well. Hope that's more clear!
CormJack
CC BY-SA 4.0
2023-04-14T09:02:35.647Z
You can try to compute a general solution for particular solutions $x(0) = x_0$, $y(0) = y_0$. Currently this is not too hard. Note that there is a relationship between your matrices:
$\begin{pmatrix} 1& \;1 \\\frac{3}{2}& -1 \end{pmatrix} \begin{pmatrix} 2& 0 \\0& \; -1 \end{pmatrix}  = \begin{pmatrix} 2& -1 \\3& \;1 \end{pmatrix}$. This implies that for any particular starting condition you will get $C = 2A$, $D=-B$. Substituting these in, you will have the same equations for $x(t),y(t)$. 

The relationship also implies that the columns of the second matrix are eigenvectors as well.
Giskard
<mathematical-economics>
Substitution VS Diagonalization for System of Differential Equations
CormJack
# Substitution VS Diagonalization for System of Differential Equations

(Asked by: username_0 on 2023-04-13)

Take the following system of differential equations.

* $x'(t) = x + 2y$
* $y'(t) = 3x + 2y$

I have used two different methods (substitution and diagonalisation) to solve the system. The working is just fine, and is included for context. I want to ask the following questions about the two different approaches.

1. Does the coefficients ABCD in the **general equations** tell me anything meaningful about the system? I ask because two different methods give two seemingly different general equations, that only converge once a general solution is provided. Is there a relationship between them? Is one more "right" than the other?
2. This question didn't actually come with a particular solutions. The solutions I use,$x(0) = 2$ , $y(0) = 1$, I invented. I then tested what they gave in the first methods and then checked the second gave the same particular solution. It did, but would this always work for picking random initial conditions?
3. The second method doesn't directly give you the Eigen Vectors which are useful for drawing the phase portraits? Would i need to take the roots from the characteristic equation and use them to find Eigen Vectors in the normal way. Or they implicit somewhere else in the second approach!.

* Thanks. Working below for context!

**Method 1: Diagonalisation**

$λ\_1 = 4$ & $λ\_2 = -1$

Eigen Vectors $V\_1 = (2,3)^T$ & $V\_2 = (-1, 1)^T$

$P = \begin{pmatrix} 2& -1 \\3& \;1 \end{pmatrix}$

* **General System 1:**
* $x(t) = 2Ae^{4t} - Be^{-t}$
* $y(t) = 3Ae^{4t} + Be^{-t}$

Working with the particular solutions.
$x(0) = 2$ , $y(0) = 1$

Solving the System: $\begin{pmatrix} 2& -1 \\3& \;1 \end{pmatrix} \begin{pmatrix} A \\B\end{pmatrix} = \begin{pmatrix} 2 \\1\end{pmatrix}$ You get: $A = \frac{3}{5}$ , $B = -\frac{4}{5}$

Which then plugging back into the system gives you

* $x(t) = \frac{6}{5}e^{4t} + \frac{4}{5}e^{-t}$
* $y(t) = \frac{9}{5}e^{4t} - \frac{4}{5}e^{-t}$

**Method 2: Substitution**

$x'(t) = x + 2y$
$y'(t) = 3x + 2y$

Through substitution we get the Linear Second Order Homogenous ODE in $x(t)$

$x''(t) - 3x'(t) - 4x(t) = 0$

The characteristic equation $(λ - 4)(λ + 1)$ gives the expected: $λ\_1 = 4 \;, λ\_2 = -1$

* **General System 2**
* $x(t) = Ce^{4t} + De^{-t}$

Then using: $y(t) = \frac{1}{2}[x'(t) - x(t)] = \frac{1}{2}[x'(t) - x(t)] = \frac{1}{2}[4Ce^{4t} - De^{-t} - (Ce^{4t} + De^{-t})]$

* $y(t) = \frac{3}{2}Ce^{4t} - De^{-t}$

So then solving the system: $\begin{pmatrix} 1& \;1 \\\frac{3}{2}& -1 \end{pmatrix} \begin{pmatrix} C \\D\end{pmatrix} = \begin{pmatrix} 2 \\1\end{pmatrix}$ $C = \frac{6}{5}$ , $D = \frac{4}{5}$

Which then plugging back into the system gives the desired result:

* $x(t) = \frac{6}{5}e^{4t} + \frac{4}{5}e^{-t}$
* $y(t) = \frac{18}{10}e^{4t} - \frac{4}{5}e^{-t} = \frac{9}{5}e^{4t} - \frac{4}{5}e^{-t}$

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: Hi! Could you please elaborate on your question 2., I am not sure I understand. If the two solution methods both work, they should come to the same conclusion. Are you asking if the people who invented them made a mistake?  
> **username_0**: hey Giskard, thanks for your response! No, the second question what I'm asking is. The original question came with no particular solutions. So I just invented those particular solutions that you see. Like I randomly just said okay x(0) = 2, y(0) = 1. And then I worked through what that would give me with the first method, and then I worked through what that would give me with the second method, and they both gave the same (I assume) correct answer.  
> **username_0**: My Question is. Will that always work. Can I always just make up a random particular x(o) and y(0), and that will always give me some meaningful particular solution. Or could I have generated a random x(0) and y(0) where perhaps the system would have broke down. A symptom of that maybe being that the two different approaches give a different particular solution or something?  
> **username_0**: The reason it was interesting to me, and why these questions are totally related. UP until the very last second where I solved for the particular solution of the second method. I had no idea if any of my working was "correct" because I had no answer sheet, and also. The general solutions for each method look very different, which I found strange and have hence asked a question on that as well. Hope that's more clear!  
> **username_1**: You can try to compute a general solution for particular solutions $x(0) = x_0$, $y(0) = y_0$. Currently this is not too hard. Note that there is a relationship between your matrices: $\begin{pmatrix} 1& \;1 \\\frac{3}{2}& -1 \end{pmatrix} \begin{pmatrix} 2& 0 \\0& \; -1 \end{pmatrix}  = \begin{pmatrix} 2& -1 \\3& \;1 \end{pmatrix}$. This implies that for any particular starting condition you will get $C = 2A$, $D=-B$. Substituting these in, you will have the same equations for $x(t),y(t)$.   The relationship also implies that the columns of the second matrix are eigenvectors as well.  

---

## 1 answers

### Answer by username_2 on 2023-04-14. Score: 1 | Accepted answer

You are dealing with **a linear system of two differential equations.**

This system has solutions that you found using two different methods, that is solving directly the system of two equations with the eigenvalue-eigen vector method, or reducing it to a second order differential equation.

**The solutions are infinite**, as they vary as the coefficients of the solution, $A$, $B$, $C$, $D$ vary.

**Point 1**

> Does the coefficients ABCD in the general equations tell me anything
> meaningful about the system?

No, the coefficients are just **arbitrary constant**, giving rise to different solutions of the system as they vary.

You could have used the same letters in both cases, for example $C$ and $D$ in the first method of solution, it is the same, they are just 'names' of arbitrary constants.

**Point 2**

You wrote:

> I then tested what they gave in the first methods and then checked the
> second gave the same particular solution. It did, but would this
> always work for picking random initial conditions?

The answer is **yes**.

No possible '*break down*' of the system, as you wrote in the comments.

Why? This is linked to a very important theorem, the ***Theorem of existence and unicity*** for the solutions of differential equations.

This is a **fundamental theorem** of the theory of differential equations.

What you made, imposing two initial conditions, is finding a solution of a so-called ***Cauchy's problem***, which for a first order differential equation is defined as follow:

$$y'(x)= f(x, y(x)) \;\;\;\; x\in I \qquad (1.1)$$

$$ y(x\_0)=0 \qquad \qquad \qquad \qquad \;(1.2)$$

where the equation in $(1.1)$ is a first order differential equation in *normal form*$^1$, $I \subseteq \mathbb{R}$, $x\_0 \in \mathbb{R}$, $f$ is a known function.

Equation $(1.1)$ together with the initial condition $(1.2)$ is called the ***Cauchy's Problem* for the differential equation $(1.1)$: that is, a differential equation of first order together with an initial condition.**

Analogously, **for a differential equation of order $n$ (in normal form), the Cauchy's problem** is:

$$ y^{(n)}= f(x, y(x), y'(x),..., y^{(n-1)}) \qquad x\in I \qquad (2)$$
$$y(x\_0)= y\_0$$
$$y'(x\_0)= y\_1$$
$$...$$
$$y^{(n-1)}= y\_{n-1}$$

where $x\_0, y\_0, y\_1, y\_{n-1}$ are given real numbers.

As you can notice, in this case there are **$n$ initial conditions,** they are about the function $y(x)$ and its derivatives, until order $n-1$.

**The Theorem of existence and unicity says that, under appropriate assumptions, the Cauchy's problems above have one and only one solution** $^2$.

Therefore, you are assured that the solution exists, it is unique, given any arbitrary initial condition, no 'break down' occurs, and obviuosly it is indipendent of the method of solution.

**REMARK**: It can be proved that any differental equation of order $n$ in normal form **can be transformed (is equivalent to) a system of linear $n$ equations of the first order, and vice versa.**

That is the reason why you can solve the system through the substitution method.

Therefore, the theorem of existence and unicity **immediately holds also for systems of first order differential equations** as your system.

---

$^1$ A differential equation is said in *normal form* if a function $f$ exists such that the equation can be written as $ y^{(n)}= f(x, y(x), y'(x),..., y^{(n-1)})$.

$^2$  One must be careful, because the assumptions of the theorem are not trivial (and also the proofs of the theorem, there are two standard proofs in the literature and both are not trivial). But those assumptions **are always satisfied in the case of linear differential equations** as in your case. Moreover, this theorem is a **local result**, holding only in an appropriate subset of $\mathbb{R^n}$.

> **Comments:**
> **username_0**: Another great answer @BakerStreet, thank you! Just to slightly rephrase my first question. Is it meaningful that I have produced two seemingly different general solutions? For example I find it weird that in an sam let's say, two completely different correct answers could be produce.  Another way of thinking about this is. the matrix that Giskard produced in the comments. Is there a theory determine what that matrix would be, that allows us to correspond between the general solution from diagonalisation vs, the general solution from substitution?  
> **username_0**: Also your remark/reminder that any normal form differential equation corresponds to a system is helpful. Does it work in reverse order always. The way I started with a system here and turned it into a single second order equation. I assume for a linear system of differential equations it must?  
> **username_0**: Also apologies I must remember to add reference numbers to my key equations, as you mentioned before!!  
> **username_2**: @ CormJack Don’t mind of the numbers beside the equations!  
> **username_2**: As for your question above, your general solutions, obtained with the two methods,  aren’t different! Don’t mind the numbers before the $e$! As there are arbitrary constants, the solutions obtained are the same. You can obtain whichever number you want before the $e$ just varying the arbitrary constants.  Thanks for your kindness.  

---

