General Solution Differential Equation

Question

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, 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.

Answer 1

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.

Answer 2

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.