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. $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{sin}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)$

Short answer

$\mathbf{tan}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)\mathbf{+}\mathbf{sec}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}$

Step-by-step solution

General form of dydx=G(ax+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 localid="1664178140389" $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{G}\left(\mathbf{ax}\mathbf{+}\mathbf{by}\right)$, then the substitution$\mathbf{z}\mathbf{=}\mathbf{ax}\mathbf{+}\mathbf{by}$transforms the equation into a separable

Evaluate the given equation

Given, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{sin}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)$.

Let us take$\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\to \mathbf{y}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{z}$

Differentiate with respect to x.

$\begin{array}{c}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{dz}}{\mathbf{dx}}\\ \mathbf{sin}\mathbf{z}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{dz}}{\mathbf{dx}}\\ \mathbf{1}\mathbf{-}\mathbf{sin}\mathbf{z}\mathbf{=}\frac{\mathbf{dz}}{\mathbf{dx}}\\ \frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{sin}\mathbf{z}}\mathbf{dz}\mathbf{=}\mathbf{dx}\end{array}$

Integrate the equation

Now, integrate on both sides.

$\begin{array}{l}\int \frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{sin}\mathbf{z}}\mathbf{dz}\mathbf{=}\int \mathbf{dx}\\ \int \frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{sin}\mathbf{z}}\mathbf{\times }\frac{\mathbf{1}\mathbf{+}\mathbf{sin}\mathbf{z}}{\mathbf{1}\mathbf{+}\mathbf{sin}\mathbf{z}}\mathbf{dz}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}\\ \int \frac{\mathbf{1}\mathbf{+}\mathbf{sin}\mathbf{z}}{\mathbf{1}\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathbf{z}}\mathbf{dz}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}\\ \int \frac{\mathbf{1}\mathbf{+}\mathbf{sin}\mathbf{z}}{{\mathbf{cos}}^{\mathbf{2}}\mathbf{z}}\mathbf{dz}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}\end{array}$

$\begin{array}{c}\int \frac{\mathbf{1}}{{\mathbf{cos}}^{\mathbf{2}}\mathbf{z}}\mathbf{dz}\mathbf{+}\int \frac{\mathbf{sin}\mathbf{z}}{\mathbf{cos}\mathbf{z}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{cos}\mathbf{z}}\mathbf{dz}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}\\ \int {\mathbf{sec}}^{\mathbf{2}}\mathbf{zdz}\mathbf{+}\int \mathbf{tan}\mathbf{z}\mathbf{sec}\mathbf{z}\mathbf{dz}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}\\ \mathbf{tan}\mathbf{z}\mathbf{+}\mathbf{sec}\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}\end{array}$

Substitute z=x-y$\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}$.

$\mathbf{tan}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)\mathbf{+}\mathbf{sec}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}$

Hence, the solution is$\mathbf{tan}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)\mathbf{+}\mathbf{sec}\left(\mathbf{x}\mathbf{-}\mathbf{y}\right)\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}$