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Why is it valid to divide by erte^{rt} when deriving the characteristic equation?

It is valid to divide by erte^{rt} because the exponential function is never zero for any real values of rr and tt. Since the factor is strictly nonzero, dividing both sides of the equation by it does not lose any solutions for rr, allowing the differential equation to reduce to the algebraic characteristic equation.

Conditions

  • The equation is of the form 0=ert[ar2+br+c]0 = e^{rt}[ar^2 + br + c].
  • rr and tt are real numbers.

Reasoning, step by step

  1. Factor out the common term erte^{rt} from the substituted equation to get 0=ert[ar2+br+c]0 = e^{rt}[ar^2 + br + c].
  2. Observe the behavior of erte^{rt} for r>0r > 0, r<0r < 0, and r=0r = 0.
  3. Note that in all these cases, the graph of erte^{rt} stays strictly above the tt-axis or equals 1, meaning it never hits zero.
  4. Conclude that because ert≠0e^{rt} \neq 0, you can safely divide both sides by it.
  5. Resulting equation is 0=ar2+br+c0 = ar^2 + br + c, which is the characteristic equation.

Example

The video sketches erte^{rt} for positive, negative, and zero rr. The speaker states, "But in all these cases, we notice that e to the rt is never zero. It's non-zero along the entire domain for all values of r."

Common misconceptions

  • Assuming that dividing by a variable expression is always safe without checking if it can be zero.
  • Thinking that erte^{rt} approaches zero for negative rr means it actually equals zero at some finite point.

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.