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I want to get a particular solution to the differential equation y''+2y'+2y=2e^x cos(x) and therefore I would like to 'complexify' the right hand side. This means that I want to write the right hand side as $q(x)e^{\alpha x}$ with $q(x)$ a polynomial. How is this possible?

The solution should be $(1/4)e^x(\sin(x)+\cos(x))$ but I cannot see that.

3 Answers 3

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The point is that (for real $x$) $2 e^x \cos(x)$ is the real part of $2 e^x e^{ix} = 2 e^{(1+i)x}$. Find a particular solution of y'' + 2 y' + 2 y = 2 e^{(1+i)x}, and its real part is a solution of y'' + 2 y' + 2 y = 2 e^x \cos(x).

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    The passage from $2 e^x \cos(x)$ to $2 e^{(1+i)x}$ could be considered as "complexifying".2011-06-29
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As $\cos x=\frac{e^{ix}+e^{-ix}}{2}$, $2e^x \cos x = e^{x+ix}+e^{x-ix}$, but that is not of the requested form. Is it close enough?

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    Find the particular solution with RHS $e^{x+ix}$, then find the particular solution with RHS $e^{x-ix}$. Your final particular solution is then a linear combination of these.2011-06-28
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I know that this might be awfully late, but I just started learning about complexification this term and thought I would put up my solution--please excuse any incorrect language I might use as it's the idea I am trying to get across.

First we start off by defining a complex analogue to your function:

eq {1}: $ z''+2z'+2z=2e^xe^{ix} $

where $z=RE(y)+i*IM(y)$

Basically, we can recover the original diffEq by extracting the real part of our complex diffEq. The next step is to use the method of undetermined coefficients to find a guess for what our particular complex solution might be. Guess:

eq {2}: $ z=Ae^{x}e^{ix} $ so that:

$ z'=Ae^{x}e^{ix}+iAe^{x}e^{ix} $ and $ z''=i2Ae^{x}e^{ix} $

Plugging this into {1}:

$ i2Ae^{x}e^{ix} + 2(Ae^{x}e^{ix}+iAe^{x}e^{ix}) + 2(Ae^{x}e^{ix})= 2e^xe^{ix} $

We can simplify by removing the common factor of $ e^{x}e^{ix} $:

$ A(4+4i)=2 => A= \frac{1}{2 + 2i}$

Convert A to complex polar form:

$ A=\frac{\sqrt[]{2}}{4}e^{-i\frac{\pi}{4}} $

Plugging this into {2}:

$ z=\frac{\sqrt[]{2}}{4}e^{-i\frac{\pi}{4}}e^{x}e^{ix} $

This can be simplified to eq {3}:

$ z=\frac{\sqrt[]{2}}{4}e^{x}e^{i(x-\frac{\pi}{4})} $

Since our particular solution should be of the form $cos(x)$, we take the real part of {3} and call that our particular x-solution:

$ x = RE(z) = \frac{\sqrt[]{2}}{4}e^{x}cos(x-\frac{\pi}{4}) $

Finally using our difference of cosine identity:

$ \frac{\sqrt[]{2}}{4}e^{x}(cos(x)\frac{\sqrt[]{2}}{2} + sin(x)\frac{\sqrt[]{2}}{2}) $ $ \frac{\sqrt[]{2}}{4}\frac{\sqrt[]{2}}{2}e^{x}(cos(x) + sin(x)) $ $ \frac{2}{8}e^{x}(cos(x) + sin(x)) $ $ \frac{1}{4}e^{x}(cos(x) + sin(x)) $

$QED$