दिया गया अवकल समीकरण है-
`ydx + x log ((y)/(x)) dy -2 xdy =0`
`impliesydx =-[x log ""((y)/(x))-2x=dy`
`impliesydx =[2x-x log ""((y)/(x))]dy.`
`implies (dy)/(dx) =(y)/(2x- x log ((y)/(x)))" "...(1)`
माना `F(x,y) =(y)/(2x-x log ((y)/(x)))`
x को `lamdax ` और y को `lamday ` से प्रतिस्थापित करने पर,
`F(lamdax, lamday) =(lamday)/(2 lamdax -lamda x log ((lamday)/(lamdax)))`
`implies F(lamdax, lamday)=(lamday)/(lamda[2x-xlog((y)/(x))])`
`implies F (lamdax, lamday)= lamda^(0)F(x,y)`
अतः `F(x,y)` एक समीकरण फलन है इसलिए दिया गया अवकल समीकरण समघाट अवकल समीकरण है.
माना `y=vx` तब `(dy)/(dx) =v+x(dy)/(dx)" "...(2)`
समी (1 ) और (2 ) से,
`v+x(dv)/(dx)=(vx)/(2x-xlog ((vx)/(x)))`
`impliesv+x (dv)/(dx)=(v)/(2-log v)`
`impliesx(dv)/(dx) =(v)/(2-log v)-v`
`implies(dv)/(dx)=(v-2v+vlog v)/(2-log v)`
`implies x(dy)/(dx) =(-v+v log v)/(2-log v)`
`impliesx(dy)/(dx) =(v(logv -1))/(2-log v)`
`implies (2-log v)/(v (log v-1))dv =(dx)/(x),`
(चरो के पृथकरण से )
समाकलन करने पर,
`int (2- log v)/(v(log v-1))dv=int (dx)/(x)`
माना `log v =t implies 1/v dv=dt`
`therefore int(2-t)/((r-1))dt =int (dx)/(x)`
`impliesint ((1)/(t-1)-1)dt =int (dx)/(x),` [भाग विधि से]
`impliesint (1)/(t-1) dt - int 1 dt =int (dx)/(x)`
`implieslog |t-1|-t=log |x| +C`
`implieslog |(log v-1)/(v)| -log |x|=C`
`implies log |(log v-1)/(vx)|=C`
`implies log |(log ((y)/x) -1)/(y)|=C,`
` [becausey= vximpliesv =(y)/(x)]`
`implies(log ((y)/(x))-1)/(y)=e^(C)`
`implieslog ((y)/(x))-1=C_(1)y,` जहाँ `C_(1)=e^(c).`
