Question

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

$\mathbf{cos}\theta \mathbf{dr}\mathbf{-}\mathbf{(}\mathbf{r}\mathbf{sin}\theta \mathbf{-}{\mathbf{e}}^{\theta }\mathbf{)}\mathbf{d}\theta \mathbf{=}\mathbf{0}$

Short answer

The solution is $r=\left(C-e^{\theta }\right)\sec \theta$

Step-by-step solution

Evaluate whether the equation is exact

Here$\mathbf{cos}\theta \mathbf{dr}\mathbf{-}\mathbf{(}\mathbf{r}\mathbf{sin}\theta \mathbf{-}{\mathbf{e}}^{\theta }\mathbf{)}\mathbf{d}\theta \mathbf{=}\mathbf{0}$

The condition for exact is$\frac{\partial \mathbf{M}}{\partial \theta }=\frac{\partial \mathbf{N}}{\partial \mathbf{r}}$.

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)}\mathbf{=}\mathbf{cos}\theta \\ \mathbf{N}\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{sin}\theta \mathbf{+}{\mathbf{e}}^{\theta } \end{gathered}$$

$\frac{\partial \mathbf{M}}{\partial \theta }\mathbf{=}\mathbf{-}\mathbf{sin}\theta \mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{r}}$

This equation is exact.

Find the value of  F(r,θ)

Here,

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)}\mathbf{=}\mathbf{cos}\theta \\ \mathbf{F}\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)}\mathbf{=}\int \mathbf{M}\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)}\mathbf{dr}\mathbf{+}\mathbf{g}\mathbf{(}\theta \mathbf{)} \\ \mathbf{=}\int \mathbf{cos}\theta \mathbf{dr}\mathbf{+}\mathbf{g}\mathbf{(}\theta \mathbf{)} \\ \mathbf{=}\mathbf{r}\mathbf{cos}\theta \mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \end{gathered}$$

Determine the value of g(θ)

$$\begin{gathered} \frac{\partial \mathbf{F}}{\partial \theta }\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)}\mathbf{=}\mathbf{N}\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)} \\ \mathbf{-}\mathbf{r}\mathbf{sin}\theta \mathbf{+}\mathbf{g}\mathbf{\text{'}}\mathbf{(}\theta \mathbf{)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{sin}\theta \mathbf{+}{\mathbf{e}}^{\theta } \\ \mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\theta } \\ \mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\theta } \end{gathered}$$

Now$\mathbf{F}\mathbf{(}\mathbf{r}\mathbf{,}\theta \mathbf{)}\mathbf{=}\mathbf{r}\mathbf{cos}\theta \mathbf{+}{\mathbf{e}}^{\theta }$

Therefore, the general solution of the differential equation is

$$\begin{gathered} \mathbf{r}\mathbf{cos}\theta \mathbf{+}{\mathbf{e}}^{\theta }\mathbf{=}\mathbf{C} \\ r=\frac{C-e^{\theta }}{\cos \theta } \\ r=\left(C-e^{\theta }\right)\sec \theta \end{gathered}$$

Hence the solution isrole="math" localid="1664173591620" $\mathit{r}\mathbf{=}\left(C-e^{\theta }\right)\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{\theta }$