一个二阶微分方程为：\[\frac{{{d^2}w}}{{d{x^2}}} - q = 0\] 边界条件为：\[w\left| {_{x = 0}} \right. = 0,w\left| {_{x = L}} \right. = 0\] 以下哪项不能作为此问题加权残值法中的试函数？
举一反三
- 5.下列函数中，在其定义域上有最大值和最小值的是（）。 A: $f(x)=\left\{ \begin{array}{*{35}{l}} \ln \left| x \right|,\ \ \ x\ne 0 \\ 0,\ \ \ \ \ \ \ \ x=0 \\ \end{array} \right.$ B: $f(x)=\ln \left( \left| x \right|+1 \right)\ x\in [-1,1]$ C: $f(x)=\ln \left| x \right|,\ \ \ x\in [-1,1]\backslash \{0\}$ D: $f(x)=\left\{ \begin{array}{*{35}{l}} \ln \left| x \right|,\ \ \ 0\lt |x|\lt 1 \\ 0,\ \ \ \ \ \ \ \ x=0 \\ \end{array} \right.$
- 曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为（ ） A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \) B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \) C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \) D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
- 下列方程中，不是全微分方程的为（ ）。 A: \(\left( {3{x^2} + 6x{y^2}} \right)dx + \left( {6{x^2}y + 4{y^2}} \right)dy = 0\) B: \({e^y}dx + \left( {x \cdot {e^y} - 2y} \right)dy = 0\) C: \(y\left( {x - 2y} \right)dx - {x^2}dy = 0\) D: \(\left( { { x^2} - y} \right)dx - xdy = 0\)
- 函数$z=\arcsin\dfrac{1}{~\sqrt{x+y}~}$的定义域为( ) A: $\left\{(x,y)\left|~x+y\geq<br/>0\right.\right\}$; B: $\left\{(x,y)\left|~x+y\geq<br/>1~\text{或}~x+y\leq<br/>-1 \right.\right\}$; C: $\left\{(x,y)\left|~x+y\geq<br/>1\right.\right\}$; D: $\left\{(x,y)\left|~x+y\geq<br/>\dfrac{4}{~\pi^2~}\right.\right\}$．
- 设\(D = \left\{ {(x,y)\left| { { x^2} + {y^2} \le 9,x \ge 0,y \ge 0} \right.} \right\}\)，则\(\int\!\!\!\int\limits_D {(x + 3y)} d\sigma = \)______