General Exponential Response Formula [ODE]

The 2nd question is

and the solution is :

My question is, how do they go from the complex form to the real form? Re(z), what is this operation.

Also, if the function had been a sin(t) instead of cos(t) in the ODE, then I would take Im(z), how do I do this operation aswell?

An example where that might apply is let's say x''+5'x+6x = sin(5t)

• Welcome to math.stackexchange. This is a well-posed first question (you showed effort, nice screenshot, ...). Do no forget to accept an answer or ask for more information if it was still unclear. Mar 17, 2015 at 0:02

$Re(x)$ gives you the real part: $Re(x) = Re(a + bi) = cos(a)$

$Im(x)$ gives you the imagniary part: $Im(x) = Re(a + bi) = sin(b)$

In your example: $x_P = Re(z_p) = Re(\frac{te^{-t}( cos(t) + i sin(t) )}{2i}) = Re(\frac{te^{-t}cos(t)}{2i} + \frac{te^{-t}sin(t)}{2}) = Re(-\frac{ite^{-t}cos(t)}{2} + \frac{te^{-t}sin(t)}{2}) = \frac{te^{-t}sin(t)}{2}$

$Im(z_p) = Im(-\frac{ite^{-t}cos(t)}{2} + \frac{te^{-t}sin(t)}{2}) = -\frac{te^{-t}cos(t)}{2}$.

let us take the example you have. $$x'' + 5x' + 6x = \sin 5t.$$ then $$x = e^{\lambda t}$$ is a solution of the homogenous equation if $$\lambda^2 + 5 \lambda + 6 = 0 \to \lambda = \frac{-5 \pm \sqrt{1}}{2}=-2, -3$$ unfortunately, this equation has no complex roots.

let us try instead $$x'' + 4x' + 6x = \sin 5t.$$ the char equation is $$\lambda^2 + 4 \lambda + 6 = 0 \to \lambda = \frac{-4 \pm \sqrt{-8}}{2}=-2\pm\sqrt 2i$$

you can verify that both $$x = e^{-2t}\cos\sqrt 2 t,\, x = e^{-2t}\cos\sqrt 2 t$$ satisfy the homogeneous equation.

these two solution can be thought as coming from the real and imaginary parts of $$e^{-2t + i\sqrt 2t} = e^{-2t}\left( \cos \sqrt 2 t + i \sin \sqrt 2 t\right).$$ i have used the eulers formula $e^{it} = \cos t + i \sin t.$