Solving ODEs with Complex Numbers: A Comprehensive Guide

In summary, the conversation is about using a specific technique to solve a particular solution in ODEs. The technique involves substituting z = ue^{2xi} and then taking the real part of z to find the solution. The conversation also discusses the motivation behind this method and how it can be applied to other similar ODEs. The conversation also touches on the idea of using undetermined coefficients and making substitutions in order to solve ODEs.
  • #1
Benny
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0
Hi, I've been working on some ODEs and I've been using all of the standard techniques. Recently, I came across some solutions to some IVP problems(I don't have the questions, only the solutions). I'm curious as to the motivation behind the follow technique. As in, why would this method be used and what is its name?

To find the particular solution, solve instead [tex]z'' + 2z' + z = e^{2xi} [/tex] and afterwards take the real part of z.

Let [tex]z = ue^{2xi} [/tex] to get [tex]z' = u'e^{2xi} + 2ie^{2xi} u[/tex] and [tex]z'' = u''e^{2xi} + 4ie^{2xi} u' - 4e^{2xi} u[/tex].

Substituting into the DE gives:

[tex]
u''e^{2xi} + 4ie^{2xi} u' - 4e^{2xi} u + 2u'e^{2xi} + 4ie^{2xi} u + ue^{2xi} = e^{2xi}
[/tex]

[tex]
u'' + \left( {4i + 2} \right)u' + \left( { - 3 + 4i} \right)u = 1
[/tex]

This is satisified by [tex]u = \frac{1}{{ - 3 + 4i}} = \frac{1}{{25}}\left( { - 3 - 4i} \right)[/tex].

Hence [tex]z = \frac{1}{{25}}\left( { - 3 - 4i} \right)e^{2xi} = \frac{1}{{25}}\left( { - 3 - 4i} \right)\left( {\cos 2x + i\sin 2x} \right)[/tex].

[tex]
{\mathop{\rm Re}\nolimits} al\left( z \right) = \frac{1}{{25}}\left( { - 3\cos 2x + 4\sin 2x} \right)
[/tex]

Hence A = -(3/25) and B = 4/25.

Particular solution is [tex]y_p = - \frac{3}{{25}}\cos 2x + \frac{4}{{25}}\sin 2x[/tex].

So general solution is [tex]y = c_1 e^{ - x} + c_2 xe^{ - x} - \frac{3}{{25}}\cos 2x + \frac{4}{{25}}\sin 2x[/tex].


I know that its hard to explain without having the original question but I would really like some help with this one. Looking at the general solution the characteristic equation is [tex]\left( {\lambda + 1} \right)^2 = 0[/tex] so the LHS the original DE is probably [tex]y'' + 2y' + 1[/tex]. The RHS is non-zero and looking at the general solution would suggest that it is of the form [tex]A\cos \left( {2x} \right) + B\sin \left( {2x} \right)[/tex] where either A or B, but not both, could be zero.

So the original ODE(again I apologise for not having the actual question with me) is porbably something like:

[tex]
y'' + 2y' + 1 = A\cos \left( {2x} \right) + B\sin \left( {2x} \right)
[/tex]

The beginning of the solution replaces y by z and sets z = uexp(2xi) so z = u(cos(2x)+isin(2x)) but what is the point in doing that? The form of the solution suggests that perhaps even undetermined coefficients would work(although its hard to say w/o the actual question) It seems that the question is the kind where you make the substitution y = a(x)b(x), where b(x) is a part of the complimentary solution. In any case if we were to solve the ODE(or any ODE where this technique of using z complex is applicable) in z instead, how do we know whether to take the real part, imaginary part or both?

I'd really like to know how to do use this technique(considering an equation in z and then taking the real or imaginary component) as it seems quite useful. So any explanations as to why one would consider an equation in z instead of y to solve an ODE of this kind would be great thanks.
 
Last edited:
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  • #2
kind of seems similar to reduction of order, is this the section of higher order differential equations you were referring to (meaning higher then 2nd order)?
 
  • #3
The section I referred to means 2nd or higher order.
 

1. What are ODEs?

ODEs, or ordinary differential equations, are mathematical equations that describe how a variable changes over time. They are used to model many real-world situations, such as population growth, motion of objects, and chemical reactions.

2. Why use complex numbers to solve ODEs?

Complex numbers allow us to represent both real and imaginary components of a solution to an ODE. This can be useful in situations where the solution involves oscillations or exponential growth/decay, which often involve complex numbers.

3. What are the steps for solving ODEs with complex numbers?

The steps for solving ODEs with complex numbers include: identifying the type of ODE (linear, non-linear, etc.), finding the general solution using complex numbers, substituting in initial conditions to find the specific solution, and checking the solution for accuracy.

4. Can complex numbers be used to solve all types of ODEs?

No, not all ODEs can be solved using complex numbers. Some ODEs may not have complex solutions, while others may require more advanced techniques such as Laplace transforms.

5. Are there any practical applications for solving ODEs with complex numbers?

Yes, there are many practical applications for solving ODEs with complex numbers. Some examples include modeling electrical circuits, analyzing vibrations in mechanical systems, and predicting chemical reactions.

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