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 that, `(x+y)dy+(x-y)dx=0`
`implies (dy)/(dx)=(y-x)/(x+y)`………`(1)`
Given differential equation is homogenous.
Let `y=vx`
`implies (dy)/(dx)=v+x(dv)/(dx)`
`implies` From equation `(1)`, `v+x(dv)/(dx)=(vx-x)/(x+vx)`
`implies v+x(dv)/(dx)=(x(v-1)/(x(1+v)impliesv+x(dv)/(dx)=(v-1)/(1+v)`
`impliesx(dv)/(dx)=(v-1)/(1+v)-v`
`impliesx(dv)/(dx)=(v-1-v-v^(2))/(1+v)`
`implies -x(dv)/(dx)=(1+v^(2))/(1+v)=((v+1))/(1+v^(2))dv=-(1)/(x)dx`
On integration,
`int((v+1))/(1+v^(2))dv=-int(dx)/(x)`
`impliesint(v)/(1+v^(2))dv+int(1)/(1+v^(2))dv=-int(dx)/(x)`
`implies(1)log|v^(2)+1|+tan^(-1)v+log|x|=C`
`implies(1)/(2)log((y^(2)+x^(2))/(x^(2)))+log|x|+tan^(-1)((y)/(x))=C`
`implies log((y^(2)+x^(2))/(x^(2)))+2log|x|+2tan^(-1)((y)/(x))=2C`
`implieslog((y^(2)+x^(2))/(x^(2)))+logx^(2)+2tan^(-1)((y)/(x))=2C`
`implies log((y^(2)+x^(2))x^(2))/(x^(2))+2tan^(-1)((y)/(x))=2C`
`implies log(x^(2)+y^(2))+2tan^(-1)((y)/(x))=A` (put `2C=A`) .........`(2)`
Given, `x=1`, `y=1`
`log2+(2xx(pi)/(4))=A`
`implies A=(pi)/(2)+log2`
put the value of `A` in equation `(2)`,
`log(x^(2)+y^(2))+2tan^(-1)((y)/(x))=(pi)/(2)+log2`
which is the required solution of the given differential equation