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Q2.6-1E

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Found in: Page 76

### Fundamentals Of Differential Equations And Boundary Value Problems

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 ${{\mathbf{y}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbf{G}}\left(\mathrm{ax}+\mathrm{by}\right)$ . ${\mathbf{2}}{\mathbf{t}}{\mathbf{×}}{\mathbf{dx}}{\mathbf{+}}\left({t}^{2}-{x}^{2}\right){\mathbf{dt}}{\mathbf{=}}{\mathbf{0}}$

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

See the 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 $\left(ax+by\right)$ 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 $\left({a}_{1}x+{b}_{1}y+{c}_{1}\right)dx+\left({a}_{2}x+{b}_{2}y+{c}_{2}\right)dy=0$

## Evaluate the given equation

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

By Evaluation,

role="math" localid="1654934940793" .........(1)

## Substitution method

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

From Equation (1),

role="math" localid="1654934950153"

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.