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

\(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ + xt = }}{{\bf{e}}^{\bf{x}}}\).

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, \(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ + xt = }}{{\bf{e}}^{\bf{x}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Evaluative the given equation is separable or linear.

\(\begin{array}{c}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ + xt = }}{{\bf{e}}^{\bf{x}}}\\\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = }}{{\bf{e}}^{\bf{x}}}{\bf{ - xt}}\end{array}\)

\(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = }}{{\bf{e}}^{\bf{x}}}{\bf{ - xt}}\)

So, the given equation is not linear because \({{\bf{e}}^{\bf{x}}}\) is a function that depends on the dependent variable x, and referring to the evaluation of the equation of the right-hand side one can’t find any product of function which depends only on x and function which depends only on t, so this equation is not separable.

Hence the given equation is neither separable nor linear.