# Form a differential equation by eliminating the arbitrary constant a from the equation. y = a sin(2x + 3).??

Jul 5, 2018

$y ' = 2 y \cot \left(2 x + 3\right)$

#### Explanation:

$y = a \sin \left(2 x + 3\right) q \quad \mathbb{A}$

Differentiate both sides wrt $x$:

$y ' = 2 a \cos \left(2 x + 3\right) q \quad \mathbb{B}$

Use $\frac{\mathbb{A}}{\mathbb{B}}$ to eliminate $a$

$\frac{\mathbb{A}}{\mathbb{B}} \implies \frac{y}{y '} = \frac{1}{2} \tan \left(2 x + 3\right) q \quad y ' = 2 y \cot \left(2 x + 3\right)$

That's a first-order linear ordinary differential equation, solvable by separation of variables.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \setminus \cot \left(2 x + 3\right)$

#### Explanation:

Given equation:

$y = a \setminus \sin \left(2 x + 3\right)$

differentiating above equation w.r.t. $x$ as follows

$\frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{d}{\mathrm{dx}} \left(a \setminus \sin \left(2 x + 3\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = a \setminus \frac{d}{\mathrm{dx}} \left(\setminus \sin \left(2 x + 3\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = a \setminus \cos \left(2 x + 3\right) \setminus \frac{d}{\mathrm{dx}} \left(2 x + 3\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = a \setminus \cos \left(2 x + 3\right) \left(2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 a \setminus \cos \left(2 x + 3\right)$

setting $a = \frac{y}{\setminus} \sin \left(2 x + 3\right)$ in above D.E. as follows

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\frac{y}{\setminus} \sin \left(2 x + 3\right)\right) \setminus \cos \left(2 x + 3\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \setminus \cot \left(2 x + 3\right)$