The particular solution of the differential equation x dy+2y dx=0, at x=2, y=1 is

Options

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

Answer

B

Explanation

We have, x dy+2y dx=0
`implies(dy)/(dy)+(2dx)/(x)=0` [dividing both sides by xy]
On integrating both sides, we get
`int(1)/(y)dy+2int(1)/(x)dx=logC`
`logy+2logx=logC" "[becausenlogm=logm^(n)]`
`thereforelogyx^(2)=logC" "[becauselogm+logn=logmn]`
`implies logyx^(2)=logC" "becauselogm+logn=logmn]`
`impliesyx^(2)=C`
When x=2 and y=1, then C=`1xx2^(2)=4`
`therefore`Particular solution is `x^(2)y=4`