Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Solve the Following Differential Equation:- ( X + Y ) D Y D X = 1 - Mathematics

Sum

Solve the following differential equation:-

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

#### Solution

We have,

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

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

Let x + y = v

$\Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1$

$\therefore \frac{dv}{dx} - 1 = \frac{1}{v}$

$\Rightarrow \frac{dv}{dx} = \frac{1}{v} + 1$

$\Rightarrow \frac{dv}{dx} = \frac{1 + v}{v}$

$\Rightarrow \frac{v}{1 + v}dv = dx$

Integrating both sides, we get

$\int\frac{v}{1 + v}dv = \int dx$

$\Rightarrow \int\frac{v + 1 - 1}{1 + v}dv = \int dx$

$\Rightarrow \int dv - \int\frac{1}{1 + v}dv = \int dx$

$\Rightarrow v - \log \left| v + 1 \right| = x - \log C$

$\Rightarrow x + y - \log \left| x + y + 1 \right| = x - \log C$

$\Rightarrow y - \log \left| x + y + 1 \right| = - \log C$

$\Rightarrow y = \log\left| x + y + 1 \right| - \log C$

$\Rightarrow y = \log\left| \frac{x + y + 1}{C} \right|$

$\Rightarrow C e^y = x + y + 1$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 66.13 | Page 147