國立臺北科技大學109學年度碩士班招生考試
系所組別:2131 電機工程系碩士班丙組
第一節 工程數學 試題(選考)
解答:(a) y=a−x⇒dydx=−1=(x+y)(x+y−2)=a(a−2)⇒a2−2a+1=0⇒a=1(b) dydx=(x+y)(x+y−2)⇒Riccati's equation⇒y=S(x)+1v=1−x+1v⇒y′=−1−v′v2=(x+y)(x+y−2)=(1+1v)(1v−1)⇒−1−v′v2=1v2−1⇒v′=−1⇒v=−x+c1⇒y=1−x+1c1−x解答:x2y″−2xy′+2y=0⇒Let y=xm, then y′=mxm−1,y″=m(m−1)xm−2⇒m(m−1)xm−2mxm+2xm=(m2−3m+2)xm=0⇒m2−3m+2=0⇒m=1,2⇒yh=c1x+c2x2Suppose that {y1=xy2=x2⇒{y′1=1y′2=2x⇒W(x,y)=|y1y2y′1y′2|=|xx212x|=x2Using variations of parameters, yp=−x∫x2⋅1x2cos1xx2dx+x2∫x⋅1x2cos(1x)x2dx⇒yp=−x∫cos1xx2dx+x2∫cos1xx3dx=−x(−sin1x)+x2(−sin1x+xcos1xx)⇒yp=−x2cos1x⇒y=yh+yp⇒y=c1x+c2x2−x2cos1x
解答:{2x1+2x2+x′2=δ(t−3)x′1−x2=0⇒{L{2x1+2x2+x′2}=L{δ(t−3)}L{x′1−x2}=0⇒{2X1(s)+2X2(s)+sX2(s)=e−3ssX1(s)−X2(s)=0⇒2X1(s)+2sX1(s)+s2X1(s)=e−3s⇒X1(s)=1s2+2s+2e−3s⇒x1(t)=L−1{X1(s)}=L−1{1s2+2s+2e−3s}=L−1{1(s+1)2+1e−3s}⇒x1(t)=H(t−3)e−t+3sin(t−3)X2(s)=sX1(s)=ss2+2s+2e−3s⇒x2(t)=L−1{X2(s)}=L−1{ss2+2s+2e−3s}⇒x2(t)=H(t−3)e−t+3(cos(t−3)−sin(t−3))
解答:A=[110−110001]⇒det
解答:\textbf{(a) }\text{Let }\cases{f_1=e^{-2x} \\f_2=e^{2x}} \text{ and }\{g_1,g_2\} \text{ denote the orthogonal basis. We apply the Gram-Schmidt process.}\\ g_1=f_1=e^{-2x} \Rightarrow \cases{f_2\cdot g_1= \int_0^1 1\,dx =1 \\ ||g_1||^2 = g_1\cdot g_1 = \int_0^1 e^{-4x}\,dx={1\over 4}(1-e^{-4})} \\ \Rightarrow g_2 =f_2-{f_2\cdot g_1\over ||g_1||^2}g_1 =e^{2x} -{4\over 1-e^{-4}}e^{-2x} \Rightarrow \{g_1,g_2\}=\bbox[red, 2pt]{\left\{e^{-2x}, e^{2x} -{4\over 1-e^{-4}}e^{-2x}\right\}} \\\textbf{(b) } f(x)=x \Rightarrow \cases{f\cdot g_1 = \int_0^1 xe^{-2x}\,dx = {1\over 4}(1-3e^{-2}) \\ f\cdot g_2 =\int_0^1 (xe^{2x}-{4x\over 1-e^{-4}}e^{-2x}) \,dx ={1\over 4}(1+e^2-{4e^2(e^2-3) \over e^4-1})\\ g_1\cdot g_1= \int_0^1 e^{-4x}dx ={1\over 4}(1-e^{-4}) \\ g_2\cdot g_2 = \int_0^1 (e^{2x}-{4\over 1-e^{-4}}e^{-2x})^2\,dx= {1-18e^4+e^8 \over 4(e^4-1)}} \\ \Rightarrow f(x) \approx{f\cdot g_1 \over g_1\cdot g_1}g_1 +{f\cdot g_2\over g_2\cdot g_2}g_2 ={1-3e^{-2}\over 1-e^{-4}}g_1+{ 1+e^2 -{4e^2(e^2-3)\over e^4-1}\over e^4-1}g_2 \\ \Rightarrow \bbox[red, 2pt]{f(x)\approx {1-3e^{-2}\over 1-e^{-4}}g_1+{e^6-3e^4+11e^2-1 \over (e^4-1)^2}g_2}
解答:\textbf{(a) }\mathbf v=\begin{bmatrix} v_1\\ v_2\\ v_3\end{bmatrix} \Rightarrow A\mathbf v = \begin{bmatrix} \cos(\alpha) & -\sin(\alpha) & 0\\ \sin(\alpha) & \cos(\alpha) & 0\\ 0& 0 & 1\end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3\end{bmatrix}= \begin{bmatrix} v_1 \cos (\alpha) -v_2\sin(\alpha)\\ v_1\sin(\alpha)+ v_2\cos(\alpha)\\ v_3\end{bmatrix} =\begin{bmatrix} u_1\\ u_2\\ u_3\end{bmatrix} =\mathbf u \\\quad \Rightarrow \begin{bmatrix} u_1\\ u_2 \end{bmatrix} =\begin{bmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{bmatrix} \begin{bmatrix} v_1\\ v_2 \end{bmatrix} \Rightarrow \mathbf u= \text{ rotating }\mathbf v \text{ about }z \text{ axis through angle }\alpha\quad \bbox[red, 2pt]{QED}\\ \textbf{(b) }A=\begin{bmatrix} \cos(\alpha) & -\sin(\alpha) & 0\\ \sin(\alpha) & \cos(\alpha) & 0\\ 0& 0 & 1\end{bmatrix} =\begin{bmatrix} B & 0\\ 0& 1\end{bmatrix} ,\text{ where }B= \begin{bmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{bmatrix} \text{ is a rotation matrix}\\ \Rightarrow A^{20} =\begin{bmatrix} B^{20} & 0\\ 0& 1\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} \cos(20\alpha) & -\sin(20\alpha) & 0\\ \sin(20 \alpha) & \cos(20\alpha) & 0\\ 0& 0 & 1\end{bmatrix}}
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解題僅供參考,碩士班歷年試題及詳解
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