Radioactive Decay

Big Idea

To solve a separable differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = f(x)g(y),$

we separate the variables and integrate on each side:

$\displaystyle \int \frac 1{g(y)}\, \mathrm dy = \int f(x)\, \mathrm dx.$

This is called the method of separable variables. Remember to include a $+C$ on at least one side of the result! You may leave your answer in implicit form if there is no $|y|$ in your expression.

Questions

Question 1. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{x}{\cos(y) + 1}.$

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{x}{\cos(y) + 1} \\\int (\cos(y)+1)\ \mathrm dy &= \int x\ \mathrm dx \\ \sin(y) + y &= \frac {x^2}{2} + C. \end{aligned}$

Question 2. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{1 + \cos(x)}{e^y}.$

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{1 + \cos(x)}{e^y} \\\int e^y \ \mathrm dy &= \int (1 + \cos(x))\ \mathrm dx \\ e^y &= x + \sin(x) + C. \end{aligned}$

Question 3. A radioactive substance decays with a decay rate of $\lambda > 0$ . Let $x \equiv x(t)$ denote the amount of a radioactive substance at time $t$ . The radioactive decay model states that $x$ decreases at a rate proportional to $x$ with proportionality constant $\lambda$ .

Compute the half-life of the substance (i.e. the time taken for the substance to decay to half of its quantity) in terms of $\lambda$ .

(Click for Solution)

Solution. The rate of change of $x$ is $\displaystyle \frac{\mathrm dx}{\mathrm dt}$ . Since $\displaystyle \frac{\mathrm dx}{\mathrm dt}$ is proportional to $x$ and decreasing,

$\displaystyle \frac{\mathrm dx}{\mathrm dt} = -\lambda x.$

By the method of separable variables,

$\begin{aligned} \frac{\mathrm dx}{\mathrm dt} &= -\lambda x \\ \int \frac 1x\, \mathrm dx &= \int -\lambda\ \mathrm dt \\ \ln |x| &= -\lambda t + C \\ |x| &= e^{-\lambda t + C} \\ x &= \pm e^{-\lambda t} \cdot e^C \\ &= Ae^{-\lambda t}, \end{aligned}$

where $A := \pm e^C$ is an arbitrary constant. Now, $x(0) = A > 0$ . Since we want to find $t$ such that $x(t) = A/2$ , we solve the equation

$\displaystyle \begin{aligned} \frac A2 &= Ae^{-\lambda t} \\ \frac 12 &= e^{-\lambda t} \\ \ln \left(\frac 12 \right) &= -\lambda t \\ -\ln 2 &= - \lambda t \\ t &= \frac{\ln 2}{\lambda}, \end{aligned}$

so that the half-life of the substance is $(\ln 2)/{\lambda}$ . See this post for more exponential models.

For Fun. See this video for goated-era music. See this game for a game named after this idea.

Question 4. Solve the differential equation

$\displaystyle \frac{\mathrm d y}{\mathrm dx} + \frac yx = \frac{\sec(xy)}{x}.$

Hint: Use the substitution $u = xy$ .

(Click for Solution)

Solution. By the substitution $u = xy$ ,

$\displaystyle \frac{\mathrm du}{\mathrm dx} = y + x \frac{\mathrm dy}{\mathrm dx}.$

Multiplying the original equation by $x$ ,

$x\displaystyle \frac{\mathrm d y}{\mathrm dx} + y = \sec(xy).$

Therefore, the substitution $u = xy$ yields

$\displaystyle \frac{\mathrm du}{\mathrm dx} = \sec(u) = \frac{1}{\cos(u)}.$

By the method of separable variables,

$\begin{aligned} \frac{\mathrm du}{\mathrm dx} &= \frac{1}{\cos(u)} \\ \int \cos(u)\, \mathrm du &= \int 1\, \mathrm dx \\ \sin(u) &= x + C \\ \sin(xy) &= x + C.\end{aligned}$

—Joel Kindiak, 17 Apr 25, 1453H

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