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Differential Equations help?

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Got a question about Student Finance? Ask the experts this week on TSR! 14-09-2014
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    Hi, I think I've either gone wrong somewhere throughout this, or I need help establishing how to get to the answer for the second part, because I don't see how I would do it :/

    Use integration to solve the differential equation dt/dx = (1/40)(x-15) where x>15, given that x=85 when t=0, expressing t in terms of x.

    My working:
    dt = (1/40)(x-15).dx
    40.dt = (x-15).dx
    integrate both sides
    40t = (x2/2)-15x+c

    I then subbed in x=85, t=0

    0=(852/2)-15(85)+c
    0=3612.5-1275+c
    0=2337.5+c
    c= -2337.5

    Making my equation 40t=(x2/2)-15x-2337.5

    The second part of the question is "Hence show that x=15+70e-t/40

    How would I do this?
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    Works if its dx/dt instead...although im not getting a minus on the t
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    (Original post by Tottenh)
    Works if its dx/dt instead...although im not getting a minus on the t
    Oops, just looked at the question and it is dx/dt!

    40.1/dt = (x-15).1/dx
    1/40.dt = 1/(x-15).dx
    1/(40t) = ln(x-15)+c
    40t = 1/(ln(x-15)+c)
    0=1/(ln60+c)
    e0=eln60+c
    1=60+c
    59=c

    I still don't think this is right though, because when I try and work it through I get this

    40t=1/(ln(x-15)+59)
    t=1/40(ln(x-15)+59)
    1/t=40(ln(x-15)+59)
    t-1/40=ln(x-15)+59
    et^-1/40=x-15+e59
    15+et^-1/40-e59=x

    :confused:
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    85-15 is 70.
    So subbing in your conditions as soon as you have integrated gives:
    ln70 = 0 + c, making c ln 70

    Keep me updated
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    Also, when you do something to each side of the equation, you do it to ALL of it. You aren't raising the c to the power e.
    But you also don't raise ln 70 to the power of e and then c, you raise them both TOGETHER.

    i.e:
    e^(ln60+c) which is defined as e^ln60 MULTIPLIED by e^c by the law of logs.

    ^ Obviously i am using your incorrect working as an example their but its just to show.
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    Is the differential equation \frac{dx}{dt}=\frac{(x-15)}{40}?
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    (Original post by Tottenh)
    85-15 is 70.
    So subbing in your conditions as soon as you have integrated gives:
    ln70 = 0 + c, making c ln 70

    Keep me updated
    Oh dear What am I doing at all.

    Laws of logs - ln(x-15) - ln70 is the same thing as ln((x-15)/70) right?

    et/40=x-15/70
    70et/40=x-15
    15+70et/40=x

    I'm thinking my teacher missed typing in a minus sign on the 1/40 initially, as it works if I stick that in.

    Making the question dx/dt=-(1/40)(x-15) with the same constraints

    1/(x-15).dx = -1/40.dt
    ln(x-15) = -t/40 +c
    ln70 = 0/40 +c
    ln70 = c
    e-t/40=x-15/70
    70e-t/40=x-15
    15+70e-t/40=x

    Thankyou very much!
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    No problem man. Nicely done.
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    (Original post by SpringNicht)
    Oh dear What am I doing at all.

    Laws of logs - ln(x-15) - ln70 is the same thing as ln((x-15)/70) right?

    et/40=x-15/70
    70et/40=x-15
    15+70et/40=x

    I'm thinking my teacher missed typing in a minus sign on the 1/40 initially, as it works if I stick that in.

    Making the question dx/dt=-(1/40)(x-15) with the same constraints

    1/(x-15).dx = -1/40.dt
    ln(x-15) = -t/40 +c
    ln70 = 0/40 +c
    ln70 = c
    e-t/40=x-15/70
    70e-t/40=x-15
    15+70e-t/40=x

    Thankyou very much!

    Just following the steps
    when I multiply both sides by 40
    I get
    40ln|70|=0+C
    therefore c=40ln|70|
    Maybe I've gone crazy
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    (Original post by arvin_infinity)
    Just following the steps
    when I multiply both sides by 40
    I get
    40ln|70|=0+C
    therefore c=40ln|70|
    Maybe I've gone crazy
    You've forgot to multiply the constant, you'd have 40ln70 = 0 +40c
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    (Original post by SpringNicht)
    You've forgot to multiply the constant, you'd have 40ln70 = 0 +40c
    +rep
    Oh it makes sense now!! that was the problem I did not know when to add the C ..I mean I multiplied it by 40 then I add the C (if you see what I mean)
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    (Original post by arvin_infinity)
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    Oh it makes sense now!! that was the problem I did not know when to add the C ..I mean I multiplied it by 40 then I add the C (if you see what I mean)
    Add it on one side as soon as you integrate

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