- #1

Benny

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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?

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.

*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].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.

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