Home » Blog » Find and solve a differential equation for the decay of a material with given properties

Find and solve a differential equation for the decay of a material with given properties

by RoRi

A given substance decays at a rate proportional to the square of the amount of the material present. At the end of one year there is 0.5 grams of the substance remaining.

  1. Create and solve a differential equation that governs the mass of the material present after t years.
  2. Find the decay constant of the material in units \operatorname{gm}^{-1} \operatorname{yr}^{-1}.

Incomplete.

One comment

  1. Evangelos says:

    From our given information, we know that the rate of decay is proportional to the square of the amount present at time t. If we let y(t) denote how much material is there at time t, our differential equation becomes

        \begin{align*} \frac{dy}{dt} &= -ky^{2} \end{align*}

    Where k is a constant

    The above differential equation is separable, so we can turn the equation into the following

        \begin{align*} \frac{-dy}{y^{2}} &= kdt \end{align*}

    Integrating from time 0 to some arbitrary time x, we get

        \begin{align*} \int_{y(0)}^{y(x)}\frac{-dy}{y^{2}} &= \int_{0}^{x}kdt \\ \end{align*}

        \begin{align*} [\frac{1}{y}]_{y(0)}^{y(x)} &= kx\\ \frac{1}{y(x)} - \frac{1}{y(0)}&= kx\\ y(x) &= \frac{1}{kx + \frac{1}{y(0)}} \end{align*}

    Now, we also know from our given information that y(1) = 1/2 g. So, we can find in terms of y(0) as follows:

        \begin{align*} (2 - \frac{1}{y(0)}) g^{-1}yrs^{-1}&= k \\ \end{align*}

    Plugging this value into our equation for y gives us:

        \begin{align*} y &= \frac{1}{(2 - \frac{1}{y(0)})x + \frac{1}{y(0)}} \end{align*}

    And once again using our given value y(1) = 1/2

        \begin{align*} \frac{1}{2} &= \frac{1}{(2 - \frac{1}{y(0)}) + \frac{1}{y(0)}} \\ y(0) &= \frac{2}{(2y(0) - 1) + 1}\\ y(0) &= \frac{1}{y(0)} \\ (y(0))^2 &= 1 \\ y(0) &= \sqrt{1} \\ &= 1 \end{align*}

    Now we can plug this value back into our equations for k and y

        \begin{align*} k &= (2 - \frac{1}{y(0)}) g^{-1}yrs^{-1} \\ &= 1 g^{-1}yrs^{-1} \end{align*}

        \begin{align*} y &= \frac{1}{kx + \frac{1}{y(0)}} \\ &= \frac{1}{x + 1} \end{align*}

    A side note: I had to find the value of k before finding the function y, which is in reverse from the way the book presented the problem. Nevertheless, the answers are as requested.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):