Undergrad Convert complex ODE to matrix form

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The discussion centers on converting a complex ordinary differential equation (ODE) into matrix form. The original ODE presented is aY'' + ibY' = 0, and an attempt was made to express it as a matrix [a, ib; -1, 1]. However, participants question the logic behind the matrix representation, particularly the choice of coefficients in the second row. It is suggested that to convert the second-order ODE into a system, one should introduce a new variable W(x) = Y'(x), leading to a different set of equations. Ultimately, the proposed matrix form does not accurately represent the original ODE without additional context or conditions.
SeM
Hi, I have the following complex ODE:

aY'' + ibY' = 0

and thought that it could be written as:

[a, ib; -1, 1]

Then the determinant of this matrix would give the form

a + ib = 0

Is this correct and logically sound?

Thanks!
 
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As far as I am aware, writing in matrix form is for when you have a system of differential equations in correspondingly multiple variables. I do not see how it can be applied to a single equation. On what would you base the choice of coefficients for the second row?
 
SeM said:
Hi, I have the following complex ODE:

aY'' + ibY' = 0

and thought that it could be written as:

[a, ib; -1, 1]

Then the determinant of this matrix would give the form

a + ib = 0

Is this correct and logically sound?

Thanks!

Hey, SeM,

I don't understand how you arrive at the second row in the matrix. Maybe, what you want to convert the second order differential equation into a system of two differential equations. This can be done by introducing
$$W(x)=Y'(x),$$
leading to the equations
$$aW'(x)+ibW(x)=0, \quad Y'(x)=W(x)$$

Indeed, both of your equations, i.e.
$$aY''(x)+ibY'(x)=0, Y''(x)-Y'(x)=0$$
if
$$a+ib=0,$$ but this is only a special case and it doesn't follow from your stated differential equation.
 

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