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MEI C4 Differential Equations

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    Obtain a particular solution to (1-e^2y)dy/dx = e^y

    Given that y=1 when x=2.
    There is no need to express y in terms of x.

    Any ideas?
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    \frac{1-e^{2y}}{e^y}dy=dx

    e^{-y}-e^ydy=dx

    \int e^{-y}-e^y\ dy=\int 1\ dx

    -e^{-y}-e^y=x+C
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    (Original post by Pheylan)
    \frac{1-e^{2y}}{e^y}dy=dx

    e^{-y}-e^ydy=dx

    \int e^{-y}-e^y\ dy=\int 1\ dx

    -e^{-y}-e^y=x+C
    carry on..
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    (Original post by ziedj)
    carry on..
    -e^{-1}-e^1=2+C

    C=-(e+\frac{1}{e}+2)

    -e^{-y}-e^y=x-(e+\frac{1}{e}+2)

    e+\frac{1}{e}+2=x+e^y+\frac{1}{e  ^y}
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    (Original post by Pheylan)
    -e^{-1}-e^1=2+C

    C=-(e+\frac{1}{e}+2)

    -e^{-y}-e^y=x-(e+\frac{1}{e}+2)

    e+\frac{1}{e}+2=x+e^y+\frac{1}{e  ^y}
    You have passed the test. One internet will be sent your way.
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    Solved the solution up to substituting the x and y values and was about to post when i scrolled down and saw Pheylan's answer :sigh:
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    (Original post by Pheylan)
    \frac{1-e^{2y}}{e^y}dy=dx

    e^{-y}-e^ydy=dx

    \int e^{-y}-e^y\ dy=\int 1\ dx

    -e^{-y}-e^y=x+C
    so is x + e^-y + e^y = 4 the particular solution?
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    (Original post by Hippysnake)
    so is x + e^-y + e^y = 4 the particular solution?
    this is:
    (Original post by Pheylan)
    -e^{-1}-e^1=2+C

    C=-(e+\frac{1}{e}+2)

    -e^{-y}-e^y=x-(e+\frac{1}{e}+2)

    e+\frac{1}{e}+2=x+e^y+\frac{1}{e  ^y}
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    (Original post by Hippysnake)
    so is x + e^-y + e^y = 4 the particular solution?
    where did 4 come from?
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    (Original post by time.to.dance)
    Solved the solution up to substituting the x and y values and was about to post when i scrolled down and saw Pheylan's answer :sigh:
    sorry :awesome:
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    Cheers dude. How about this one?

    x^2 dy/dx - y^2 =0?

    I end up with y = x + c but that seems a bit too easy...
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    \frac{dy}{dx}=\frac{y^2}{x^2}

    y^{-2}dy=x^{-2}dx

    \int y^{-2}\ dy=\int x^{-2}\ dx

    -y^{-1}=-x^{-1}+C

    -\frac{1}{y}=-\frac{1}{x}+C

    -x=-y+Cxy

    y=x+Cxy
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    (Original post by Pheylan)
    \frac{dy}{dx}=\frac{y^2}{x^2}

    y^{-2}dy=x^{-2}dx

    \int y^{-2}\ dy=\int x^{-2}\ dx

    -y^{-1}=-x^{-1}+C

    -\frac{1}{y}=-\frac{1}{x}+C

    -x=-y+Cxy

    y=x+Cxy
    I got x/1+cx?

    Ah well, I have one more question?

    At time t seconds the rate of increase in the concentration of flesh eating bugs in a controlled environment is proportional to the concentration C of bugs present. Initially C= 100 bugs and after 2 seconds there are five times as many.

    Write down a differential equation connection dC/dt, C and t and hence find an expression for C in terms of t.

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