Question

Question: In Problems 7-16, obtain the general solution to the equation.

$\frac{\mathbf{dr}}{\mathbf{d\theta }}\mathbf{+}\mathbf{tan\theta }\mathbf{=}\mathbf{sec\theta }$

Short answer

The general solution to the given equation is $\mathrm{r}=\mathrm{sin\theta }+\mathrm{Ccos\theta }$.

Step-by-step solution

Method for solving linear equations

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor by $\mathit{\mu }\left(x\right)$the formula .
• Multiply the equation in standard form by $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

$$\begin{gathered} \mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

• Integrate the last equation and solve for y by dividing by $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$to obtain$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{[}{\mathbf{\int }}_{}^{}\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{c}\mathbf{]}$ . Here C is an arbitrary constant.

 Step 2: Solve the given equation

Given that,

$\frac{\mathrm{dr}}{\mathrm{d\theta }}+\mathrm{tan\theta }=\mathrm{sec\theta }$

Calculate the integrating factor of $\mu \left(x\right)$.

Where $\mathrm{P}(\theta )=tan\theta$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\theta \right)d\theta }\\ & =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{tan\theta d\theta }}\\ & =& {\mathrm{e}}^{\ln \left|\mathrm{sec\theta }\right|}\\ & =& \mathrm{sec\theta }\end{array}$

Simplification method

Multiply $\mu \left(x\right)$in equation (2)

$\begin{array}{rcl}\mathrm{sec\theta }\frac{\mathrm{dr}}{\mathrm{d\theta }}+\mathrm{rsec\theta tan\theta }& =& {\sec }^2\mathrm{\theta }\\ \frac{\mathrm{d}}{\mathrm{d\theta }}\left[\mathrm{rsec\theta }\right]& =& {\sec }^2\mathrm{\theta }\end{array}$

Integrating both sides.

role="math" localid="1664197341858" $\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{d\theta }}\left[\mathrm{rsec\theta }\right]& =& {\int }_{}^{}{\sec }^2\mathrm{\theta }\\ \mathrm{rsec\theta }& =& \mathrm{tan\theta }+\mathrm{C}\\ \mathrm{r}& =& \frac{\mathrm{tan\theta }}{\mathrm{sec\theta }}+\frac{\mathrm{C}}{\mathrm{sec\theta }}\\ \mathrm{r}& =& \mathrm{sin\theta }+\mathrm{Ccos\theta }\end{array}$

So, the solution is$\begin{array}{rcl}\mathbf{r}& \mathbf{=}& \mathbf{sin\theta }\mathbf{+}\mathbf{Ccos\theta }\end{array}$.