Turn on thread page Beta
    • Thread Starter
    Offline

    19
    ReputationRep:
    So I had this question which I got from the new spec sample paper. There is this differential equation which you have to solve.

    \[\frac{dn}{dt}=0.1n(1-\frac{n}{50})\]

    It says show that he solution general solution to the differential equation can be written in the form where A is an arbitrary positive constant.

    \[n=\frac{50A}{e^{-0.1t}+A}\]

    You have to find the general solution to it. I have tried multiple ways and tried to see if any work but with no luck. Need someone to help me out here
    Offline

    12
    ReputationRep:
    Post your working out.

    Also note:
    If d is a constant then e^{d} is also a constant.
    Offline

    13
    ReputationRep:
    You can seperate it into:
    \int \frac{10}{n(1-\frac{n}{50})}dy=\int dx
    then integrate after using partial fractions on the LHS.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by Edgemaster)
    You can seperate it into:
    \int \frac{10}{n(1-\frac{n}{50})}dy=\int dx
    then integrate after using partial fractions on the LHS.
    That is exactly what I was thinking. Anyway, I did the following and still can't seem to get the answer. Here is what I did

    \[\frac{dn}{dt}=0.1n-\frac{0.1n^{2}}{50}\]

    \[500\int \frac{dn}{50n-n^{2}} = \int dt\]

    \[10\left [ ln (n) + ln (1-n) \right ] = t + c\]

    Somehow can't get the answer because either I have done it wrong or I can't evaluate and simplify.
    Offline

    12
    ReputationRep:
    (Original post by y.u.mad.bro?)
    That is exactly what I was thinking. Anyway, I did the following and still can't seem to get the answer. Here is what I did

    \[\frac{dn}{dt}=0.1n-\frac{0.1n^{2}}{50}\]

    \[500\int \frac{dn}{50n-n^{2}} = \int dt\]

    \[10\left [ ln (n) + ln (1-n) \right ] = t + c\]

    Somehow can't get the answer because either I have done it wrong or I can't evaluate and simplify.
    It appears you either did the partial fractions incorrectly or the integrating incorrectly.
    Offline

    13
    ReputationRep:
    (Original post by y.u.mad.bro?)
    That is exactly what I was thinking. Anyway, I did the following and still can't seem to get the answer. Here is what I did

    \[\frac{dn}{dt}=0.1n-\frac{0.1n^{2}}{50}\]

    \[500\int \frac{dn}{50n-n^{2}} = \int dt\]

    \[10\left [ ln (n) + ln (1-n) \right ] = t + c\]

    Somehow can't get the answer because either I have done it wrong or I can't evaluate and simplify.
    Can you show us the complete working for partial fractions?
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by Edgemaster)
    Can you show us the complete working for partial fractions?
    \[\frac{1}{n(50-n)}=\frac{A}{n}+\frac{B}{50-n}\]

    \[n=50, B=\frac{1}{50}\]
    \[n=0, A=\frac{1}{50}\]

    \[\frac{1}{n(50-n)}=\frac{1}{50n}+\frac{1}{50(50-n)}\]
    Offline

    12
    ReputationRep:
    (Original post by y.u.mad.bro?)
    \[\frac{1}{n(50-n)}=\frac{A}{n}+\frac{B}{50-n}\]

    \[n=50, B=\frac{1}{50}\]
    \[n=0, A=\frac{1}{50}\]

    \[\frac{1}{n(50-n)}=\frac{1}{50n}+\frac{1}{50(50-n)}\]

    Remember that you have 500 outside of the integral. Multiplying this through gives:

    \displaystyle\int \frac{10}{n}+ \frac{10}{50-n}\ dn
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by IWantIPods)
    Remember that you have 500 outside of the integral. Multiplying this through gives:

    \displaystyle\int \frac{10}{n}+ \frac{10}{50-n}\ dn
    I did that if you look at my first working out I posted. I just don't know what I've done wrong here or whats wrong in the integration.
    Offline

    12
    ReputationRep:
    (Original post by y.u.mad.bro?)
    I did that if you look at my first working out I posted. I just don't know what I've done wrong here or whats wrong in the integration.
    \frac{10}{50-n} = -10\times \frac{-1}{50-n} = -10\frac{f'(n)}{f(n)}


    Integrating this gives:

     -10\ln(f(n))
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by IWantIPods)
    \frac{10}{50-n} = -10\times \frac{-1}{50-n} = -10\frac{f'(n)}{f(n)}


    Integrating this gives:

     -10\ln(f(n))
    Ah, I see what I did wrong there. Let me redo and see what I get.
    • Thread Starter
    Offline

    19
    ReputationRep:
    nvm. Still not there. I get somewhere like

    \[t + c= ln(n)^{10}-ln(50-n)^{10}\]

    \[e^{t}+c = n^{10}-(50-n)^{10}\]

    \[(e^{t})^{0.1}+c=2n-50\]

    I still cant get the answer -_-' Something else I am missing?
    Offline

    12
    ReputationRep:
    (Original post by y.u.mad.bro?)
    nvm. Still not there. I get somewhere like

    \[t + c= ln(n)^{10}-ln(50-n)^{10}\]

    \[e^{t}+c = n^{10}-(50-n)^{10}\]

    \[(e^{t})^{0.1}+c=2n-50\]

    I still cant get the answer -_-' Something else I am missing?
    Your transition from line 1 to line 2 is incorrect.

    I think you should start with it in the form:

    t+c=10\ln(n)-10\ln(50-n)

    then using the fact that:

    m\ln(a)-m\ln(b)=m\ln(\frac{a}{b})

    then anti-log both sides of the equation.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by IWantIPods)
    Your transition from line 1 to line 2 is incorrect.

    I think you should start with it in the form:

    t+c=10\ln(n)-10\ln(50-n)

    then using the fact that:

    m\ln(a)-m\ln(b)=m\ln(\frac{a}{b})

    then anti-log both sides of the equation.
    I did do that the first time around but didn't get the answer so I used the other one which was again wrong.

    \[ln(\frac{n}{50-n})=\frac{t+c}{10}\]

    \[\frac{n}{50-n}=e^{\frac{t+c}{10}}\]

    but I can't seem to simplify this to the required answer.
    Offline

    12
    ReputationRep:
    (Original post by y.u.mad.bro?)
    I did do that the first time around but didn't get the answer so I used the other one which was again wrong.

    \[ln(\frac{n}{50-n})=\frac{t+c}{10}\]

    \[\frac{n}{50-n}=e^{\frac{t+c}{10}}\]

    but I can't seem to simplify this to the required answer.
    Let A=e^{\frac{c}{10}}
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by IWantIPods)
    Let A=e^{\frac{c}{10}}
    I did this but I don't think it makes a big difference.

    \[n=50e^{\frac{t}{10}}e^{\frac{c}{  10}}-ne^{\frac{1}{10}}e^{\frac{c}{10}  }\]

    \[n(1+e^{\frac{1}{10}}e^{\frac{c}{  10}})=50e^{\frac{t}{10}}e^{\frac  {c}{10}}\]

    Therefore, \[n=\frac{50e^{\frac{t}{10}}e^{\fr  ac{c}{10}}}{(1+e^{\frac{1}{10}}e  ^{\frac{c}{10}})}\]
    Offline

    12
    ReputationRep:
    (Original post by y.u.mad.bro?)
    I did this but I don't think it makes a big difference.

    \[n=50e^{\frac{t}{10}}e^{\frac{c}{  10}}-ne^{\frac{1}{10}}e^{\frac{c}{10}  }\]

    \[n(1+e^{\frac{1}{10}}e^{\frac{c}{  10}})=50e^{\frac{t}{10}}e^{\frac  {c}{10}}\]

    Therefore, \[n=\frac{50e^{\frac{t}{10}}e^{\fr  ac{c}{10}}}{(1+e^{\frac{1}{10}}e  ^{\frac{c}{10}})}\]
    Look at what the question is asking you to prove. The numerator is 50A where A is a constant. Just divide the numerator and the denominator by e^{\frac{t}{10}}.

    Also, in the denominator, you should have e^{\frac{t}{10}} not e^{\frac{1}{10}}.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by IWantIPods)
    Look at what the question is asking you to prove. The numerator is 50A where A is a constant. Just divide the numerator and the denominator by e^{\frac{t}{10}}.

    Also, in the denominator, you should have e^{\frac{t}{10}} not e^{\frac{1}{10}}.
    My bad. I mistyped that but I see what you mean. Didn't read the question properly.
 
 
 

2,650

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.