Chapter 7.2, Problem 27E

a.

To determine

To draw: A direction field for the differential equation.

Explanation of Solution

Given information:

Differential equation: $R\frac{dQ}{dt}+\frac{1}{C}Q=E\left(t\right)$

  $R=5\Omega$ , $C=0.05F$ , $E=60V$

Concept used :

If $\frac{dy}{dx}=F\left(x,y\right)$ , $F\left(x,y\right)$ gives the slope of the solution curve at point $\left(x,y\right)$ . If small line segments with slope $F\left(x,y\right)$ are drawn on several points $\left(x,y\right)$ , the results are called direction field.

Graph:

  $R\frac{dQ}{dt}+\frac{1}{C}Q=E\left(t\right)$

  $R=5\Omega$ , $C=0.05F$ , $E=60V$

  $\begin{array}{l}5\frac{dQ}{dt}+\frac{1}{0.05}Q=60\\ \Rightarrow 5\frac{dQ}{dt}+20Q=60\\ \Rightarrow \frac{dQ}{dt}+4Q=12\\ \Rightarrow \frac{dQ}{dt}=12-4Q\end{array}$

  

Interpretation: The graph is the direction field of the differential equation $\frac{dQ}{dt}=12-4Q$

b.

To determine

To find: The limiting value of the charge.

Answer to Problem 27E

The limiting value of the charge is $3$ Coulomb.

Explanation of Solution

Given information:

  

It is clear from the direction field that all the solutions approaches to the value $3$ .

i.e. $\underset{t\to \infty }{\lim }Q\left(t\right)=3$

c.

To determine

To find: Whether there are any equilibrium solution or not.

Answer to Problem 27E

  $Q\left(t\right)=3$is an equilibrium solution.

Explanation of Solution

Given information:

  $\frac{dQ}{dt}=12-4Q$

  

Concept used :

If for a particular solution , $\frac{dQ}{dt}$ be zero, then that solution will be equilibrium solution.

Calculation:

From the direction field it appears that $Q=3$ is an equilibrium solution of $\frac{dQ}{dt}=12-4Q$ .

  $\begin{array}{c}\frac{dQ}{dt}=12-4Q\\ =12-4\times 3\\ =12-12\\ =0\end{array}$

d.

To determine

To draw: The graph of the solution when $Q\left(0\right)=0C$ using direction field.

Explanation of Solution

Given information:

  $Q\left(0\right)=0\text{ C}$

  

Graph :

  $Q\left(0\right)=0\text{ C}$

  

Interpretation:

The graph shows the particular solution when $Q\left(0\right)=0\text{ C}$ and hence it passes through origin.

e.

To determine

To estimate: The charge after half a second using Euler’s method.

Answer to Problem 27E

The charge after half a second is $2.7672\text{ C}$

Explanation of Solution

Given information:

  $\frac{dQ}{dt}=12-4Q$ ; $Q\left(0\right)=0\text{ C}$

  $h=0.1$

Formula used :

Euler’s Method: Approximate values for the solution of the initial-value problem $y^{\prime }=F\left(x,y\right)$ , $y\left(x_0\right)=y_0$ with the step size $h$ , at $x_n=x_{n-a}+h$ .

  $y_n=y_{n-1}+hF\left(x_{n-1},y_{n-1}\right)n=1,2,\mathrm{3.....}$

Calculation:

  $h=0.1$

  $\frac{dQ}{dt}=12-4Q$ and $Q\left(0\right)=0\text{ C}$

Here, $Q^{\prime }=F\left(Q,t\right)=12-4Q$

  $Q_0=0$

  $\begin{array}{l}{Q}_1=Q_0+hF\left(Q_0,t_0\right)=0+0.1\times 12=1.2\\ Q_2=Q_1+hF\left(Q_1,t_1\right)=1.2+0.1\times 7.2=1.92\\ Q_3=Q_2+hF\left(Q_2,t_2\right)=1.92+0.1\times 4.32=2.352\\ Q_4=Q_3+hF\left(Q_3,t_3\right)=2.352+0.1\times 2.592=2.612\\ Q_5=Q_4+hF\left(Q_4,t_4\right)=2.612+0.1\times 1.552=2.7672\end{array}$

  $Q\left(0.5\right)=2.7672$

Therefore, the charge after half a second is $2.7672\text{ C}$