# Thread: Differential Equation + Taylor Polynomial + Euler's Method

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

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

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

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

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

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

Thank you to anyone can help!

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

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

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

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

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

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

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

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

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

c) This is basic application of formula.

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

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

Bobak