Why are the numbers switched around in this partial differential problem?

[Kajan thana]

TL;DR

changing signs with given equality.

I am going through some proofs for Damping oscillations in relation to partial differentials. Can someone help on why the numbers are switched around after giving inequality condition? Please see the images for better clarity. The highlighted characters that gets switched around.

Thank you in advance.

 

[pasmith]

I feel there's some missing context.

It looks like you're triny to solve $$m\ddot x + 2c \dot x + m\omega^2 x = 0.$$ The solution to this is $$x(t) = Ae^{\lambda_{+}t} + Be^{\lambda_{-}t}$$ where $$\lambda_{\pm} = - \frac cm \pm \sqrt{\frac{c^2}{m^2} - \omega^2}.$$ Now iif $c^2 \geq m^2 \omega^2$ then that's fine as it stands, because the quantity under the square root is positive and you'll get a real result.

Otherwise, the quantity under the square root is negative and you get a complex result, which you can write as $$\pm\sqrt{ \frac{c^2}{m^2} - \omega^2} = \pm\sqrt{-\left(\omega^2 - \frac{c^2}{m^2}\right)} = \pm i \sqrt{\omega^2 - \frac{c^2}{m^2}}$$ where $i^2 = -1$. Thus if $c^2 < m^2 \omega^2$ you have $$\lambda_{\pm} = -\frac{c}{m} \pm i \sqrt{ \omega^2 - \frac{c^2}{m^2}}.$$ For some reason the factor of $i$ in front of the square root is missing from what you have posted.

 

[Kajan thana]

I feel there's some missing context.

It looks like you're triny to solve $$m\ddot x + 2c \dot x + m\omega^2 x = 0.$$ The solution to this is $$x(t) = Ae^{\lambda_{+}t) + Be^{\lambda_{-}t$$ where $$\lambda_{\pm} = - \frac cm \pm \sqrt{\frac{c^2}{m^2} - \omega^2}.$$ Now iif $c^2 \geq m^2 \omega^2$ then that's fine as it stands, because the quantity under the square root is positive and you'll get a real result.

Otherwise, the quantity under the square root is negative and you get a complex result, which you can write as $$\pm\sqrt{ \frac{c^2}{m^2} - \omega^2} = \pm\sqrt{-\left(\omega^2 - \frac{c^2}{m^2}\right)} = \pm i \sqrt{\omega^2 - \frac{c^2}{m^2}}$$ where $i^2 = -1$. Thus if $c^2 < m^2 \omega^2$ you have $$\lambda_{\pm} = -\frac{c}{m} \pm i \sqrt{ \omega^2 - \frac{c^2}{m^2}}.$$ For some reason the factor of $i$ in front of the square root is missing from what you have posted.

Thank you Pasmith, a small copying error that led to this misunderstanding. Thank you again for taking the time to point this out to me and explaining it clearly. You are a superstar.