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Q2.6-1E

Expert-verifiedFound in: Page 76

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form ${{\mathbf{y}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbf{G}}{\left(\mathrm{ax}+\mathrm{by}\right)}$ . ${\mathbf{2}}{\mathbf{t}}{\mathbf{\times}}{\mathbf{dx}}{\mathbf{+}}{\left({t}^{2}-{x}^{2}\right)}{\mathbf{dt}}{\mathbf{=}}{\mathbf{0}}$**

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

**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 $\left(ax+by\right)$ where and 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 $P\left(x\right)$ and $Q\left(x\right)$ are continuous on interval $\left(a,b\right)$ and localid="1654934017689" $n$ 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 $\left({a}_{1}x+{b}_{1}y+{c}_{1}\right)dx+\left({a}_{2}x+{b}_{2}y+{c}_{2}\right)dy=0$

Given, $2txdx+\left({t}^{2}-{x}^{2}\right)dt=0$

By Evaluation,

role="math" localid="1654934940793" $\begin{array}{rcl}2txdx+\left({t}^{2}-{x}^{2}\right)dt& =& 0\\ 2txdx& =& -\left({t}^{2}-{x}^{2}\right)dt\\ \frac{dx}{dt}& =& \frac{{t}^{2}-{x}^{2}}{2tx}\\ & =& -\frac{1}{2}\left(\frac{t}{x}-\frac{x}{t}\right)\\ & & \\ & =& \frac{1}{2}\left(\frac{x}{t}-\frac{1}{\frac{x}{t}}\right)\\ & & \\ & & \end{array}$.........(1)

Let us consider $v=\frac{x}{t}$

From Equation (1),

role="math" localid="1654934950153" $\begin{array}{rcl}\frac{{x}^{2}-{t}^{2}}{2tx}& =& G\left(\frac{x}{t}\right)\\ & =& \frac{1}{2}\left(\frac{x}{t}-\frac{1}{\frac{x}{t}}\right)\\ G\left(v\right)& =& \frac{1}{2}\left(v-\frac{1}{v}\right)\end{array}$

So, the given equation is homogeneous.

It can be rewritten as,

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{{x}^{2}-{t}^{2}}{2tx}\\ & =& \frac{x}{2t}-\frac{t}{2x}\\ \frac{dx}{dt}+\left(-\frac{1}{2t}\right)x& =& \left(-\frac{t}{2}\right){x}^{-1}\end{array}$

Comparing the found equation with Bernoulli general form of equation. It seems that the given equation is Bernoulli equation as well.

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

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