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Math Help - Calculus Problem

  1. #1
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    Calculus Problem

    I am stumped on these last parts of my hw due...

    dy/dx=x^2y, f(0)=1....find f(.2)



    f(0)=7, f'(0)=-4,f''(0)=1,f'''(0)=6

    Find the 3rd degree Taylor polynomial and use that to approximate f(1.3)
    g(x)=f(x^2)...what's the 4th degree taylor
    h(x)= the integral from 0 to x of f(t)dt...find the 4th degree taylor

    Can anyone help me on any of these?
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  2. #2
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    Quote Originally Posted by sfgiants13 View Post
    I am stumped on these last parts of my hw due...

    dy/dx=x^2y, f(0)=1....find f(.2)



    f(0)=7, f'(0)=-4,f''(0)=1,f'''(0)=6

    Find the 3rd degree Taylor polynomial and use that to approximate f(1.3)
    g(x)=f(x^2)...what's the 4th degree taylor
    h(x)= the integral from 0 to x of f(t)dt...find the 4th degree taylor

    Can anyone help me on any of these?
    for the first equation can be solved by seperating variables

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

    Now we integrate both sides to get

    \ln|y|=\frac{1}{3}x^3+c \iff y=e^{\frac{1}{3}x^3+c }=e^{\frac{1}{3}x^3}e^{e^{c}}=Ae^{\frac{1}{3}x^3}

    f(x)=Ae^{\frac{1}{3}x^3}

    f(0)=1=Ae^{\frac{1}{3}0^3} \iff 1=A

    f(x)=e^{\frac{1}{3}x^3}

    f(0.2)=e^{\frac{1}{3}(0.2)^3}=e^{\frac{1}{3}\cdot \frac{1}{125}}=e^{\frac{1}{375}}


    for the next one the 3rd degree taylor polynomial is given by

    T_3(x)=\frac{f(0)}{0!}+\frac{f'(0)}{1!}x+\frac{f''  (0)}{2!}x^2+\frac{f'''(0)}{3!}x^3

    T_3(x)=7-4x+\frac{1}{2}x^2+x^3

    Now just plug in 1.3 for x

     <br />
g(x)=f(x^2) \approx T(x^2)<br />

    See what you can do from here

    Hint remember degree means highest exponent.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sfgiants13 View Post
    I am stumped on these last parts of my hw due...

    dy/dx=x^2y, f(0)=1....find f(.2)



    f(0)=7, f'(0)=-4,f''(0)=1,f'''(0)=6

    Find the 3rd degree Taylor polynomial and use that to approximate f(1.3)
    g(x)=f(x^2)...what's the 4th degree taylor
    h(x)= the integral from 0 to x of f(t)dt...find the 4th degree taylor

    Can anyone help me on any of these?
    This is a seperable differential equation

    so \frac{dy}{dx}=x^2y\Rightarrow{\frac{dy}{y}=x^2dx}

    Integrating we get

    \ln|y|=\frac{x^3}{3}+C

    So we solve for y getting

    y=\pm{C}{e^{\frac{x^3}{3}}}

    Now using the initial condition

    1=Ce^{\frac{0^3}{3}}\Rightarrow{C=1}

    so y=e^{\frac{x^3}{3}}

    -----------------------------------------------------------------------
    For the second one since all the values given are evaluated at zero this is a maclaurin sereis

    so

    f(x)\approx{f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+\frac  {f'''(0)x^3}{3!}}


    Or in this case

    f(x)\approx{ }7-4x+\frac{x^2}{2}+x^3

    Now to approximate f(x^2)

    we just substitute

    f(x^2)\approx{ }7-4x^2+\frac{x^4}{2}+x^6



    So assuming that f(t) converges on the interval of integration

    \int_0^{x}f(t)dt=\int_0^{x}\bigg[7-4t+\frac{t^2}{2}+t^6\bigg]dt=\bigg[7t-2t^2+\frac{t^3}{6}+\frac{t^7}{7}\bigg]\bigg|_0^{x}=7x-2x^2+\frac{x^3}{6}+\frac{x^7}{7}
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