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Q2.6 - 7E

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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 ${{\mathbf{y}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbf{G}}\left(\mathrm{ax}+\mathrm{by}\right)$.${\mathbf{cos}}\left(x+y\right){\mathbf{dy}}{\mathbf{=}}{\mathbf{sin}}\left(x+y\right){\mathbf{dx}}$

The given equation is the form of ${y}^{\text{'}}=G\left(ax+by\right)$

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, $\mathrm{cos}\left(x+y\right)dy=\mathrm{sin}\left(x+y\right)dx$.

By evaluating

$\begin{array}{rcl}\mathrm{cos}\left(x+y\right)dy& =& \mathrm{sin}\left(x+y\right)dx\\ \frac{dy}{dx}& =& \frac{\mathrm{sin}\left(x+y\right)}{\mathrm{cos}\left(x+y\right)}\\ & =& \mathrm{tan}\left(x+y\right)\end{array}$

It seems that the given equation is ${y}^{\text{'}}=G\left(ax+by\right)$

Therefore, the given equation is the form of ${{y}}^{{\text{'}}}{=}{G}\left(ax+by\right)$