Question

In problem $\mathbf{1}\mathbf{-}\mathbf{6}$, determine whether the differential equation is separablerole="math" localid="1654775979001" $\mathbf{(}\mathit{x}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{2}}\mathbf{)}\mathit{d}\mathit{y}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{}\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$.

Short answer

The differential equation $(x{y}^2+3y^2)dy-2xdx=0$is separable.

Step-by-step solution

Concept of Separable Differential Equation

A first-order ordinary differential equation$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{f}\left(x,y\right)$is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions$\mathit{g}\left(x\right)$ that is a function of $\mathit{x}$alone and$\mathit{h}\left(y\right)$ that is a function of alone.

Mathematically, the equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{f}\left(x,y\right)$is separable when$\mathit{f}\left(x,y\right)\mathbf{=}\mathit{g}\left(x\right)\mathbf{+}\mathit{h}\left(y\right)$ .

Determining whether the equation is Separable or not 

The given equation is

$\left(x{y}^2+3y^2\right)dy-2xdx=0..........\left(1\right)$

Equation (i) can be written as

$$\begin{gathered} y^2\left(x+3\right)dy-2xdx=0 \\ \frac{dy}{dx}=\frac{2x}{y^2\left(x+3\right)} \end{gathered}$$

The function in the right – hand side of equation (1) is

$$\begin{gathered} f\left(x,y\right)=\frac{2x}{y^2\left(x+3\right)} \\ =\frac{2x}{x^2+3}\frac{1}{y^2} \end{gathered}$$

This function can be written as a product of two functions $g\left(x\right)$and $h\left(y\right)$defined as,

$$\begin{gathered} g\left(x\right)=\frac{2x}{x^2+3} \\ =\frac{1}{y^2} \end{gathered}$$

Therefore, the given differential equation is separable.