In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{G}\left(\mathrm{ax}+\mathrm{by}\right)$.

$\left(y^3-{\mathrm{\theta y}}^2\right)\mathbf{d\theta }\mathbf{+}\mathbf{2}{\mathbf{\theta }}^{\mathbf{2}}\mathbf{ydy}\mathbf{=}\mathbf{0}$

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 $\frac{dy}{dx}=f\left(x,y\right)$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 P9x) 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 to transform the original equation into a new equation that we could solve. In some cases, we must transform bothandinto new variables, sayand. This is the situation for equations with linear coefficients-that is, equations of the form

.$\left(a_{1}x+b_{1}y+c_{1_{}}\right)dx+\left(a_{2}x+b_{2}y+c\right)dy=0$

Given, $\left({\mathbf{y}}^{\mathbf{3}}\mathbf{-}\theta {\mathbf{y}}^{\mathbf{2}}\right)\mathbf{d}\theta \mathbf{+}\mathbf{2}{\theta }^{\mathbf{2}}\mathbf{ydy}\mathbf{=}\mathbf{0}$$\left(y^3-\theta y^2\right)d\theta +2{\theta }^{2}ydy=0$

By evaluating

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

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

Then,

role="math" localid="1655118354092" $\begin{array}{rcl}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{-}{\left(\frac{\mathbf{y}}{\theta }\right)}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{y}}{\theta }\right)& \mathbf{=}& \mathbf{G}\left(\frac{\mathbf{y}}{\theta }\right)\\ & \mathbf{=}& \mathbf{G}\left(\mathbf{v}\right)\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{-}{\mathbf{v}}^{\mathbf{2}}\mathbf{+}\mathbf{v}\right)\end{array}$

So, the given equation is homogeneous.

Now, rewriting the equation,

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

It seems that the given equation is Bernoulli.

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