`2xy+y^(2)-2x^(2)(dy)/(dx)=0 , y = 2` यदि x = 1

Explanation

`2xy+y^(2)-2x^(2)(dy)/(dx)=0`
`implies (dy)/(dx)=(2xy+y^(2))/(2x^(2))" ....(1)"`
यह समघातीय अवकल समीकरण है।
माना y = vx
`implies (dy)/(dx)=v+x(dv)/(dx)`
समीकरण (1) से,
`v+x(dv)/(dx)=(2x^(2)v+v^(2)x^(2))/(2x^(2))=v+(1)/(2)v^(2)`
`implies x(dv)/(dx)=(1)/(2)v^(2)`
`implies (2dv)/(v^(2))=(dx)/(x)`
`implies 2int(1)/(v^(2))dv=int(dx)/(x)`
`implies -(2)/(v)+c=logx`
`implies -(2x)/(y)+c=logx`
`implies logx+(2x)/(y)=c" ....(2)"`
दिया है कि x = 1 पर y = 2
`therefore log 1+(2xx1)/(2)=c`
`implies c = 1`
अतः दी गई समीकरण का विशिष्ट हल `log x + (2x)/(y)=1`