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

\({{\bf{x}}^{\bf{2}}}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ + sinx - y = 0}}\).

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, \({{\bf{x}}^{\bf{2}}}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ + sin}}\;{\bf{x - y = 0}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Evaluative the given equation is separable or linear.

\(\begin{array}{c}{{\bf{x}}^{\bf{2}}}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ + sin}}\;{\bf{x - y = 0}}\\{{\bf{x}}^{\bf{2}}}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = - sin}}\;{\bf{x + y}}\\\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}\frac{{{\bf{ - sin}}\;{\bf{x + y}}}}{{{{\bf{x}}^{\bf{2}}}}}\\{\bf{ = - }}\frac{{{\bf{sin}}\;{\bf{x}}}}{{{{\bf{x}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{y}}}{{{{\bf{x}}^{\bf{2}}}}}\\\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ - }}\frac{{\bf{1}}}{{{{\bf{x}}^{\bf{2}}}}}{\bf{y = - }}\frac{{{\bf{sin}}\;{\bf{x}}}}{{{{\bf{x}}^{\bf{2}}}}}\end{array}\)

Where \({\bf{P}}\left( {\bf{x}} \right){\bf{ = - }}\frac{{\bf{1}}}{{{{\bf{x}}^{\bf{2}}}}}\) and \({\bf{Q}}\left( {\bf{x}} \right){\bf{ = - }}\frac{{{\bf{sin}}\;{\bf{x}}}}{{{{\bf{x}}^{\bf{2}}}}}\).

\(\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = - }}\frac{{{\bf{sin}}\;{\bf{x}}}}{{{{\bf{x}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{y}}}{{{{\bf{x}}^{\bf{2}}}}}\)

So, the given equation is linear and the referring to evaluation the equation of right-hand side we can’t found any product of function which depends only on x and function which depends only on y, so this equation is not separable.

Hence the given equation is linear.