Question

Question: In Example 2 the decay constant for isotope ${\mathbf{RA}}_{\mathbf{1}}$was $\mathbf{k}\mathbf{=}\mathbf{10}\mathbf{/}\mathbf{sec}$, which expresses itself in the exponent of the rate term $\mathbf{50}{\mathbf{e}}^{\mathbf{-}\mathbf{10}\mathbf{t}}\mathbf{kg}\mathbf{/}\mathbf{sec}$. When the decay constant for ${\mathbf{RA}}_{\mathbf{2}}$is $\mathbf{k}\mathbf{=}\mathbf{5}\mathbf{/}\mathbf{sec}$, we see that in formula (14) for y the term$\left(\frac{185}{4}\right){\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}$ eventually dominates (has greater magnitude for t large).

1. Redo Example 2 taking $\mathbf{k}\mathbf{=}\mathbf{20}\mathbf{/}\mathbf{sec}$. Now which term in the solution eventually dominates?
2. Redo Example 2 taking $\mathbf{k}\mathbf{=}\mathbf{10}\mathbf{/}\mathbf{sec}$.

Short answer

(a) The initial value of the equation is $\mathrm{y}=5{{\mathrm{e}}^{-10\mathrm{t}}}_{+}35{\mathrm{e}}^{-20\mathrm{t}}$and the term$5{\mathrm{e}}^{-10\mathrm{t}}$ is eventually dominates.

(b) The initial value of the equation is $y=10{\mathrm{e}}^{-10\mathrm{t}}\left(5\mathrm{t}+4\right)$.

Step-by-step solution

Definitions or concepts that to be used 

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.

 Step 2(a): Given data and find the equation if k=20/sec

Given that, RAwas$10/\sec$andthe exponent of the rate term $50{\mathrm{e}}^{-10\mathrm{t}}\mathrm{kg}/\sec$.

That is,

$\frac{\mathrm{dy}}{\mathrm{dt}}=50{\mathrm{e}}^{-10\mathrm{t}}-\mathrm{ky}........\left(1\right)$

To find the solution if $\mathrm{k}=20/\sec$.

Substitute the value of k in equation (1).

Then,

$\frac{\mathrm{dy}}{\mathrm{dt}}=50{\mathrm{e}}^{-10\mathrm{t}}-20y$

Let us create an initial value problem,

$\frac{\mathrm{dy}}{\mathrm{dt}}+20y=50{\mathrm{e}}^{-10\mathrm{t}},\mathrm{y}\left(0\right)=40...........\left(2\right)$

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

Where $P\left(t\right)=20$.

Then,

role="math" localid="1664177218727" $\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{t}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{t}\right)\mathrm{dt}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}20\mathrm{dt}}\\ & =& {\mathrm{e}}^{20\mathrm{t}}\end{array}$

 Step 3: Using Simplification method

Multiply $\mu \left(t\right)$in equation (1)

role="math" localid="1664176976032" $$\begin{gathered} e^{20t}\frac{\mathrm{dy}}{\mathrm{dt}}+20e^{20t}y=50e^{20t}{\mathrm{e}}^{-10\mathrm{t}} \\ \frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{20\mathrm{t}}\mathrm{y}\right]=50{\mathrm{e}}^{10\mathrm{t}} \end{gathered}$$

Integrating both sides.

role="math" localid="1664176264971" $$\begin{gathered} {\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{20\mathrm{t}}\mathrm{y}\right]dt={\int }_{}^{}50{\mathrm{e}}^{10\mathrm{t}}\mathrm{dt} \\ {\mathrm{e}}^{20\mathrm{t}}\mathrm{y}=\frac{50}{10}{\mathrm{e}}^{10\mathrm{t}}+\mathrm{C} \\ \mathrm{y}=5\frac{{\mathrm{e}}^{10\mathrm{t}}}{{\mathrm{e}}^{20\mathrm{t}}}+\frac{\mathrm{C}}{{\mathrm{e}}^{20\mathrm{t}}} \\ \mathrm{y}=5{\mathrm{e}}^{-10\mathrm{t}}+{\mathrm{Ce}}^{-20\mathrm{t}} \end{gathered}$$

So, the solution is $\mathrm{y}=5{\mathrm{e}}^{-10\mathrm{t}}+{\mathrm{Ce}}^{-20\mathrm{t}}......\left(3\right)$

 Step 4: Find the initial value

Given that, $y\left(0\right)=40$.

Then, $\mathrm{t}=0$and $y=40$.

Substitute the value in equation (3) to get the value of C.

$$\begin{gathered} \mathrm{y}=5{\mathrm{e}}^{-10\mathrm{t}}+{\mathrm{Ce}}^{-20\mathrm{t}} \\ 4=5{\mathrm{e}}^{-10\left(0\right)}+{\mathrm{Ce}}^{-20\left(0\right)} \\ 40=5+\mathrm{C} \\ \mathrm{C}=35 \end{gathered}$$

Substitute the value of C in equation (2).

$y=5{\mathrm{e}}^{-10\mathrm{t}}+35e^{-20\mathrm{t}}$

So, the term $\mathbf{5}{\mathbf{e}}^{\mathbf{-}\mathbf{10}\mathbf{t}}\mathbf{}$eventually dominates.

 Step 5(b): Find solution of the equation if k=10/sec

To find the solution if $\mathrm{k}=10/\sec$.

Substitute the value of k in equation (1).

Then,

$\frac{\mathrm{dy}}{\mathrm{dt}}=50{\mathrm{e}}^{-10\mathrm{t}}-10y$

Let us create an initial value problem,

$\frac{\mathrm{dy}}{\mathrm{dt}}+10y=50{\mathrm{e}}^{-10\mathrm{t}}y\left(0\right)=40.......\left(4\right)$

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

Where $\mathrm{P}(t)=10$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{t}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{t}\right)\mathrm{dt}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}10\mathrm{dt}}\\ & =& {\mathrm{e}}^{10\mathrm{t}}\end{array}$

Integrating and evaluating part

Multiply $\mathrm{\mu }\left(\mathrm{t}\right)$in equation (1)

$$\begin{gathered} e^{10t}\frac{\mathrm{dy}}{\mathrm{dt}}+20e^{10t}y=50e^{10t}{\mathrm{e}}^{-10\mathrm{t}} \\ \frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{10\mathrm{t}}\mathrm{y}\right]=50 \end{gathered}$$

Integrating both sides.

$$\begin{gathered} {\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{10\mathrm{t}}\mathrm{y}\right]dt={\int }_{}^{}50\mathrm{dt} \\ {\mathrm{e}}^{10\mathrm{t}}\mathrm{y}=50t+\mathrm{C} \\ \mathrm{y}=\frac{50t}{{\mathrm{e}}^{10\mathrm{t}}}+\frac{\mathrm{C}}{{\mathrm{e}}^{10\mathrm{t}}} \\ \mathrm{y}=5t{\mathrm{e}}^{-10\mathrm{t}}+{\mathrm{Ce}}^{-10\mathrm{t}} \end{gathered}$$

So, the solution is $\mathbf{y}\mathbf{=}\mathbf{5}{\mathbf{te}}^{\mathbf{-}\mathbf{10}\mathbf{t}}\mathbf{}\mathbf{+}{\mathbf{Ce}}^{\mathbf{-}\mathbf{10}\mathbf{t}}$

Find the initial value

Given that, $y\left(0\right)=40$.

Then, $\mathrm{t}=0$and $y\left(0\right)=40$.

Substitute the value in equation (3) to get the value of C.

$$\begin{gathered} \mathrm{y}=50{\mathrm{te}}^{-10\mathrm{t}}+{\mathrm{Ce}}^{-10\mathrm{t}} \\ 40=50\left(0\right){\mathrm{te}}^{-10\left(0\right)}+{\mathrm{Ce}}^{-10\left(0\right)} \\ 40=\mathrm{C} \end{gathered}$$

Substitute the value of C in equation (2).

$$\begin{gathered} \mathrm{y}=50{\mathrm{te}}^{-10\mathrm{t}}+40{\mathrm{e}}^{-10\mathrm{t}} \\ \mathrm{y}=10{\mathrm{e}}^{-10\mathrm{t}}\left(5\mathrm{t}+4\right) \end{gathered}$$

Therefore, the solution is $\mathbf{y}\mathbf{=}\mathbf{10}{\mathbf{e}}^{\mathbf{-}\mathbf{10}\mathbf{t}}\left(5t+4\right)$.