In problems identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{G}\mathbf{(}\mathbf{ax}\mathbf{+}\mathbf{by}\mathbf{)}$.${\left(y-4x-1\right)}^{\mathbf{2}}\mathit{d}\mathit{x}\mathbf{-}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

If the right-hand side of the equation $\frac{dy}{dx}=f\left(x,y\right)$can be expressed as a function of the ratio$\frac{y}{x}$ alone, then we say the equation is homogeneous.

Equations of the form$\frac{dy}{dx}=G\left(ax+by\right)$

When the right-hand side of the equation can be expressed as a function of the combination , where a and b are constants, that is $\frac{dy}{dx}=G\left(ax+by\right)$,then the substitution z = ax + by

transforms the equation into a separable one.

A first-order equation that can be written in the form$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)y^n$ , where P(x) and Q(x) are continuous on an interval (a,b) and n is a real number, is called a Bernoulli equation.

We have used various substitutions for y to transform the original equation into a new equation that we could solve. In some cases, we must transform both x and y into new variables, say u and v . This is the situation for equations with linear coefficients-that is, equations of the form

$\left(a_{1}x+b_{1}y+c_1\right)dx+\left(a_{2}x+b_{2}y+c_2\right)dy=0$

Given, ${\left(y-4x-1\right)}^{2}dx-dy=0$.

By Evaluating,

$$\begin{gathered} {\left(y-4x-1\right)}^{2}dx-dy=0 \\ \frac{dy}{dx}={\left(y-4x-1\right)}^2 \end{gathered}$$

Let u=y-4x. Then, differentiate it to find the value of $\frac{du}{dx}$.

$$\begin{gathered} \frac{du}{dx}=\frac{dy}{dx}-4 \\ \frac{dy}{dx}=\frac{du}{dx}+4 \end{gathered}$$

Now,

$$\begin{gathered} \frac{du}{dx}+4={\left(u-1\right)}^2 \\ \frac{du}{dx}=\left(u^2-2u+1\right)-4 \\ =u^2-2u-3 \\ \frac{du}{u^2-2u-3}=dx........\left(1\right) \end{gathered}$$

Integrating Equation (1),

$$\begin{gathered} \int \frac{du}{u^2-2u-3}=\int dx \\ \ln \left[{\left(\frac{3-u}{u+1}\right)}^{\frac{1}{4}}\right]=x+C \\ \frac{3-u}{u+1}=C{e}^{4x} \\ -1+\frac{4}{u+1}=C{e}^{4x} \\ \frac{4}{u+1}=C{e}^{4x}+1 \\ u+1=\frac{4}{C{e}^{4x}+1} \end{gathered}$$

Substitute .

It seems that the given equation is linear coefficient.

Therefore, the given equation is the form of linear coefficient.