Chapter 7.3, Problem 52E

(a)

To determine

To Find: The solution for the differential equation.

Answer to Problem 52E

The required expression as $y={\left(cx\right)}^{\frac{1}{\theta }}$ .

Explanation of Solution

Given:

The given equation is,

  $\frac{dy}{dx}=\frac{1}{\theta }\frac{y}{x}$

Calculation:

The given differential is written as,

  $\theta \frac{dy}{y}=\frac{dx}{x}$

Then, solve as,

  $\begin{array}{l}\theta \ln y=\ln x+\ln c\\ \ln y=\frac{1}{\theta }\left(\ln \left(cx\right)\right)\\ y={\left(cx\right)}^{\frac{1}{\theta }}\end{array}$

Thus, the required expression as $y={\left(cx\right)}^{\frac{1}{\theta }}$ .

(b)

To determine

To Find: The situation when $\theta =1$ and when $\theta \to \infty$ .

Answer to Problem 52E

The value of $y=cx$ for $\theta =1$ and $y=1$ for $\theta \to \infty$ .

Explanation of Solution

Calculation:

Consider the given expression as,

  $y={\left(cx\right)}^{\frac{1}{\theta }}$

Consider for $\theta =1$ ,

  $\begin{array}{c}y={\left(cx\right)}^{\frac{1}{1}}\\ =cx\end{array}$

Consider for $\theta \to \infty$ ,

  $\begin{array}{c}y={\left(cx\right)}^{\frac{1}{\theta \to \infty }}\\ =1\end{array}$

The value of $y=cx$ for $\theta =1$ and $y=1$ for $\theta \to \infty$ .