Question

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.

$\frac{\mathbf{dy}}{\mathbf{d\theta }}\mathbf{=}\frac{\mathbf{\theta }\mathbf{}\mathbf{sec}\left(\frac{y}{\theta }\right)\mathbf{+}\mathbf{y}}{\mathbf{\theta }}$

Short answer

Homogeneous equation for the given equation is$\mathbf{y}\mathbf{=}\theta {\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{ln}\left|\theta \right|\mathbf{+}\mathbf{C}\right)$

Step-by-step solution

General form of Homogeneous equation

If 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 ratio $\frac{y}{x}$alone, then we say the equation is homogeneous.

Evaluate the given equation

Given, $\frac{dy}{d\theta }=\frac{\theta \sec \left(\frac{y}{\theta }\right)+y}{\theta }$

Evaluate it.

$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{d}\theta }\mathbf{=}\frac{\theta \mathbf{sec}\left(\frac{y}{\theta }\right)}{\theta }\mathbf{+}\frac{\mathbf{y}}{\theta } \\ \frac{\mathbf{dy}}{\mathbf{d}\theta }\mathbf{=}\mathbf{sec}\left(\frac{\mathbf{y}}{\theta }\right)\mathbf{+}\frac{\mathbf{y}}{\theta } \end{gathered}$$

Substitution method

Let us take $v=\frac{y}{x}$

Then $y=vx$

By Differentiating,

$\begin{array}{rcl}\frac{\mathbf{dy}}{\mathbf{d}\theta }& \mathbf{=}& \mathbf{v}\mathbf{+}\theta \frac{\mathbf{dv}}{\mathbf{d}\theta }\\ \mathbf{sec}\mathbf{v}\mathbf{+}\mathbf{v}& \mathbf{=}& \mathbf{v}\mathbf{+}\theta \frac{\mathbf{dv}}{\mathbf{d}\theta }\\ & & \end{array}$

$\begin{array}{rcl}\mathbf{sec}\mathbf{v}& \mathbf{=}& \theta \frac{\mathbf{dv}}{\mathbf{d}\theta }\\ \frac{\mathbf{1}}{\mathbf{sec}\mathbf{v}}\mathbf{dv}& \mathbf{=}& \frac{\mathbf{1}}{\theta }\mathbf{d}\theta \\ & & \end{array}$

Integrate the equation

Now, integrate on both sides.

$\begin{array}{rcl}\int \frac{\mathbf{1}}{\mathbf{sec}\mathbf{v}}\mathbf{dv}& \mathbf{=}& \int \frac{\mathbf{1}}{\theta }\mathbf{d}\theta \\ \int \mathbf{cos}\mathbf{v}\mathbf{dv}& \mathbf{=}& \mathbf{ln}\left|\theta \right|\mathbf{+}\mathbf{C}\\ \mathbf{sin}\mathbf{v}& \mathbf{=}& \mathbf{ln}\left|\theta \right|\mathbf{+}\mathbf{C}\end{array}$

Substitute $v=\frac{y}{\theta }$

$\begin{array}{rcl}\mathbf{sin}\left(\frac{\mathbf{y}}{\theta }\right)& \mathbf{=}& \mathbf{ln}\left|\theta \right|\mathbf{+}\mathbf{C}\\ \frac{\mathbf{y}}{\theta }& \mathbf{=}& {\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{ln}\left|\theta \right|\mathbf{+}\mathbf{C}\right)\\ \mathbf{y}& \mathbf{=}& \theta {\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{ln}\left|\theta \right|\mathbf{+}\mathbf{C}\right)\end{array}$

Therefore, Homogeneous equation for the given equation is $\mathbf{y}\mathbf{=}\theta {\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{ln}\left|\theta \right|\mathbf{+}\mathbf{C}\right)$.