The differential equations , find the particular solution satisfying the given condition: `(x + y) dy + (x – y) dx = 0; y = 1` when `x = 1`

Explanation

given differential equation is
`(x + y) dy + (x – y) dx = 0`
`dy/dx = (y-x)/ (x+y).......(i)`
put `y= vx`
`dy/dx = v+ x (dv)/dx......(ii)`
from equation (i) and (ii)
`v+ x (dv)/dx. = (y-x)/ (x+y) = (vx-x)/(x+vx) = ((v-1)x)/((v+1)x) = (v-1)/(v+1)`
`x(dv)/(dx) = (v-1)/(v+1) - v = (v-1-v^2-v)/(v+1)`
`(v+1)/(1+v^2) dv = - (dx)/x => int 1/ (1+v^2) dv + int v/ (1+v^2) dv = - int dx/x`

`tan^-1(v) + 1/2 int( 2v)/(1+v^2) dv = -log|x| +c`
`tan^-1(v) +1/2 log|1+v^2| =-log|x| +c`
`tan^-1(y/x) + 1/2 log|1+(y/x)^2| = -log|x| +c.....(iii)`
it is given thant `y=1` when `x=1` putting these values in equation (iii) , we have
`tan^-1(1) + 1/2 log|1+(1)^2| = -log|1| +c`
`pi/4 + (1/2)log2 = 0+c`
`c= pi/4+logsqrt(2)`
putting the value of c in equation (iii)
`tan^-1(y/x) + 1/2 log|1+(y/x)^2| = -log|x| + pi/4+logsqrt(2)`
`tan^-1(y/x) + 1/2 log|1+(y/x)^2|+log|x| = pi/4+logsqrt(2)`
or
`tan^-1(y/x) + log sqrt(x^2 + y^2) =pi/4+logsqrt(2)`
this is a given required solution