Question

Use the method discussed under “Equations of the Form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$” to solve problems 17-20.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}{\left(\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}$

Short answer

Equation of the form of$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$ for the given equation is$\mathbf{y}\mathbf{=}\mathbf{tan}\left(\mathbf{x}\mathbf{+}\mathbf{C}\right)\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{2}$.

Step-by-step solution

General form of dydx=Gax+by

Equations of the form$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$

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

Evaluate the given equation

Given, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}{\left(\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}$

Let us take $$\begin{gathered} \mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y} \\ \mathbf{y}\mathbf{=}\mathbf{z}\mathbf{-}\mathbf{x} \end{gathered}$$

Differentiate with respect to x.

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{dx}}& \mathbf{=}& \frac{\mathbf{dz}}{\mathbf{dx}}\mathbf{-}\mathbf{1}\\ {\left(\mathbf{z}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{dz}}{\mathbf{dx}}\mathbf{-}\mathbf{1}\\ {\left(\mathbf{z}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}& \mathbf{=}& \frac{\mathbf{dz}}{\mathbf{dx}}\\ \frac{\mathbf{1}}{{\left(\mathbf{z}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{dz}& \mathbf{=}& \mathbf{dx}\end{array}$

Integrate the equation

Now, integrate on both sides.

$$\begin{gathered} \int \frac{\mathbf{1}}{{\left(\mathbf{z}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{dz}\mathbf{=}\mathbf{dx} \\ \int \frac{\mathbf{1}}{{\left(\mathbf{z}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{dz}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C} \end{gathered}$$

Evaluate $\int \frac{\mathbf{1}}{{\left(\mathbf{z}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{dz}$separately.

Let; $\begin{array}{rcl}\mathrm{v}& =& \mathrm{z}+2\\ \frac{\text{dv}}{\text{dz}}& =& 1\\ \mathrm{dz}& =& \mathrm{dv}\\ & & \end{array}$

$\begin{array}{rcl}\int \frac{\mathbf{1}}{{\mathbf{v}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{dv}& \mathbf{=}& \mathbf{x}\mathbf{+}\mathbf{C}\\ {\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{v}& \mathbf{=}& \mathbf{x}\mathbf{+}\mathbf{C}\\ & & \end{array}$

Substitute $\mathbf{v}\mathbf{=}\mathbf{z}\mathbf{+}\mathbf{2}$and $\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}$

$\begin{array}{rcl}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{z}\mathbf{+}\mathbf{2}\right)& \mathbf{=}& \mathbf{x}\mathbf{+}\mathbf{C}\\ {\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{2}\right)& \mathbf{=}& \mathbf{x}\mathbf{+}\mathbf{C}\\ \mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{2}& \mathbf{=}& \mathbf{tan}\left(\mathbf{x}\mathbf{+}\mathbf{C}\right)\\ \mathbf{y}& \mathbf{=}& \mathbf{tan}\left(\mathbf{x}\mathbf{+}\mathbf{C}\right)\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{2}\end{array}$

Therefore, Equation of the form of$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$ for the given equation isrole="math" localid="1655207486825" $\mathbf{y}\mathbf{=}\mathbf{tan}\left(\mathbf{x}\mathbf{+}\mathbf{C}\right)\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{2}$.