# Solve the differential equation : $(x+y)dy + (x-y) dx=0$

$\begin{array}{1 1}\text{a.$ tan^{-1} \frac {y}{x} + \frac {1}{2} log t | 1 + \frac {y}{x}| = log x + c$} \\ \text{b.$ tan^{-1} \frac {y}{x} + \frac {1}{2} log t | 1 + \frac {y^2}{x^2}| = log x + c$} \\ \text{c.$ tan^{-1} \frac {y}{x} + \frac {1}{2} log t | 1 + \frac {y^2}{x^2}| = - log x + c$} \\ \text{d.$ tan^{-1} \frac {y}{x} + \frac {1}{2} log t | 1 + \frac {y}{x}| = - log x + c$} \end{array}$

Given : $(x+y)dy + (x-y) dx=0$
$(x + y) dy = -(x - y) dx$
$\frac {dy}{dx} = \frac {-(x - y)}{(x + y)}$
$\frac {dy}{dx} = \frac {y - x}{x + y}$
$\frac {dy}{dx} = \frac {\frac{y}{x} -1}{1 + \frac {y}{x}}$
$y = vx \implies \frac {y}{x} = v$
$\frac {dy}{dx} = v + \frac {dv}{dx} x$
$v + \frac {dv}{dx} x = \frac {v-1}{1 + v} \implies \frac {dv}{dx}x = \frac {v-1}{1+v} - v$
$\frac {dv}{dx} x = \frac {v - 1 - v (1 + v)}{1 + v} \implies \frac {v - 1 - v - v^2}{1 + v}$
$\frac {dv}{dx} x = \frac {-1 - v^2}{1 + v}$
$\frac {1+v}{1+v^2} dv = \frac {- dx}{x}$
$\int \frac {1}{1+v^2} dv + \int \frac { v dv}{ 1+v^2} = -\int \frac {dx}{x}$
$tan^{-1} v + \int \frac {vdv}{1+v^2} = - log x + c$
$1 + v^2 = t$
$dt = 2 vdv$
$vdv = \frac {dt}{2}$
$tan^{-1} v + \int \frac{dt}{2t} = - log x + c$
$tan^{-1} v + \frac {1}{2} log t = - log x + c$
$tan^{-1} \frac {y}{x} + \frac {1}{2} log t | 1 + v^2 | = - log x + c$
$tan^{-1} \frac {y}{x} + \frac {1}{2} log t | 1 + \frac {y^2}{x^2}| = - log x + c$
answered Feb 16, 2017