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Here are two questions regarding 2nd order linear DE's:

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Regarding (5), I believe the only requirement for the roots of the CE are that they be real, and negative. If they are complex, we may write the solution without trigonometric functions, and no oscillation will occur.

Regarding (6)a, I believe we need the equation $\frac{dF}{dt} = \frac{1}{10} (70 -F(t))$, which solved yields $F(t) = c_1 e^{-\frac{t}{10}} + 70$ with $t$ in hours. Now I'm not sure what to do now, as using $y(0) = 70$ yields $c_1 = 0$, which naturally isn't correct. Any suggestions helpful.

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You can have damped harmonic motion. The quadratic equation gives a leading term of $-\frac{b}{2a}$ which is negative when the ratio of $b$ and $a$ is positive. Therefore you get a negative $\lambda$ when $\frac{b}{a} > 0$. If there are complex roots $b^2 < 4ac$ and $\frac{b}{a} > 0$ then you get damped harmonic motion (complex solutions with a leading, negative exponential). If $b^2 > 4ac$, then you can exponential decay when $\frac{b}{a} < 0$. This happens when either all of the coefficients are positive or all are negative and $b^2 > 4ac$ (note that this means $ac > 0$).

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  • $\begingroup$ Although so essentially I'm only concerned with $\frac{b}{a} > 0$, and the last bit about $b^2 > 4ac$? What about (6)? $\endgroup$ Apr 9, 2014 at 2:08
  • $\begingroup$ Well actually you also need $ac > 0$ which means that $a$, $b$, and $c$ must all have the same sign (when $b^2 > ac$). When you get complex roots ($b^2 < 4ac$) then it doesn't matter as long as you add the imaginary part to a negative real part. $\endgroup$
    – Jared
    Apr 9, 2014 at 2:19
  • $\begingroup$ thank you, do you have anything regarding 6? $\endgroup$ Apr 9, 2014 at 2:54

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