`(x-y)dy-(x+y)dx=0`

Explanation

`(x-y)dy-(x+y)dx=0`
`implies (x-y)dy=(x+y)dx`
`implies (dy)/(dx)=(x+y)/(x-y)" ....(1)"`
यह एक समघातीय अवकल समीकरण है। माना y = v x
`implies (dy)/(dx)=v+x.(dv)/(dx)`
समीकरण (1) में रखने पर,
`v+x(dv)/(dx)=(x+vx)/(x-vx)`
`=(1+v)/(1-v)`
`implies x(dv)/(dx)=(1+v)/(1-v)-v`
`=(1+v-v+v^(2))/(1-v)`
`=(1+v^(2))/(1-v)`
`implies (1-v)/(1+v^(2))dv=(dx)/(x)`
`implies int(1)/(1+v^(2))dv-int(v)/(1+v^(2))dv=int(dx)/(x)`
`implies tan^(-1)v-(1)/(2)log(1+v^(2))=logx+c`
`impliestan^(-1).(y)/(x)-(1)/(2)log(1+(y^(2))/(x^(2)))=logx+c`
`implies tan^(-1).(y)/(x)-(1)/(2)[log(x^(2)+y^(2))-logx^(2)]=logx+c`
`implies tan^(-1).(y)/(x)-(1)/(2)log(x^(2)+y^(2))+log x = log x+c`
`implies tan^(-1).(y)/(x)-(1)/(2)log(x^(2)+y^(2))=c`