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Determine the values of $r$ for which the differential equation $y'+8y=0$ has solutions of the form $y=e^{rt}$.

We have never done a problem like this in class. We've only done direction lines and separable differential equations.

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If you plug $y = e^{rt}$ into the given differential equation, you get $$re^{rt} + 8e^{rt} = 0,$$ or $$(r + 8)e^{rt} = 0. \tag{$\ast$}$$ Now $e^{rt} \neq 0$ for all $t$, so you can divide both sides of $(\ast)$ by $e^{rt}$ to get $$r + 8 = 0.$$ This tells us that $r = -8$.

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I think this is the most appropriate answer to the question. – Tunococ Jan 25 '13 at 3:37

This equation is separable:

$$\frac{dy}{dt} = 8 y \implies \frac{dy}{y} = -8 \, dt$$

Now integrate both sides:

$$\int \frac{dy}{y} = -8 t + C$$

where $C$ is a constant. The integral on the left is a natural log:

$$\log{y} = -8 t + C \implies y = e^{-8 t+C} = A e^{-8 t} $$

so $r=-8$.

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$$y'/y=(\log(y))'\implies y'/y=-8 = (\log(y))'.$$ Integrate both sides to get $$-8x + C=\log(y)\implies y(x)=e^{-8x+C}=C_0 e^{-8x}.$$ Setting $C_0=1,$ we see that $r=-8.$

You can easily turn this into a separable equation and avoid the trick $y'/y=(\log(y))'.$ Here's your equation: $$\frac{\mathrm dy}{\mathrm dx}=-8y\\\implies\int\frac{\mathrm dy}{y}=-\int 8x\,\mathrm dx$$ and I think you know what to do next: find the solution of the from $C_0 e^{kt}$ and compare it to $e^{rt}.$

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Hint: $y^\prime+8y=0$ $$\frac{y'}{y}=-8$$ Integrate both sides to get $$\ln y=-8t+c$$ $y=\exp(c)\exp(-8t)$

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