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Expert-verified 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 ${{\mathbf{y}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbf{G}}\left(\mathrm{ax}+\mathrm{by}\right)$.$\left({\mathrm{ye}}^{-2x}+{y}^{3}\right){\mathbf{dx}}{\mathbf{-}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{2}\mathbf{x}}{\mathbf{dy}}{\mathbf{=}}{\mathbf{0}}$

The given equation is the form of Bernoulli Equation.

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 $ax+by$, 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, $\left(y{e}^{-2x}+{y}^{3}\right)dx-{e}^{-2x}dy=0$

By Evaluating,

role="math" localid="1655112438036" $\begin{array}{rcl}\left(y{e}^{-2x}+{y}^{3}\right)dx-{e}^{-2x}dy& =& 0\\ \frac{dy}{dx}& =& \frac{y{e}^{-2x}+{y}^{3}}{{e}^{-2x}}\\ & =& \frac{y{e}^{-2x}}{{e}^{-2x}}+\frac{{y}^{3}}{{e}^{-2x}}\\ \frac{dy}{dx}+\left(-\frac{{e}^{-2x}}{{e}^{-2x}}\right)y& =& \left(\frac{1}{{e}^{-2x}}\right){y}^{3}\\ \frac{dy}{dx}+P\left(x\right)y& =& Q\left(x\right){y}^{n}\end{array}$

It seems that the given equation is Bernoulli.

Therefore, the given equation is the form of Bernoulli Equation. ### Want to see more solutions like these? 