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Question
A manufacturing company has found that the cost C of operating and maintaining the equipment is related to the length ’m’ of intervals between overhauls by the equation `"m"^2 "dC"/"dm" + 2"mC"` = 2 and c = 4 and when = 2. Find the relationship between C and m
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Solution
`"m"^2 "dC"/"dm" + 2"mC"` = 2
÷ each term by m2
`"c"/"dm" + (2"mc")/"m"^2 = 2/"m"^2`
`"dc"/"dm" + (2"c")/"m" = 2/"m"^2`
This is a first order linear differential equation of the form
`"dc"/"dm" + "Pc"` = Q
Where P = `2"m"` and Q = `2/"m"^2`
`int "Pdm" = 2 int 1/"m" "dm" =2 log "m" = log "m"^2`
I.F = `"e"^(int"Pdm")`
= `"e"^(log"m"^2)`
= m2
General solution is
C(I.F) = `int "Q" xx ("I.F") "dm" + "k"`
C(m2) = `int 2/"m"^2 xx ("m"^2) "dm" + "k"`
C(m2) = `int 2"dm" + "k"`
Cm2 = 2m + k .......(1)
When C = 4 and m = 2, we have
(4)(2)2 = 2(2) + k
16 = 4 + k = 12
Equation (1)
Cm2 = 2m + 12
Cm2 = 2(m + 6)
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