The differential equations , find the particular solution satisfying the given condition:
`x^2dy+(x y+y^2)dx=0;`y = 1 when x = 1

Explanation

given differential equation is
`x^2dy+(x y+y^2)dx=0`
`(dy)/(dx) = -((xy + y^2)) / (x^2) ......(i)`
Let `y =vx`
diff on both side
` (dy)/(dx) = v + x (dv)/(dx)`
from (i) , we have
` v + x (dv)/(dx) = -((x^2v + x^2y^v)) / (x^2) = -v - v^2`
`x (dv)/(dx) = -(2v + v^2)`
Now
`int (dv)/(2v + v^2) = - int (dx)/x`
`int (dv) / ((v+1)^2 -1) = - int (dx)/x`
`1/2 log|(v+1-1)/(v+1+1)| = - log|x| + logc`
`log|v/ (v+2)|^(1/2) = log|c/x|`
=> `v/(v+2) = c^2/ x^2`
`y/ (y+2x) = c^2/ x^2`
`x^2 y = c^2 ( y + 2x) ....(ii)`

Given `y=1` when `x=1`
`1 = c^2 (3)`
`c^2 = 1/3`
put the value of c in equation (ii)
we get
`3x^2y = y+2x`