दर्शाइए कि अवकल समीकरण `2ye^((x)/(y))dx+(y-2xe^((x)/(y)))dy = 0` समघातीय है और यदि x = 0 जब `y = 1` दिया हुआ है तो इस समीकरण का विशिष्ट हल ज्ञात कीजिए |

Explanation

`because " " 2ye^((x)/(y))dx+(y-2xe^((x)/(y)))dy = 0`
या `" " 2ye^((x)/(y))dx-(2xe^((x)/(y))-y)dy = 0`
या `" " 2ye ((x)/(d))dx = (2xe ^((x)/(y)) - y) dy`
या `" " (dx)/(dy) = (2xe^((x)/(y))-y)/(2ye^((x)/(y)))`
अतः यह एक समाघाती अवकल समीकरण है |
अब माना x = vy हो, तब `" " (dx)/(dy) = v + y(dv)/(dy)`
`therefore " " v +y (dv)/(dy) = (2vye^(v) - y)/(2ye^(v)) = (2ve^(v) -1)/(2e^(v))`
`" " y""(dv)/(dy) = (2ve^(v) -1)/(2e^(v)) -v = (1)/(2e^(v))`
या `" " 2e^(v) dv = (dy)/(y)`
या `" " 2fe^(v)dv = -f""(dy)/(y)`
`rArr " " 2e^(v) = -log |y| + c`
अब `v = (x)/(y)` रखने पर,
`2e^((x)/(y)) + log |y| =c `
अब `x = 0` और `y = 1` रखने पर,
`2e^(0) + log 1 = c`
`c= 2`
अतः `" " 2e^((x)/(y)) + log|y| = 2.`