Question

Use the method discussed under “Bernoulli Equations” to solve problems 21-28

$\frac{\mathbf{dr}}{\mathbf{d\theta }}\mathbf{=}\frac{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{r\theta }}{{\mathbf{\theta }}^{\mathbf{2}}}$

Short answer

Equation of the form of Bernoulli equation for the given equation is $\mathbf{r}\mathbf{=}\frac{{\theta }^{\mathbf{2}}}{\mathbf{C}\mathbf{-}\theta }$.

Step-by-step solution

General form of Bernoulli equation

Bernoulli’s equation

A first-order equation that can be written in the form$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\left(\mathbf{x}\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(\mathbf{x}\right){\mathbf{y}}^{\mathbf{n}}$, where P(x) and Q(x) are continuous on an interval and is a real number, is called a Bernoulli equation.

Evaluate the given equation

Given, $\frac{\mathbf{dr}}{\mathbf{d}\theta }\mathbf{=}\frac{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{r}\theta }{{\theta }^{\mathbf{2}}}$.

Evaluate it.

$\begin{array}{c}\frac{\mathbf{dr}}{\mathbf{d}\theta }\mathbf{=}\frac{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{r}\theta }{{\theta }^{\mathbf{2}}}\\ \mathbf{=}\frac{{\mathbf{r}}^{\mathbf{2}}}{{\theta }^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{2}\mathbf{r}}{\theta }\\ \frac{\mathbf{dr}}{\mathbf{d}\theta }\mathbf{-}\frac{\mathbf{2}}{\theta }\mathbf{r}\mathbf{=}{\mathbf{r}}^{\mathbf{2}}{\theta }^{\mathbf{-}\mathbf{2}}\end{array}$

Compare with the general form of the Bernoulli equation.

n = 2, $\mathbf{P}\left(\theta \right)\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\theta }$, and $\mathbf{Q}\left(\theta \right)\mathbf{=}{\theta }^2$.

Now, divide by , we get,

${\mathbf{r}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{dr}}{\mathbf{d}\theta }\mathbf{-}\frac{\mathbf{2}}{\theta }{\mathbf{r}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\theta }^{\mathbf{-}\mathbf{2}}······\left(1\right)$

Substitute $\mathbf{v}\mathbf{=}{\mathbf{r}}^{\mathbf{-}\mathbf{1}}$.

Differentiate with respect to $\theta$.

$\begin{array}{c}\frac{\mathbf{dv}}{\mathbf{d}\theta }\mathbf{=}\mathbf{-}{\mathbf{r}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{dr}}{\mathbf{d}\theta }\\ \mathbf{-}\frac{\mathbf{dv}}{\mathbf{d}\theta }\mathbf{=}{\mathbf{r}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{dr}}{\mathbf{d}\theta }\end{array}$

Substitute it on equation (1)

$\begin{array}{l}-\frac{\mathbf{dv}}{\mathbf{d}\theta }-\frac{2}{\theta }\mathbf{v}\mathbf{=}{\theta }^{-2}\\ \frac{\mathbf{dv}}{\mathbf{d}\theta }\mathbf{+}\frac{2}{\theta }\mathbf{v}\mathbf{=}-{\theta }^{-2}······\left(2\right)\end{array}$

Integrate the equation

Now, integrate the first. Where $\mathbf{P}\left(\theta \right)\mathbf{=}\frac{2}{\theta }$.

$\begin{array}{c}\int \mathbf{P}\left(\theta \right)\mathbf{d}\theta \mathbf{=}\mathbf{2}\int \frac{\mathbf{1}}{\theta }\mathbf{d}\theta \\ \mathbf{=}\mathbf{2}\mathbf{ln}\left|\theta \right|\end{array}$

Then,

$\begin{array}{c}\mu \left(\theta \right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\theta \right)\mathbf{d}\theta }\\ \mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{ln}\left|\theta \right|}\\ \mathbf{=}{\theta }^{\mathbf{2}}\end{array}$

Multiply $\mu \left(\theta \right)$with equation (2).

$\begin{array}{c}{\theta }^{\mathbf{2}}\frac{\mathbf{dv}}{\mathbf{d}\theta }\mathbf{+}\frac{\mathbf{2}}{\theta }{\theta }^{\mathbf{2}}\mathbf{v}\mathbf{=}\mathbf{-}\mathbf{1}\\ {\theta }^{\mathbf{2}}\frac{\mathbf{dv}}{\mathbf{d}\theta }\mathbf{+}\mathbf{2}\theta \mathbf{v}\mathbf{=}\mathbf{-}\mathbf{1}\\ \frac{\mathbf{d}}{\mathbf{d}\theta }\left[{\theta }^{\mathbf{2}}\mathbf{v}\right]\mathbf{=}\mathbf{-}\mathbf{1}\end{array}$

Integrate both sides,

$\begin{array}{c}\int \frac{\mathbf{d}}{\mathbf{d}\theta }\left[{\theta }^{\mathbf{2}}\mathbf{v}\right]\mathbf{d}\theta \mathbf{=}\int \mathbf{-}\mathbf{1}\mathbf{d}\theta \\ {\theta }^{\mathbf{2}}\mathbf{v}\mathbf{=}\mathbf{-}\theta \mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{v}\mathbf{=}\frac{\mathbf{C}\mathbf{-}\theta }{{\theta }^{\mathbf{2}}}\end{array}$

Substitution method

Substitute $\mathbf{v}\mathbf{=}{\mathbf{r}}^{\mathbf{-}\mathbf{1}}$.

$\begin{array}{c}{\mathbf{r}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{C}\mathbf{-}\theta }{{\theta }^{\mathbf{2}}}\\ \mathbf{r}\mathbf{=}\frac{{\theta }^{\mathbf{2}}}{\mathbf{C}\mathbf{-}\theta }\end{array}$

Hence the solution is$\mathbf{r}\mathbf{=}\frac{{\theta }^{\mathbf{2}}}{\mathbf{C}\mathbf{-}\theta }$