Question

In problems $\mathbf{1}\mathbf{-}\mathbf{8}$identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form$\mathit{y}\mathbf{=}\mathit{G}\left(ax+b\right)$.$\mathbf{cos}\left(x+y\right)\mathbf{dy}\mathbf{}\mathbf{=}\mathbf{}\mathbf{sin}\left(x+y\right)\mathbf{dx}$

Short answer

The given equation is the form of$\mathrm{y}=\mathrm{G}(\mathrm{ax}+\mathrm{b})$.

Step-by-step solution

General form of homogeneous, Bernoulli, linear coefficients of the form of  y=G(ax+b)

• Homogeneous equation

If the right-hand side of the equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$can be expressed as a function of the ratio $\frac{\mathrm{y}}{\mathrm{x}}$alone, then we say the equation is homogeneous.

Equations of the form$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{G}(\mathrm{ax}+\mathrm{by})$.

When the right-hand side of the equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$can be expressed as a function of the combination $\mathrm{ax}+\mathrm{by}$, where $\mathrm{a}$and $\mathrm{b}$are constants, that is, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{G}(\mathrm{ax}+\mathrm{by})$then the substitution $\mathrm{z}=\mathrm{ax}+\mathrm{by}$transforms the equation into a separable one.

• Bernoulli’s equation

A first-order equation that can be written in the form $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{P}\left(\mathrm{x}\right)\mathrm{y}=\mathrm{Q}\left(\mathrm{x}\right){\mathrm{y}}^{\mathrm{n}}$, where$\mathrm{P}(\mathrm{x})$and $\mathrm{Q}\left(\mathrm{x}\right)$are continuous on an interval $(\mathrm{a},\mathrm{b})$and $\mathrm{n}$is a real number, is called a Bernoulli equation.

• Equation of Linear coefficients

We have used various substitutions for$\mathrm{y}$ to transform the original equation into a new equation that we could solve. In some cases, we must transform both$\mathrm{x}$and$\mathrm{y}$into new variables, say$\mathrm{u}$and$\mathrm{v}$. This is the situation for equations with linear coefficients-that is, equations of the form$\left({\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1\right)\mathrm{dx}+\left({\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\right)\mathrm{dy}=0$

.

Evaluate the given equation.

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

By Evaluating,

role="math" localid="1654852982147" $\begin{array}{rcl}\cos (\mathrm{x}+\mathrm{y})\mathrm{dy}& =& \sin (\mathrm{x}+\mathrm{y})\mathrm{dx}\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{\sin \left(\mathrm{x}+\mathrm{y}\right)}{\cos \left(\mathrm{x}+\mathrm{y}\right)}\\ & =& \tan \left(\mathrm{x}+\mathrm{y}\right)\end{array}$

It seems that the given equation is role="math" localid="1654852988301" $\mathrm{y}\text{'}=\mathrm{G}\left(\mathrm{ax}+\mathrm{by}\right)$.

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