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Math Help - Integration qs and checks

  1. #1
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    Integration qs and checks

    Here a few questions and my answers where i have done. check means i want a yes or no as to my answer =P

    Here goes:

    Q1: 2xy(y')=y^2-9 x not=0

    a) find steady state solution...

    no answer..help

    b) find the general solution:

    y'=(y^2-9)/(2xy) ? check

    c) Find the particular solution when y=5 and x=4

    Is the answer 2/5 check

    Q2: The half life of morphine in the human bloodstream is 3 hours.

    (a) If initially there is 0.4 mg of morphine in the bloodstream, find an equation for the amount of morphine in the bloodstream at any time t.

    M=morphine amount, t=time in hours

    M=(cuberoot(0.5)^t)*)0.4 ? check

    (b) When does the amount drop below 0.01 mg?

    15.96 hours? check

    Q3: Find a function f(x) such that the point (1, 2) is on the graph of y=f(x), the slope of the tangent line at (1, 2) is 3 and f''(x)=x-1

    I got f(x)=(x^3)/6-(x^2)/2+3.5x-1/1/6 check please...(1/1/6 means 1 and 1 sixth)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    Here a few questions and my answers where i have done. check means i want a yes or no as to my answer =P

    Here goes:

    Q1: 2xy(y')=y^2-9 x not=0

    a) find steady state solution...

    no answer..help

    b) find the general solution:

    y'=(y^2-9)/(2xy) ? check

    c) Find the particular solution when y=5 and x=4

    Is the answer 2/5 check

    Q2: The half life of morphine in the human bloodstream is 3 hours.

    (a) If initially there is 0.4 mg of morphine in the bloodstream, find an equation for the amount of morphine in the bloodstream at any time t.

    M=morphine amount, t=time in hours

    M=(cuberoot(0.5)^t)*)0.4 ? check

    (b) When does the amount drop below 0.01 mg?

    15.96 hours? check

    Q3: Find a function f(x) such that the point (1, 2) is on the graph of y=f(x), the slope of the tangent line at (1, 2) is 3 and f''(x)=x-1

    I got f(x)=(x^3)/6-(x^2)/2+3.5x-1/1/6 check please...(1/1/6 means 1 and 1 sixth)

    f''(x)=x-1\Rightarrow{f'(x)=\frac{x^2}{2}-x+C}

    We know that [tex]f'(1)=3

    So

    3=\frac{1}{2}-1+C\Rightarrow{C=\frac{7}{2}}

    So now

    f'(x)=\frac{x^2}{2}-x+\frac{7}{2}\Rightarrow{f(x)=\frac{x^3}{6}-\frac{x^2}{2}-\frac{7x}{2}+C}


    we know f(1)=2

    So

    2=\frac{1}{6}-\frac{1}{2}+\frac{7}{2}+C\Rightarrow{C=\frac{-5}{3}}

    \therefore{f(x)=\frac{x^3}{6}-\frac{x^2}{2}+\frac{7x}{2}-\frac{5}{3}}
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  3. #3
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    are you sure about you working for C in f(1)=2, and there were others(hehe work slave work)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    are you sure about you working for C in f(1)=2, and there were others(hehe work slave work)
    You are right, it should be -7/6...well I am going to bed. I am sure someone else will help you.

    Good luck
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  5. #5
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    OK thanks for all your help.

    Anyone else want to step in and look at Q1 and Q2?
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by deragon999 View Post
    Here a few questions and my answers where i have done. check means i want a yes or no as to my answer =P

    Here goes:

    Q1: 2xy(y')=y^2-9 x not=0

    a) find steady state solution...

    no answer..help
    Rewrite as

    2xy\frac{dy}{dx}=y^2-9 \iff \frac{y}{y^2-9}dy=\frac{1}{2x}dx

    Now we integrate both sides to get

    \int \frac{y}{y^2-9}dy= \int \frac{1}{2x}dx

    \frac{1}{2}\ln|y^2-9|=\frac{1}{2} \ln|x|+C

    y^2-9 = Ae^{x} \iff y=\sqrt{Ae^x+9}
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