Comments by "MC116" (@angelmendez-rivera351) on "Solving a Separable Differential Equation dS/dr = kS" video.

  1. Actually, one could be far more careful here. dS/dr = k·S implies S = 0, or 1/S·dS/dr = k. Consider f(S) = ln(–S) + A iff S < 0 and f(S) = ln(S) + B iff S > 0. f'(S) = 1/S. Hence 1/S·dS/dr = d[f(S)]/dr = k, which implies ln(–S) = k·r – A iff S < 0, and ln(S) = k·r – B iff S > 0. With this, we have S = 0, or S = –exp(–A)·exp(k·r) iff S < 0, or S = exp(–B)·exp(k·r) iff S > 0. S = 0 = 0·exp(k·r) = λ·exp(k·r) iff λ = 0, while S = –exp(–A)·exp(k·r) = λ·exp(k·r) iff λ < 0, and S = exp(–B)·exp(k·r) = λ·exp(k·r) iff λ > 0. Therefore, S = λ·exp(k·r).
    1