Find the particular solution of the differential equation satisfying the given conditions: `x^2dy+(x y+y^2)dx=0; y=1\ ` when `x=1.`

Explanation

`x^(2)dy+(xy+y^(2))dx=0`
`implies (dy)/(dx)+(xy+y^(2))/(x^(2))=0` …..`(1)`
Let `y=vx`
`implies (dy)/(dx)=v+(xdv)/(dx)`
From equation `(1)`,
`v+x(dv)/(dx)+(vx^(2)+v^(2)x^(2))/(x^(2))=0`
`implies x(dv)/(dx)+2v+v^(2)=0`
`implies x(dv)/(dx)=-v(v+2)`
`implies (dv)/(v(v+2))=-(dx)/(x)`
`implies int(2)/(v(v+2))dv=-int(2)/(x)dx`
`implies int((1)/(v)-(1)/(v+2))dv=-int(2)/(x)dx`
`implies logv=log(v+2)=-logx+logc` ltbgt `implies log(v)/(v+2)=log(c )/(x^(2))`
`implies (v )/(v+2)=(c )/(x^(2))`
`implies (y//x)/((y)/(x)+2)=(c )/(x^(2))`
`implies (y)/(2x+y)=(c )/(x^(2))`
`implies x^(2)y=x(2x+y)`
Givn, `x=1 `at `y=1`
`:. 1=c(2+1)`
`implies c=(1)/(3)`
Therefore, `x^(2)y=(1)/(3)(2x+y)`
`implies 3x^(2)y=2x+y`