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प्रश्न
Solve the following differential equation:
y log y = (log y2 - x) `"dy"/"dx"`
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उत्तर
y log y = (log y2 - x) `"dy"/"dx"`
∴ `1/("dy"/"dx") = (log "y"^2 - "x")/("y" log "y")`
∴ `"dx"/"dy" = (2 log "y - x")/("y" log "y")`
∴ `"dx"/"dy" = 2/"y" - "x"/("y" log "y")`
∴ `"dx"/"dy" + "x"/("y" log "y") = 2/"y"` .....(1)
This is the linear differential equation of the form
`"dx"/"dy" + "Px" = "Q"` where P = `1/("y" log "y") and "Q" = 2/"y"`
∴ I.F. = `"e"^(int "P dy") = "e"^(int 1/("y" log "y")"dy")`
`= "e"^(int (1//"y")/(log "y")"dy") = "e"^(log |log "y"|)` = log y
∴ the solution of (1) is given by
`"x" * ("I.F.") = int "Q" * ("I.F.") "dy" + "c"`
∴ `"x" * log "y" = int 2/"y" * log "y" "dy" + "c"`
∴ `(log "y") * "x" = 2 int (log "y")/"y" "dy" + "c"`
Put log y = t
∴ `1/"y" "dy" = "dt"`
∴ `("log y")*"x" = 2 int "t" "dt" + "c"`
∴ x log y = `2 * "t"^2/2 + "c"`
∴ x log y = (log y)2 + c
This is the general solution.
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