Question

Question: Radioactive Decay. In Example 2 assume that the rate at which${\mathbf{RA}}_{\mathbf{1}}$ decays into ${\mathbf{RA}}_{\mathbf{2}}$is $\mathbf{40}{\mathbf{e}}^{\mathbf{-}\mathbf{20}\mathbf{t}}\mathbf{}\mathbf{kg}\mathbf{/}\mathbf{sec}$and the decay constant for ${\mathbf{RA}}_{\mathbf{2}}$is $\mathbf{k}\mathbf{=}\mathbf{5}\mathbf{/}\mathbf{sec}$. Find the mass of ${\mathbf{RA}}_{\mathbf{2}}$for $\mathbf{t}\mathbf{\ge }\mathbf{0}$if initially $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{kg}$.

Short answer

$\mathrm{y}=\left(\frac{38}{3}\right){\mathrm{e}}^{-5\mathrm{t}}-\left(\frac{8}{3}\right){\mathrm{e}}^{-20\mathrm{t}}$

Step-by-step solution

Method for solving linear equations

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor by $\mathbf{\mu }\left(x\right)$the formula .
• Multiply the equation in standard form by $\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

$$\begin{gathered} \mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

• Integrate the last equation and solve for y by dividing by$\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ to obtain$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{[}{\mathbf{\int }}_{}^{}\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{c}\mathbf{]}$ . Here C is an arbitrary constant.

Solve the given equation

Given that, ${\mathrm{RA}}_1$decays into ${\mathrm{RA}}_2$is and the decay constant for $40e^{-20t}\mathrm{kg}/\sec$is $\mathrm{k}=5/\sec$

That is, $\frac{\mathrm{dA}}{\mathrm{dt}}=40{\mathrm{e}}^{-20\mathrm{t}}\mathrm{kg}/\sec ;\mathrm{k}=5/\sec$.

To find the mass of ${\mathrm{RA}}_2$for$t\ge 0$if initially$y\left(0\right)=10\mathrm{kg}$

Let us create an initial value problem,

$\frac{\mathrm{dy}}{\mathrm{dt}}+5\mathrm{y}=40{\mathrm{e}}^{-20\mathrm{t}},\mathrm{y}\left(0\right)=10\mathrm{kg}......\left(1\right)$

Calculate the integrating factor of $\mu \left(t\right)$.

Where role="math" localid="1664172714912" $\mathrm{P}\left(t\right)=5$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{t}\right)& =& {\mathrm{e}}^{\int \mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{\int 5\mathrm{tdt}}\\ & =& {\mathrm{e}}^{5\mathrm{t}}\end{array}$

 Step 3: Simplification method

Multiply in equation (1)

$$\begin{gathered} e^{5t}\frac{\mathrm{dy}}{\mathrm{dt}}+5e^{5t}\mathrm{y}=40e^{5t}{\mathrm{e}}^{-20\mathrm{t}} \\ {\mathrm{e}}^{5\mathrm{t}}\frac{\mathrm{dy}}{\mathrm{dt}}+5{\mathrm{e}}^{5\mathrm{t}}\mathrm{y}=40{\mathrm{e}}^{-15\mathrm{t}} \\ \frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{5\mathrm{t}}\mathrm{y}\right]=40{\mathrm{e}}^{-15\mathrm{t}} \end{gathered}$$

Integrating both sides.

$$\begin{gathered} {\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{5\mathrm{t}}\mathrm{y}\right]dt={\int }_{}^{}40{\mathrm{e}}^{-15\mathrm{t}}\mathrm{dt} \\ {\mathrm{e}}^{5\mathrm{t}}\mathrm{y}=-\frac{40}{15}{\mathrm{e}}^{-15\mathrm{t}}+\mathrm{C} \\ \mathrm{y}=-\frac{8}{3}\frac{{\mathrm{e}}^{-15\mathrm{t}}}{{\mathrm{e}}^{5\mathrm{t}}}+\frac{\mathrm{C}}{{\mathrm{e}}^{5\mathrm{t}}} \\ \mathrm{y}=-\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+{\mathrm{Ce}}^{-5\mathrm{t}} \end{gathered}$$

So, the solution is $\mathrm{y}=-\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+{\mathrm{Ce}}^{-5\mathrm{t}}.......\left(2\right)$

Find the initial value

Given that, $y\left(0\right)=10\mathrm{kg}$.

Then, t = 0 and y = 10.

Substitute the value in equation (2) to get the value of C.

role="math" localid="1664173608266" $\begin{array}{rcl}\mathrm{y}& =& -\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+{\mathrm{Ce}}^{-5\mathrm{t}}\\ 10& =& -\frac{8}{3}{\mathrm{e}}^{-20\left(0\right)}+{\mathrm{Ce}}^{-5\left(0\right)}\\ 10& =& -\frac{8}{3}+\mathrm{C}\\ \mathrm{C}& =& \frac{38}{3}\end{array}$

Substitute the value of C in equation (2).

$\begin{array}{rcl}\mathrm{y}& =& -\frac{8}{3}{\mathrm{e}}^{-20\mathrm{t}}+\frac{38}{3}{e}^{-5\mathrm{t}}\end{array}$

So, the solution is$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \mathbf{-}\frac{\mathbf{8}}{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{20}\mathbf{t}}\mathbf{+}\frac{\mathbf{38}}{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{}\end{array}$