The particular solution of the differential equation xdy + 2 y dx = 0, when x = 2,

Options

A. `x y = 4`
B. `x^(2)y = 4`
C. `xy^(2) = 4`
D. `x^(2)y^(2) = 4`

Answer

b

Explanation

We have, `xdy + 2y dx = 0`
`rArr (dy)/(dy) + (2dx)/(x) = 0`
[dividing both sides xy]
On interating both sides , we get
`int1/ydy +2int(1)/(x) dx = logC`
`rArr log y +2logx = log C [ :' n log m = log m^(n)]`
`:. log yx^(2) = log C`
`[ :' log m+ log n = log mn] `
`rArr yx^(2) = C`
When `x = 2` and ` y = 1`.
then `C = 1 xx 2^(2) = 4`
`:.` Particular solution is `x^(2)y = 4`.