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    Hi , I'm not too sure what to do for the below question.


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    Do I put x=3 and y=0 for the first equation, then x=18 and y=0? I can't seem to understand on how to start as I can't find any examples.
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    (Original post by kerrypop)
    Hi , I'm not too sure what to do for the below question.


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    Do I put x=3 and y=0 for the first equation, then x=18 and y=0? I can't seem to understand on how to start as I can't find any examples.
    Put y=0 and arrange the equation
    \displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt

    Integrate and find x=f(t)
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    (Original post by ztibor)
    Put y=0 and arrange the equation
    \displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt

    Integrate and find x=f(t)
    Thanks for the reply

    I still don't really understand what is going on. If I integrate it I should get my answer, what does that mean in terms of the question though?
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    (Original post by kerrypop)
    Thanks for the reply

    I still don't really understand what is going on. If I integrate it I should get my answer, what does that mean in terms of the question though?
    For the LHS

    With partial fractions

    \displaystyle \frac{A}{x}+\frac{B}{12-x}=\frac{12A-Ax+Bx}{x(12-x)}=\frac{1}{x(12-x)}
    So
    A=B and 12A=1 -> A=B=1/12

    \frac{1}{12} \int \left ( \frac{1}{x}+\frac{1}{12-x}\right ) dx=\frac{1}{12}\int dt

    \ln \left |\frac{x}{12-x}\right |=t+C

    \frac{x}{12-x}=c\cdot e^{t}

    You will get the value of c from the initial conditions t=0 -> x_0=3 or 18
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    (Original post by ztibor)
    For the LHS

    With partial fractions

    \displaystyle \frac{A}{x}+\frac{B}{12-x}=\frac{12A-Ax+Bx}{x(12-x)}=\frac{1}{x(12-x)}
    So
    A=B and 12A=1 -> A=B=1/12

    \frac{1}{12} \int \left ( \frac{1}{x}+\frac{1}{12-x}\right ) dx=\frac{1}{12}\int dt

    \ln \left |\frac{x}{12-x}\right |=t+C

    \frac{x}{12-x}=c\cdot e^{t}

    You will get the value of c from the initial conditions t=0 -> x_0=3 or 18
    Thanks, that has been really helpful

    I can see by using \frac{dx}{dt}= x\left(1- \frac{x}{12}\right)

    I can get to the same answer as you do.

    Could you please explain how you started with \displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt as my algebra isn't that great and I can't see how you got it.
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    (Original post by kerrypop)
    Thanks, that has been really helpful

    I can see by using \frac{dx}{dt}= x\left(1- \frac{x}{12}\right)

    I can get to the same answer as you do.

    Could you please explain how you started with \displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt as my algebra isn't that great and I can't see how you got it.
    Separation of variables.
 
 
 
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