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Question
Solcve: `x ("d"y)/("d"x) = y(log y – log x + 1)`
Solution
Given that: `x ("d"y)/("d"x) = y(log y – log x + 1)`
⇒ `x ("d"y)/("d"x) = y[log(y/x) + 1]`
⇒ `("d"y)/("d"x) = y/x[log(y/x) + 1]`
Since, it is a homogeneous differential equation.
∴ Put y = vx
⇒ `("d"y)/("d"x) = "v" + x * "dv"/"dx"`
∴ `"v" + x * "dv"/"dx" = "vx"/x[log("vx"/x) + 1]`
⇒ `"v" + x * "dv"/"dx" = "v"[log "v" + 1]`
⇒ `x * "dv"/"dx" = "v"[log "v" + 1] - "v"`
⇒ `x * "dv"/"dx"` = v ....[log v + 1 – 1]
⇒ `x * "dv"/"dx" = "v" * log "v"`
⇒ `"dv"/("v"log"v") = "dx"/x`
Integrating both sides, we get
`int "dv"/("v"log"v") = int "dx"/x`
Put log v = t on L.H.S.
`1/"v" "dv"` = dt
∴ `int "dt"/"t" = int "dx"/x`
`log|"t"| = log|x| + log"c"`
⇒ `log|log "v"| = log x"c"`
⇒ log v = xc
⇒ `log(y/x)` = xc
Hence, the required solution is `log(y/x)` = xc.
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