# (1 + X) Y Dx + (1 + Y) X Dy = 0 - Mathematics

Sum

(1 + xy dx + (1 + yx dy = 0

#### Solution

We have,

(1 + xy dx + (1 + yx dy = 0

$\frac{dy}{dx} = - \frac{y\left( 1 + x \right)}{x\left( 1 + y \right)}$

$\Rightarrow \left( \frac{1 + y}{y} \right)dy = - \left( \frac{1 + x}{x} \right)dx$

$\Rightarrow \left( \frac{1}{y} + y \right)dy = - \left( \frac{1}{x} + 1 \right)dx$

Integrating both sides, we get

$\int\left( \frac{1}{y} + 1 \right)dy = - \int\left( \frac{1}{x} + 1 \right)dx$

$\Rightarrow \int\frac{1}{y}dy + \int dy = - \int\frac{1}{x}dx - \int dx$

$\Rightarrow \log\left| y \right| + y = - \log\left| x \right| - x + C$

$\Rightarrow \log\left| xy \right| + y + x = C$

$\Rightarrow x + y + \log\left| xy \right| = C$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 27 | Page 145