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Math Help - find the maximum rate of change of x

  1. #1
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    find the maximum rate of change of x

    dx/dt = -4x-2x^2
    find the find the maximum rate of change of x (t)

    i know its could be very easy but i just have no ideas at all


    ---------------------------------------

    give the values of n and C for which y=Cx^n is a solution of the equation x^3 dy/dx -10 x^2 y =0

    the answer is, C=0 and n could be any no. , but why?


    thx
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  2. #2
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    Quote Originally Posted by Hikari Clover View Post
    dx/dt = -4x-2x^2
    find the find the maximum rate of change of x (t)

    i know its could be very easy but i just have no ideas at all


    ---------------------------------------

    give the values of n and C for which y=Cx^n is a solution of the equation x^3 dy/dx -10 x^2 y =0


    thx
    If you want the maximum rate of change, find the derivative of the rate of change, set it equal to 0, solve for x.

    Then take the second derivative of the rate of change, and check that it is negative at the value of x you found earlier.
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  3. #3
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    kool just realised that i did this in highschool, but not sure what 'rate of change' really was.

    how about the second question,

    give the values of n and C for which y=Cx^n is a solution of the equation x^3 dy/dx -10 x^2 y =0
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    It's a first order linear DE, and it's also separable.

    So you can use the separation of variables method, or the integrating factor method, to find the solution.
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    Quote Originally Posted by Prove It View Post
    It's a first order linear DE, and it's also separable.

    So you can use the separation of variables method, or the integrating factor method, to find the solution.

    yeh i did , got ln y = 10 ln x + constant
    how do i find that constant and make it into the y =C x^n form?
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    Quote Originally Posted by Hikari Clover View Post
    yeh i did , got ln y = 10 ln x + constant
    how do i find that constant and make it into the y =C x^n form?
    It's actually

    \ln{|y|} = 10\ln{|x|} + c

    \ln{|y|} = \ln{|x|^{10}} + c

    \ln{|y|} - \ln{|x|^{10}} = c

    \ln{\left|\frac{y}{x^{10}}\right|} = c

    \left|\frac{y}{x^{10}}\right| = e^c

    \frac{y}{x^{10}} = \pm e^c

    \frac{y}{x^{10}} = C (since \pm e^c is a constant, we can write it as another constant)

    y = Cx^{10}.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    It's actually

    \ln{|y|} = 10\ln{|x|} + c

    \ln{|y|} = \ln{|x|^{10}} + c

    \ln{|y|} - \ln{|x|^{10}} = c

    \ln{\left|\frac{y}{x^{10}}\right|} = c

    \left|\frac{y}{x^{10}}\right| = e^c

    \frac{y}{x^{10}} = \pm e^c

    \frac{y}{x^{10}} = C (since \pm e^c is a constant, we can write it as another constant)

    y = Cx^{10}.


    ohh i got it
    but how come my solution says C=0 and n=any number , it just makes no sense
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