# Differential Equation + Taylor Polynomial + Euler's Method

• May 31st 2008, 07:40 AM
bakanagaijin
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!
• May 31st 2008, 09:03 AM
bobak
Quote:

Originally Posted by bakanagaijin
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