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Question
The solution of the equation (2y – 1)dx – (2x + 3)dy = 0 is ______.
Options
`(2x - 1)/(2y + 3)` = k
`(y + 1)/(2x - 3)` = k
`(2x + 3)/(2y - 1)` = k
`(2x - 1)/(2y - 1)` = k
Solution
The solution of the equation (2y – 1)dx – (2x + 3)dy = 0 is `(2x + 3)/(2y - 1)` = k.
Explanation:
The given differential equation is (2y – 1)dx – (2x + 3)dy = 0
⇒ (2x + 3)dy = (2y – 1)dx
⇒ `("d"y)/(2y - 1) = ("d"x)/(2x + 3)`
Integrating both sides, we get
`int ("d"y)/(2y - 1) = int ("d"x)/(2x + 3)`
⇒ `1/2 log|2y - 1| = 1/2 log |2x + 3| + log"c"`
⇒ `log|2y - 1| = log|2x + 3| + 2 log "c"`
⇒ `log|2y - 1| - log|2x + 3| = log "c"^2`
⇒ `log|(2y - 1)/(2x + 3)| = log "c"^2`
⇒ `(2y - 1)/(2x + 3) = "c"^2`
⇒ `(2x + 3)/(2y - 1) = 1/"c"^2`
⇒ `(2x + 3)/(2y - 1)` = k
Where k = `1/"c"^2`.
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