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biology

Evaluate the following integrals. A sketch of the region of integration may be useful. $\int _ { - 2 } ^ { 2 } \int _ { 1 } ^ { 2 } \int _ { 1 } ^ { e } \frac { x y ^ { 2 } } { z } d z d x d y$

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engineering

Find a general solution. Check your answer by substitution. y''+6y'+8.96y=0

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biology

Like bird feathers, mammalian hair probably evolved from ____. A. teeth. B. scales. C. claws. D. setae.

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engineering

The vertical displacement u(x, t) of an infinitely long string is determined from the initial-value problem

\begin{array}{l}{a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}}, \quad-\infty<x<\infty, t>0} \\ {u(x, 0)=f(x),\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=g(x)}\end{array}

. This problem can be solved without separating variables. (a) Show that the wave equation can be put into the form $\partial^{2} u / \partial \eta \partial \xi=0$ by means of the substitutions $\xi=x+a t$ and $\eta=x-a t$ . (b) Integrate the partial differential equation in part (a), first with respect to $\eta$ and then with respect to $\xi$ , to show that $u(x, t)=F(x+a t)+G(x-a t)$ , where F and G are arbitrary twice differentiable functions, is a solution of the wave equation. Use this solution and the given initial conditions to show that $F(x)=\frac{1}{2} f(x)+\frac{1}{2 a} \int_{x_{0}}^{x} g(s) d s+c$ and $G(x)=\frac{1}{2} f(x)-\frac{1}{2 a} \int_{x_{0}}^{x} g(s) d s-c$ , where $x_0$ is arbitrary and c is a constant of integration. (c) Use the results in part (b) to show that $u(x, t)=\frac{1}{2}[f(x+a t)+f(x-a t)]+\frac{1}{2 a} \int_{x-a t}^{x+a t} g(s) d s$ . Note that when the initial velocity g(x) = 0 we obtain $u(x, t)=\frac{1}{2}[f(x+a t)+f(x-a t)],-\infty<x<\infty$ . The last solution can be interpreted as a superposition of two traveling waves, one moving to the right (that is, $\left.\frac{1}{2} f(x-a t)\right)$ and one moving to the left $\left(\frac{1}{2} f(x+a t)\right)$ . Both waves travel with speed a and have the same basic shape as the initial displacement f (x). The form of u(x, t) given in (14) is called d’Alembert’s solution.

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Product Knowledge

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