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

\(\left( {{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}} \right)\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = yt - y}}\).

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\(\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ + P}}\left( {\bf{x}} \right){\bf{y = Q}}\left( {\bf{x}} \right)\).

Given that, \(\left( {{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}} \right)\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = yt - y}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Evaluative the given equation is separable or linear.

Case (1):

\(\begin{array}{c}\left( {{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}} \right)\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = yt - y}}\\\left( {{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}} \right)\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = y}}\left( {{\bf{t - 1}}} \right)\\\left( {{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}} \right)\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ - y}}\left( {{\bf{t - 1}}} \right){\bf{ = 0}}\\\left( {{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}} \right)\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ + y}}\left( {{\bf{ - t + 1}}} \right){\bf{ = 0}}\end{array}\)

Case (2):

\(\begin{array}{c}\left( {{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}} \right)\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = y}}\left( {{\bf{t - 1}}} \right)\\\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = y}}\frac{{{\bf{t - 1}}}}{{{{\bf{t}}^{\bf{2}}}{\bf{ + 1}}}}\end{array}\)

Where referring to case (1) and (2). We found that the given equation is linear and separable.

Hence, the given equation is both linear and separable.