Question

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

$\mathbf{\theta dy}\mathbf{-}\mathbf{yd\theta }\mathbf{=}\sqrt{\mathbf{\theta yd\theta }}$

Short answer

The given equation is the form of homogeneous and Bernoulli equation.

Step-by-step solution

General form of homogeneous, Bernoulli, linear coefficients of the form of y'=Gax+by

• Homogeneous equation

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.

• Bernoulli’s equation

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.

• Equation of Linear coefficients

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

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

Evaluate the given equation

Given,

By Evaluating,

$\begin{array}{rcl}\theta dy-yd\theta & =& \sqrt{\theta yd\theta }\\ \theta dy& =& \sqrt{\theta yd\theta }+yd\theta \\ & =& \left(\sqrt{\theta y}+y\right)d\theta \\ \frac{dy}{d\theta }& =& \frac{\sqrt{\theta y}+y}{\theta }\\ & =& \frac{\sqrt{y}}{\sqrt{\theta }}+\frac{y}{\theta }\\ & =& \sqrt{\frac{y}{\theta }+}\frac{y}{\theta }\end{array}$

Substitution method

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

$\begin{array}{rcl}\sqrt{\frac{y}{\theta }}+\frac{y}{\theta }& =& G\left(\frac{Y}{\theta }\right)\\ & =& G\left(v\right)\\ & =& \sqrt{v}+v\end{array}$

So, the given equation is homogeneous.

Now rewritten the same equation,

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

It seems that the given equation is Bernoulli.

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