`(x-y)(dy)/(dx)=(x+2y)`
`implies(dy)/(dx)=(x+2y)/(x-y)" ....(1)"`
माना `f(x, y)=(x+2y)/(x-y)`
`f(lambdax, lambday)=(lambdax+2lambday)/(lambdax-lambday)=(lambda(x+2y))/(lambda(x-y))`
`implies = (x+2y)/(x-y)=f(x,y)`
`therefore` दी गई अवकल समीकरण समघातीय है।
माना `y=vx`
`implies (dy)/(dx)=v+x(dv)/(dx)`
समीकरण (1) में रखने पर
`v+x(dv)/(dx)=(x+2vx)/(x-vx)=(1+2v)/(1-v)`
`implies x(dv)/(dx)=(1+2v)/(1-v)-v`
`=(1+2v-v+v^(2))/(1-v)`
`=(1+v+v^(2))/(1-v)`
`implies (v-1)/(v^(2)+v+1)dv=-(dx)/(x)`
`implies int(v-1)/(v^(2)+v+1)dv=-int(dx)/(x)`
`impliesint((2v+1)-3)/(v^(2)+v+1)dv=-2int(dx)/(x)`
`=int(2v+1)/(v^(2)+v+1)dv-3int(1)/(v^(2)+v+1)dv=-2logx+c`
`implieslog(v^(2)+v+1)-3int(1)/((v+(1)/(2))^(2)+((sqrt(3))/(2))^(2))dv=-2logx+c`
`implies log(v^(2)+v+1)+logx^(2)`
`-3.(1)/(sqrt(3)//2)tan^(-1).(v+(1)/(2))/(sqrt(3)//2)=c`
`implieslog(x^(2)v^(2)+vx^(2)+x^(2))-2sqrt(3)tan^(-1).(2v+1)/(sqrt(3))=c`
`implieslog(y^(2)+xy+x^(2))-2sqrt(3)tan^(-1).(2y+x)/(xsqrt(3))=c`
