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.
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