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Math Help - Differential Equation + Taylor Polynomial + Euler's Method

  1. #1
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    Differential Equation + Taylor Polynomial + Euler's Method

    Consider the differential equation \frac{dy}{dx} = 5x^2 - \frac{6}{y-2} for y \neq 2. Let y=f(x) be the particular solution to this differential equation with the initial condition f(-1) = -4.

    a. Evaluate \frac{dy}{dx} and \frac{d^2 y}{dx^2} at (-1, -4)

    b. Is it possible for the x-axis to be tangent to the graph of f at some point? Explain why or why not.

    c. Find the second-degree Taylor polynomial for f about x=-1.

    d. Use Euler's method, starting at x=-1 with two steps of equal size, to approximate f(0). Show the work that leads to your answer.

    Thank you to anyone can help!
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  2. #2
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    Quote Originally Posted by bakanagaijin View Post
    Consider the differential equation \frac{dy}{dx} = 5x^2 - \frac{6}{y-2} for y \neq 2. Let y=f(x) be the particular solution to this differential equation with the initial condition f(-1) = -4.

    a. Evaluate \frac{dy}{dx} and \frac{d^2 y}{dx^2} at (-1, -4)

    b. Is it possible for the x-axis to be tangent to the graph of f at some point? Explain why or why not.

    c. Find the second-degree Taylor polynomial for f about x=-1.

    d. Use Euler's method, starting at x=-1 with two steps of equal size, to approximate f(0). Show the work that leads to your answer.

    Thank you to anyone can help!
    a) to find  \left( \frac{dy}{dx} \right)_{-1} use the formula given \left( \frac{dy}{dx} \right)_{-1}= 5(-1)^2 - \frac{6}{y_{-1}-2}

    to find  \left( \frac{d^2y}{dx^2} \right)_{-1} first find \frac{d^2 y}{dx^2} by differentiating \frac{dy}{dx} with respect to x.

    \frac{d^2 y}{dx^2} = 10x + \frac{6}{(y-2)^2} \frac{dy}{dx} you use value of  \left( \frac{dy}{dx} \right)_{-1} to find  \left( \frac{d^2y}{dx^2} \right)_{-1}

    b) If the x-axis is to be a tangent of the curve f(x) you require that \frac{dy}{dx}=0 and y=0 for some value of x. so you must determine whether 0=5x^2 - \frac{6}{0-2} has any real solutions.

    c) This is basic application of formula.

    f(x-1) = f(-1)+ (x-1)f'(-1) +\frac{(x-1)^2}{2!}f''(-1) use your values form part a) to do this.

    d) Like part c) this is just application of the formula \left( \frac{dy}{dx} \right)_{0} \approx  \frac{y_{1} -y_{0}}{h} where h = 0.5

    Bobak
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