Question

In Problems \(1 - 6\), determine whether the given equation is separable, linear, neither, or both.

\({\bf{3r = }}\frac{{{\bf{dr}}}}{{{\bf{d\theta }}}}{\bf{ - }}{{\bf{\theta }}^{\bf{3}}}\).

Short answer

The given equation is a linear equation but not separable.

Step-by-step solution

General form of separable and linear equations

Separable equation: A first-order equation is separable if it can be written in the form \(\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = g}}\left( {\bf{x}} \right){\bf{p}}\left( {\bf{y}} \right)\).

Linear equation: A first-order equation is linear if it can be written in the form\(\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ + P}}\left( {\bf{x}} \right){\bf{y = Q}}\left( {\bf{x}} \right)\).

Evaluate the given equation

Given that, \({\bf{3r = }}\frac{{{\bf{dr}}}}{{{\bf{d}}\theta }}{\bf{ - }}{\theta ^{\bf{3}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Evaluative the given equation is separable or linear.

\(\begin{array}{c}{\bf{3r = }}\frac{{{\bf{dr}}}}{{{\bf{d}}\theta }}{\bf{ - }}{\theta ^{\bf{3}}}\\\frac{{{\bf{dr}}}}{{{\bf{d}}\theta }}{\bf{ - }}{\theta ^{\bf{3}}}{\bf{ = 3r}}\\\frac{{{\bf{dr}}}}{{{\bf{d}}\theta }}{\bf{ - 3r = }}{\theta ^{\bf{3}}}\\\frac{{{\bf{dr}}}}{{{\bf{d}}\theta }}{\bf{ = }}{\theta ^{\bf{3}}}{\bf{ + 3r}}\end{array}\)

Identifying method

Where \(\frac{{{\bf{dr}}}}{{{\bf{d}}\theta }}{\bf{ - 3r = }}{\theta ^{\bf{3}}}\) and \(\frac{{{\bf{dr}}}}{{{\bf{d}}\theta }}{\bf{ = }}{\theta ^{\bf{3}}}{\bf{ + 3r}}\)

Let,

\(\begin{array}{l}{\bf{P}}\left( \theta \right){\bf{ = - 3}}\\{\bf{Q}}\left( \theta \right){\bf{ = }}{\theta ^{\bf{3}}}\end{array}\)

So, the given equation is linear and refers to the evaluation of the equation on the right-hand side. One can’t find any product of function that depends only on r and function that depends only on \(\theta \), so this equation is not separable.

Hence, the given equation is a linear equation but not separable.