`2xydx+(x^(2)+2y^(2))dy=0` को हल कीजिए।

Explanation

दिया है : `2xydx+(x^(2)+2y^(2))dy=0`
`implies(dy)/(dx)=-(2xy)/(x^(2)+2y^(2))" "......(1)`
`implies(dy)/(dx)=-(2xy)/(x^(2)+2y^(2))" "......(1)`
समीकरण (1) में y = vx अर्थात `(dy)/(dx)=-(2x(vx))/(x^(2)+2v^(2)x^(2))=-(2v)/(1+2v^(2))`
`impliesx(dv)/(dx)=-(2v)/(1+2v^(2))-v=-((3v+2v^(2)))/(1+2v^(2))`
`implies(1+2v^(2))/(3v+2v^(2))dv+(dx)/(x)=0`
दोनों ओर का समाकलन करने पर
`int(1+2v^(2))/(3v+2v^(3))dv+int(dx)/(x)=0`
`implies(1)/(3)int(dt)/(t)+log|x|=logc_(1)`
[माना `t=3v+2v^(3)implies(dt)/(3)=(1+2v^(2))dv]`
`implies(1)/(3)log|t|+log|x|=logc_(1)`
`implies|t|^(1//3).x=c_(1)`
`impliestx^(3)=c_(1)^(3)=c`
`implies(3v+2v^(3))x^(3)=c`
`implies3vx^(3)+2v^(3)x^(3)=c`
पुनः `v=(y)/(x)` रखने पर
`3x^(2)y+2y^(3)=c`
यही दी गयी समीकरण का अभीष्ट हल है।