Question

In Problems, 1-20 determine whether the given differential equation is exact. If it is exact, solve it.

$\mathbf{(}\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{y}\mathbf{)}\mathit{d}\mathit{x}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{y}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The answer is$cosxsiny-In\left|cosx\right|=c$

Step-by-step solution

Given information

The given equation is$(tanx-sinxsiny)dx+cosxcosydy=0$

Compare the given equation with $Mdx+Ndy=0$.

$$\begin{gathered} M(x,y)=tanx-sinxsiny \\ N(x,y)=cosxcosy \end{gathered}$$

Condition for exactness

An equation of the form $Mdx+Ndy=0$, is said to be exact, if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$.

$$\begin{gathered} \frac{\partial M}{\partial y}=-sinxcosy \\ \frac{\partial N}{\partial x}=-sinxcosy \end{gathered}$$

As a result, the equation is exact,$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Next, we have to find the solution of the given equation. Where$f(x,y)=cosxsiny+g\left(x\right)$

Integrate

Now, consider the equation $f(x,y)=cosxsiny+g\left(x\right)$.

Take $\frac{\partial f}{\partial x}$, then$\frac{\partial f}{\partial x}=-sinxsiny+g\text{'}\left(x\right)$

Since then,$M(x,y)=\frac{\partial f}{\partial x}$, We have notice that$g\text{'}\left(x\right)=tanx$

$$\begin{gathered} g\text{'}\left(x\right)=tanx \\ g\left(x\right)=-In\left|cosx\right| \end{gathered}$$

So, the result is$cosxsiny-In\left|cosx\right|=c$