Question

In problem $\mathbf{1}\mathbf{-}\mathbf{6}\mathbf{,}$determine whether the differential equation is separable$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{4}{\mathit{y}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{+}\mathbf{1}$.

Short answer

The differential equation$\frac{d_y}{dx}=4y^2-3y+1$ is separable.

Step-by-step solution

Concept 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)$$\mathit{g}\left(x\right)$that is a function of $\mathit{x}$ alone and $\mathit{h}\left(y\right)$that is a function of $\mathit{y}$alone.

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

Determining whether the equation is Separable or not

The given equation is

$\frac{d}{dx}=4y^2-3y+1........\left(1\right)$

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

$$\begin{gathered} f\left(x,y\right)=4y^2-3y+1 \\ =\left(1\right)\left(4y^2-3y+1\right) \\ =g\left(x\right)h\left(y\right) \end{gathered}$$

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

$\begin{array}{l}g\left(x\right)=1\\ h\left(y\right)=4y^2-3y+1\end{array}$

Therefore, the given differential equation is separable.