Determine whether each of the following differential equations is or is not separable, and, if it is separable, rewrite the equation in the form dy/dx=f(x) g(y).
\qquad \textbf{a}) \quad \frac{dy}{dx}=xy-3x-2y+6
\qquad \textbf{b})\quad \frac{dy}{dx}=\sin \left( x+y \right)
\qquad \textbf{c}) \quad y\frac{dy}{dx}=e^{x-3y^2}

Solution:

Part A

\begin{align*}
\frac{dy}{dx} & = xy-3x-2y+6 \\
\frac{dy}{dx} & = \left( xy-3x \right)-\left( 2y-6 \right)\\
\frac{dy}{dx} & = x\left( y-3 \right)-2\left( y-3 \right)\\
\frac{dy}{dx} & = \left( x-2 \right)\left( y-3 \right)
\end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.

Part B

\begin{align*}
\frac{dy}{dx} & = \sin\left( x+y \right) \\
\frac{dy}{dx} & = \sin\left( x \right)\cos\left( y \right) +\cos\left( x \right)\sin\left( y \right)\\
\end{align*}

Since F(x,y) is not factorable in the form f(x) g(y), the given differential equation is not separable.

Part C

\begin{align*}
y \frac{dy}{dx} & = e^{x-3y^2}\\
y \frac{dy}{dx} & = \frac{e^x}{e^{3y^2}}\\
\frac{dy}{dx} & =\frac{e^x}{y e^{3y^2}} \\
\frac{dy}{dx} & = e^x \left( \frac{1}{ye^{3y^2}} \right) \\
\end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.