`because" " (x-y)""(dy)/(dx) = x + 2y`
`therefore " " (dy)/(dx) = (x+2y)/(x-y)`
अतः यह एक समघाती अवकल समीकरण है |
मान लीजिए `" " y = vx`
`(dy)/(dx) = v + x(dv)/(dx)`
`therefore " " v+x(dv)/(dx) = (x+2vx)/(x-vx) = (1+2v)/(1-v)`
`x""(dv)/(dx) = (1+2v)/(1-v) -v = (1+2v - v + v^(2))/(1-v)`
` = (v^(2) + v + 1)/(1-v)`
`(v-1)/(v^(2) + v + 1) dv = (dx)/(x)`
`(1)/(2) f""(2(v-1))/(v^(2) + v +1)dv = -f""(dx)/(x)`
`rArr " " (1)/(2) f""(2v + 1-3)/(v^(2) + v+ 1)dv = -log |x|+c`
`rArr (1)/(2)f""(2v+1)/(v^(2) + v + 1)dv = (3)/(2) f""(1)/(v^(2) + v + 1) dv = -log |x| + c`
`rArr (1)/(2)log|v^(2)+v+1|-(3)/(2)f""(1)/((v+(1)/(2))^(2)+((sqrt(3))/(2))^(2))dv=-log|x|+c`
` rArr (1)/(2) log |v^(2) + v +1 | = (3)/(2) xx (2)/(sqrt(3)) tan^(-1) ((2v +1 )/(sqrt(3))) = -log |x|+c`
`rArr (1)/(2) log |v^(2) + v + 1| +log |x| = sqrt(3) tan ^(-1) ((2v +1)/(sqrt(3))) +c`
अब `v = (y)/(x)` रखने पर,
`(1)/(2)log|(y^(2))/(x^(2))+(y)/(x)+1|+(1)/(2)logx^(2) = sqrt(3)tan^(-1)""(((2y)/(x)+1)/(sqrt(3)))+c`
`rArr (1)/(2)log|((y^(2))/(x^(2))+(y)/(x)+1)x^(2)|=sqrt(3)tan^(-1)((2y+3)/(sqrt(3)x))+c`
`rArr log|y^(2)+xy+x^(2)|=2sqrt(3)tan^(-1)((2y+x)/(sqrt(3)))+2c_(1)`
`rArr log|y^(2)+xy+x^(2)|=2sqrt(3)tan^(-1)((2y+x)/(sqrt(3)))+2c` (जहाँ `2C_(1) = C`)
