Question

A find the Orthogonal trajectory of the family of
Curves e^-x cosy + xy=alpha
where alpha is a real constant
in the xy plane​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $2{e}^{-x}sin(y) +(x^2- y^2) =\beta$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

☆ɢɪᴠᴇɴ☆

• Curves $e^{-x}cos(y) + xy=\alpha.$

☆ᴛᴏ ғɪɴᴅ☆

• Orthogonal curves to the given curves.

☆ᴄᴀʟᴄᴜʟʟᴀᴛɪᴏɴ☆

The given curves are

• $e^{-x}cos(y) + xy=\alpha$

Differentiating w. r. t. x, we get

$\implies - e^{-x}cos(y) - e^{-x}sin(y)\frac{dy}{dx}$

$\:\:\:\:\:\:\:\:\:\:\:+ y + x \frac{dy}{dx} = 0$

$\implies [x - e^{-x}sin(y)] \frac{dy}{dx}$

$\:\:\:\:\:\:\:\:\:\:= e^{-x}cos(y) - y$

$\implies \frac{dy}{dx} = \frac{e^{-x}cos(y) - y}{(x - e^{-x}sin(y))}$

$\implies m_1 = \frac{e^{-x}cos(y) - y}{x - e^{-x}sin(y)}$

If $m_2$ is the slope of the orthogonal curves, then

$\implies m_1\times m_2 = - 1$

$\implies m_2 = - \frac{1}{m_1}$

$\implies m_2 = - \frac{x - e^{-x}sin(y)}{e^{-x}cos(y) - y}$

$\implies \frac{dy}{dx} = - \frac{x - e^{-x}sin(y)}{e^{-x}cos(y) - y}$

$\implies (e^{-x}cos(y) - y) dy$

$\:\:\:\:\:\:\:\:\:\:= - (x - e^{-x}sin(y))dx$

$\implies xdx - ydy - e^{-x}sin(y)dx$

$\:\:\:\:\:\:\:\:\:\:+e^{-x}cos(y)dy=0$

$\implies xdx - ydy +d(e^{-x}sin(y)) =0$

$\implies \int[xdx - ydy$

$\:\:\:\:\:\:\:\:\:\:+d(e^{-x}sin(y))] =C$

$\implies \int xdx - \int ydy$

$\:\:\:\:\:\:\:\:\:\:+ \int d(e^{-x}sin(y))=C$

$\implies \frac{x^2}{2} - \frac{y^2}{2}+ e^{-x}sin(y) =C$

$\implies 2e^{-x}sin(y) +(x^2 - y^2) =\beta$