Complex Replacement

Question

Use the complex replacement to find a particular solution to the differential equation $$x'' + 2x' + 2x = e^{-t}\cos(2t).$$

I have been able to solve a particular solution for $x'' + 2x' + 2x = \cos(2t)$, which is $\sin(2*t)/5 - \cos(2*t)/10$. For $x'' + 2x' + 2x = e^{-t}$, I got a particular solution of $e^{-t}$ itself. I'm not sure if either (or both) of my particular solutions are wrong, but when I multiplied them together to get my final answer, I was marked wrong. Could someone please help me understand how to solve this problem?

Answer 1

To make life easier, start with $x(t)=e^{-t} z(t)$ to make the equation $$z''+z=\cos(2t)$$ $$z''+z=0 \implies z=c_1 \sin(t)+c_2 \cos(t)$$ and the particular solution of the complete equation is just $\frac 1 3 \cos(2t)$.

Just go back to $x$.

Answer 2

Both of your answers are correct. Your mistake is to multiply them to get the final answer. If you want to integrate this differential equation $$x'' + 2x' + 2x = e^{-t}\cos(2t).$$ Then use Claude's trick or use $\cos t= \frac {e^{it}+e^{-it}}{2}$ $$2x'' + 4x' + 4x = e^{-t(2i+1)}+e^{t(2i-1)}.$$

Using complex replacement $$z'' + 2z' + 2z = e^{t(2i-1)}$$ $$P(s) =s^2+2s+2 \implies P(2i-1)=-3 $$ $$z_p=\frac {e^{t(2i-1)}}{-3} $$ $$y_p=\Re (z_p)=-\frac 1 3 e^{-t} \cos(2t)$$ Finally, $$ \implies y=e^{-t}\left(c_1\cos t +c_2 \sin t -\frac 1 3 \cos(2t)\right)$$

Answer 3

$$x'' + 2x' + 2x = e^{-t}\cos(2t)$$ $$\implies (D^2+2D+2)x= e^{-t}\cos(2t)\qquad\text{where} \quad D\equiv \dfrac{d}{dt}~.$$

For particular integral (P.I.),

P.I.$~=\dfrac{1}{D^2+2D+2}~e^{-t}\cos(2t)$

$~~~~~~=e^{-t}~\dfrac{1}{(D-1)^2+2(D-1)+2}~\cos(2t)~~~~~~~~$ (using note $1$, option $2$)

$~~~~~~=e^{-t}~\dfrac{1}{D^2+1}~\cos(2t)$

$~~~~~~=e^{-t}~\dfrac{1}{-(2)^2+1}~\cos(2t)~~~~~~~~$ (using note $2$, option $2$)

$~~~~~~=-\dfrac{1}{3}~e^{-t}~\cos(2t)$

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Note$~ 1:$

For the Particular Integral (i.e., P.I.) there are some general rules

$1.$ $\frac{1}{D + a} \phi (x) = e^{-ax}\int e^{ax}\phi(x)$

$2.$ $\frac{1}{f(D)} e^{ax} \phi(x) = e^{ax}\frac{1}{f(D+a)} \phi(x)$

$3.$ $\frac{1}{f(D)} x^{n} \sin ax = $Imaginary part of $e^{iax}\frac{1}{f(D+ia)} x^n$

$4.$ $\frac{1}{f(D)} x^{n} \cos ax = $Real part of $e^{iax}\frac{1}{f(D+ia)} x^n$

$5.$ $\frac{1}{f(D)} x^{n} (\cos ax + i\sin ax) = \frac{1}{f(D)} x^n e^{iax}=e^{iax}\frac{1}{f(D+ia)} x^n$

Note$~ 2:$ For the Particular Integral (i.e., P.I.) of trigonometric functions you have to follow the following rules:

If $f(D)$ can be expressed as $\phi(D^2)$ and $\phi(-a^2)\neq 0$, then

$1.$ $\frac{1}{f(D)} \sin ax=\frac{1}{\phi(D^2)} \sin ax = \frac{1}{\phi(-a^2)} \sin ax$

$2.$ $\frac{1}{f(D)} \cos ax=\frac{1}{\phi(D^2)} \cos ax = \frac{1}{\phi(-a^2)} \cos ax$

Note: If $f(D)$ can be expressed as $\phi(D^2)=D^2+a^2$, then $\phi(-a^2)= 0$.

$1.$ $\frac{1}{f(D)} \sin ax =\frac{1}{\phi(D^2)} \sin ax=x\frac{1}{\phi'(D^2)} \sin ax= x \frac{1}{2D} \sin ax= -\frac{x}{2a} \cos ax$.

$2.$ $\frac{1}{f(D)} \cos ax =\frac{1}{\phi(D^2)} \cos ax=x\frac{1}{\phi'(D^2)} \cos ax= x \frac{1}{2D} \cos ax= \frac{x}{2a} \sin ax$.

where $\phi'(D^2)\equiv\frac{d}{dD}\phi(D^2)$