In problems identify $\mathbf{1}\mathbf{-}\mathbf{18}$(do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{G}\left(\mathrm{ax}+\mathrm{by}\right)$.$\left(y^3-\theta y^2\right)\mathit{d}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}{\mathit{\theta }}^{\mathbf{2}}\mathbf{ydy}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$.

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

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

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

A first-order equation that can be written in the form role="math" localid="1654860894714" $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{P}\left(\mathrm{x}\right)\mathrm{y}=\mathrm{Q}\left(\mathrm{x}\right){\mathrm{y}}^{\mathrm{n}}$, where $\mathrm{P}\left(\mathrm{x}\right)$and $\mathrm{Q}\left(\mathrm{x}\right)$are continuous on an interval $(\mathrm{a},\mathrm{b})$and $\mathrm{n}$is a real number, is called a Bernoulli equation.

We have used various substitutions for $\mathrm{y}$to transform the original equation into a new equation that we could solve. In some cases, we must transform both $\mathrm{x}$and $\mathrm{y}$into new variables, say role="math" localid="1654861132519" $\mathrm{u}$and $\mathrm{v}$. This is the situation for equations with linear coefficients-that is, equations of the form

$\left({\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1\right)\mathrm{dx}+\left({\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\right)\mathrm{dy}=0$

.

Given, $({\mathrm{y}}^3-{\mathrm{\theta y}}^2)d\theta +2{\theta }^2\mathrm{ydy}=0$.

By Evaluating,

$\begin{array}{rcl}({\mathrm{y}}^3-\theta {\mathrm{y}}^2)\mathrm{d}\theta +2{\theta }^{\mathit{2}}\mathrm{ydy}& =& 0\\ 2{\theta }^2\mathrm{ydy}& =& -({\mathrm{y}}^3-\theta {\mathrm{y}}^2)\mathrm{d}\theta \\ \frac{\mathrm{dy}}{\mathrm{d}\theta }& =& \frac{-{\mathrm{y}}^3+\theta {\mathrm{y}}^2}{2{\theta }^2\mathrm{y}}\\ & =& \frac{-{\mathrm{y}}^2}{2{\theta }^2}+\frac{\mathrm{y}}{2\theta }\\ & =& -\frac{1}{2}{\left(\frac{\mathrm{y}}{\theta }\right)}^2+\frac{1}{2}\left(\frac{\mathrm{y}}{\theta }\right)\\ & & \end{array}$

Let us consider $\mathrm{v}=\frac{\mathrm{y}}{\theta }$.

Then,

$\begin{array}{rcl}\frac{1}{2}\left(-{\left(\frac{\mathrm{y}}{\theta }\right)}^2+\frac{\mathrm{y}}{\theta }\right)& =& \mathrm{G}\left(\frac{\mathrm{y}}{\mathrm{\theta }}\right)\\ & =& \mathrm{G}\left(\mathrm{v}\right)\\ & =& \frac{1}{2}\left(-{\mathrm{v}}^2+\mathrm{v}\right)\end{array}$

So, the given equation is homogeneous.

Now, rewriting the equation,

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{d}\theta }& =& \frac{-{\mathrm{y}}^2}{2{\theta }^2}+\frac{\mathrm{y}}{2\theta }\\ \frac{\mathrm{dy}}{\mathrm{d}\theta }+\left(-\frac{1}{2\theta }\right)\mathrm{y}& =& \left(-\frac{1}{2{\mathrm{\theta }}^2}\right){\mathrm{y}}^2\\ \frac{\mathrm{dy}}{\mathrm{d\theta }}+\mathrm{P}\left(\mathrm{\theta }\right)\mathrm{y}& =& \mathrm{Q}\left(\mathrm{\theta }\right){\mathrm{y}}^{\mathrm{n}}\end{array}$

It seems that the given equation is Bernoulli.

Therefore, the given equation is the form of homogenous and Bernoulli equation.