1.

Consider the differential equation `ydx-(x+y^(2))dy=0`. If for `y=1, x` takes value 1, then value of x when y = 4, isA. 64B. 9C. 16D. 36

Answer» Correct Answer - C
We have, `ydx-(x+y)^(2)dy=0`
`rArr" (dx)/(dy)+(-(1)/(y))x=y" ...(i)"`
This is a linear differential equation with `"I.F."=e^(int(1)/(y)dy)=(1)/(y)`
Multiplying both sides of (i) by `"I.F."=(1)/(y)` and integrating with respect to y, we get
`(x)/(y)=int yxx(1)/(y)dy+C`
`rArr" "(x)/(y)=y+C" ...(ii)"`
It is given that y=1 when x = 1. Putting x = 1, y = 1 in (ii), we get C = 0. Putting C = 0 in (ii), we get `x=y^(2)`. When y = 4 this equation gives x = 16.


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