# Find the particular solution satisfying the given condition $(x+y)dy+(x-y)dx=0;\;y=1;\;when\;x=\;1$

$\begin{array}{1 1} (A)\;2\tan^{-1}(\large\frac{y}{x}) +\log[x^2+y^2] =\large\frac{ \pi}{2} + \log 2. \\(B)\;2\tan^{-1}(\large\frac{y}{x}) - \log[x^2+y^2] =\large\frac{ \pi}{2} + \log 3. \\ (C)\;2\tan^{-1}(\large\frac{y}{x}) + \log[x^2-y^2] =\large\frac{ \pi}{2} + \log 2. \\ (D)\;2\tan^{-1}(\large\frac{y}{x}) + \log[x+y^2] =\large\frac{ \pi}{2} + \log 2.\end{array}$

Toolbox:
• To solve homogenous differnetial equation put $y = vx$ and $\large\frac{dy}{dx}$$= v+x\large\frac{dv}{dx} Step 1: Let us rearrange and write the equation as (x+y)dy = -(x-y)dx \large\frac{dy}{dx }= -\frac{(x-y)}{(x+y)} Using the information in the tool box we get, v+x\large\frac{dv}{dx }= \frac{-(x-vx)}{x+vx) } Taking the common factor x and cancelling on the RHS we get v+x\large\frac{dv}{dx} = \frac{(1-v)}{(1+v)} Bringing v from LHS to the RHS we get, x\large\frac{dv}{dx }= \frac{(1-v-v+v^2)}{(1+v)} x\large\frac{dv}{dx }=\frac{ (v^2+1)}{(1+v)} Seperating the variables we get, \large\frac{(1+v)dv}{(v^2+1) }= -\frac{ dx}{x} Step 2: Integrating on both sides \int\large\frac{(1+v)dv}{(v^2+1) }= -\int\frac{ dx}{x} \int\large\frac{1}{(v^2+1) }+ \int\large\frac{ v}{(v^2+1) }=- \int\large\frac{ dx}{x} \tan^-1(v) + \large\frac{1}{2}$$[\log(1+v^2)] = - \log x + C$
Substituting $v = \large\frac{y}{x}$ we get,
$\tan^-1(\large\frac{y}{x}) +\large\frac{1}{2}$$[\log(1+(y^2/x^2)] = - \log x + C 2\tan^{-1}(\large\frac{y}{x}) + [\large\frac{\log(x^2 + y^2)}{x^2}] =$$- 2\log x + C$
$2\tan^-1(\large\frac{y}{x}) +\large\frac{\log[x^2 + y^2]}{x^2} . $$x^2= C 2\tan^-1(\large\frac{y}{x} )$$+ \log[x^2+y^2] = C$
Step 3:
Given $x = 1$ and $y = 1$, by substituting we get
$2\tan^{-1}(1) + log[1+1] = C$
But $\tan^-1(1) =\large\frac{ \pi}{4}$
$2.(\large\frac{\pi}{4}) $$+ \log 2 = C C = \large\frac{\pi}{2}$$ + \log 2$
Substituting for $C$ we get,