The exponential decay law
dy
dt
= −λy, λ > 0
may be used to model decay of a radioactive isotope, or any other quantity that is
decreasing at a rate proportional to its own magnitude. This is just the exponential
growth model with a negative growth rate k = −λ. As is often done, I have put the
sign into the differential equation so that the parameter λ is positive. In this question
take the unit of time to be years. λ is called the decay rate.
(a) By substitution into the differential equation verify that (as expected) y(t) = ce−λt
is a solution of the exponential decay law for any constant c.

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- anonymous

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- anonymous

The problem says to show by substitution, so just plug \[y(t)=Ce^{-\lambda t}\] and its derivative into the differential equation \[\frac{dy}{dt}=-\lambda y\] and show that it is true.

- anonymous

the part of the question i needed help with got cut off .. idk if you'll still be able to help
(b) At what time T (expressed in terms of λ) will y(t) be half of its initial value y(0) = c?
T is called the half life of the substance.
(c) For the solution obtained in (a), show that y(t0 + T) = 1
2y(t0), no matter what t0
is. In other words y(t) decreases by half its magnitude over any time interval of
length T.
(d) If T = 6000 years, what is λ (expressed as a decimal)?

- anonymous

For (b), since you know y(0)=C and y(t)=Ce^(-lambda * t), then y(t) will be half that value when y(t)=C/2. Thus, just solve \[\frac{C}{2}=Ce^{-\lambda T}\] for T.
For (c), you are just showing that \[y(t_0+T)=\frac{1}{2}y(t_0)\] which you can solve by just using the value for T calculated in part (b).
For (d), since you know that \[\frac{1}{2}=e^{-\lambda \cdot 6000}\] (as T=6000), just solve for lambda.

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## More answers

- anonymous

in b when solving for T, you take ln of both sides right? but on the right hand side of the equation do you include C in the ln calculation of do you pull it out infront?

- anonymous

Just start off by dividing both sides by C, then there is no need to worry about it

- anonymous

oh alright, thanks!

- anonymous

so the answer would be -ln(1/2) over lambda?

- anonymous

if your still there can i ask you another question?

- anonymous

Mhm, that's the answer. Feel free to ask away.

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