91Ó°ÊÓ

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 $\frac{dy}{dx}=f\left(x,y\right)$can be expressed as a function of the combination $ax+by$, 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,$\left(y{e}^{-2x}+y^3\right)dx-e^{-2x}dy=0$

By Evaluating,

role="math" localid="1655112438036" $\begin{array}{rcl}\left(y{e}^{-2x}+y^3\right)dx-e^{-2x}dy& =& 0\\ \frac{dy}{dx}& =& \frac{y{e}^{-2x}+y^3}{e^{-2x}}\\ & =& \frac{y{e}^{-2x}}{e^{-2x}}+\frac{y^3}{e^{-2x}}\\ \frac{dy}{dx}+\left(-\frac{e^{-2x}}{e^{-2x}}\right)y& =& \left(\frac{1}{e^{-2x}}\right)y^3\\ \frac{dy}{dx}+P\left(x\right)y& =& Q\left(x\right)y^n\end{array}$

It seems that the given equation is Bernoulli.

Therefore, the given equation is the form of Bernoulli Equation.