Tag Archives: Separation of Variables

Determining if a given Differential Equation is Separable or Not

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).
a)dydx=xy−3x−2y+6\qquad \textbf{a}) \quad \frac{dy}{dx}=xy-3x-2y+6
b)dydx=sin⁡(x+y)\qquad \textbf{b})\quad \frac{dy}{dx}=\sin \left( x+y \right)
c)ydydx=ex−3y2\qquad \textbf{c}) \quad y\frac{dy}{dx}=e^{x-3y^2}

Solution:

Part A

dydx=xy−3x−2y+6dydx=(xy−3x)−(2y−6)dydx=x(y−3)−2(y−3)dydx=(x−2)(y−3)\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

dydx=sin⁡(x+y)dydx=sin⁡(x)cos⁡(y)+cos⁡(x)sin⁡(y)\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

ydydx=ex−3y2ydydx=exe3y2dydx=exye3y2dydx=ex(1ye3y2)\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.