115年雲林科大電機碩士班-工程數學詳解

國立雲林科技大學115學年度碩士班招生考試

系所：電機系 科目：工程數學(1)

解答:$$\textbf{(a) }{dy\over dx}={3ye^x+ 4y \sin xy+2x^2\over -3e^x-4x \sin xy} \Rightarrow (3ye^x+ 4y \sin xy+2x^2)dx+ (3e^x+4x\sin xy)dy=0\\ \quad \text{Let } \cases{M(x,y)=3ye^x+ 4y \sin xy+2x^2\\ N(x,y)= 3e^x+4x\sin xy} \Rightarrow \cases{M_y=3e^x+ 4\sin xy+4xy\cos xy \\ N_x=3e^x+4\sin xy+4xy\cos xy} \\ \quad \Rightarrow M_y=Nx \Rightarrow \text{Exact} \Rightarrow \Psi(x,y) =\int (3ye^x+ 4y \sin xy+2x^2)dx= \int (3e^x+4x\sin xy)dy \\ \Rightarrow \Psi(x,y)= 3ye^x-4\cos xy+{2\over 3}x^3+ \phi(y)= 3e^xy-4\cos xy+\rho(x) \\ \Rightarrow \Psi(x,y)= \bbox[red, 2pt]{3ye^x-4\cos xy+{2\over 3}x^3=C} \\\textbf{(b) } x'+2x= 4y \Rightarrow {dx\over dy}+2x=4y \Rightarrow {dx\over dy}e^{2y} +2xe^{2y} =4ye^{2y} \Rightarrow \left( xe^{2y} \right)'=4ye^{2y} \\ \quad \Rightarrow xe^{2y}= \int 4ye^{2y} \,dy = 2ye^{2y}-e^{2y}+C \Rightarrow \bbox[red, 2pt]{x=2y-1+Ce^{-2y}} \\\textbf{(c) } y''-4y'+4y=0 \Rightarrow r^2-4r+4=0 \Rightarrow (r-2)^2=0 \Rightarrow r=2 \Rightarrow y_h=c_1e^{2x}+ c_2xe^{2x} \\\quad y_p=(Ax+B)e^{-x} \Rightarrow y_p'= (-Ax+A-B)e^{-x} \Rightarrow y_p''  =(Ax-2A+B)e^{-x} \\ \quad \Rightarrow y_p''-4y_p'+4y_p= (9Ax-6A+9B)e^{-x} = 3xe^{-x} \Rightarrow \cases{9A=3\\ -6A+9B=0} \Rightarrow \cases{A=1/3\\ B=2/9} \\ \quad \Rightarrow y_p= \left( {1\over 3}x+{2\over 9} \right)e^{-x} \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{2x}+ c_2xe^{2x} +\left( {1\over 3}x+{2\over 9} \right)e^{-x}}$$

解答:$$\textbf{(a) } L^{-1} \left\{ {s+5\over (s+1)^2(s+3)} \right\} =L^{-1} \left\{ -{1\over 2}\cdot {1\over s+1} +{2\over (s+1)^2} +{1\over 2}\cdot {1\over s+3} \right\} \\\quad = \bbox[red, 2pt]{-{1\over 2}e^{-t}+2te^{-t}+{1\over 2}e^{-3t}} \\\textbf{(b) } L\{5+4t^2+2\sin 3t\} = {5\over s}+4\cdot {2\over s^3}+2\cdot {3\over s^2+3^2} = \bbox[red, 2pt]{{5\over s}+  {8\over s^3}+  {6\over s^2+9}} \\ \textbf{(c) }L^{-1} \left\{ {2s+5\over s^2+7} +{3s+1\over s^2-4s+13} \right\} =L^{-1  } \left\{ {2s+5\over s^2+7} \right\} + L^{-1} \left\{ {3s+1\over (s-2)^2+3^2} \right\} \\=L^{-1} \left\{ 2\cdot {s\over s^2+7}+{5\over \sqrt 7}\cdot {\sqrt 7\over s^2+7} \right\} +L^{-1} \left\{3\cdot {s-2\over (s-2)^2+9} +{7\over 3} \cdot {3\over (s-2)^2+9}\right\} \\ =2\cos(\sqrt 7t)+{5\over \sqrt 7}\sin(\sqrt 7t)+3e^{2t} \cos(3t)+ {7\over 3} e^{2t}\sin(3t) \\= \bbox[red, 2pt]{2\cos(\sqrt 7t)+{5\over \sqrt 7}\sin(\sqrt 7t)+e^{2t} \left( 3\cos(3t)+ {7\over 3} \sin(3t) \right) }$$

解答:$$\mathcal{L}\{y''\} - 4\mathcal{L}\{y'\} + 4\mathcal{L}\{y\} = \mathcal{L}\{e^x\} \Rightarrow s^2Y(s)-sy(0)-y'(0)-4(sY(s)-y(0))+4Y(s)={1\over s-1} \\ \Rightarrow s^2Y(s)-4sY(s)+4Y(s)-s+2={1\over s-1} \Rightarrow Y(s) = \frac{1}{(s-1)(s-2)^2} + \frac{1}{s-2} =  \frac{1}{s-1} + \frac{1}{(s-2)^2} \\ \Rightarrow y(x) =L^{-1} \left\{ \frac{1}{s-1} + \frac{1}{(s-2)^2}  \right\} =e^x+xe^{2x}  \Rightarrow \bbox[red, 2pt]{y(x) =e^x+xe^{2x} }$$

解答:$$\det(A-\lambda I) =-\lambda(\lambda-2) (\lambda-3)=0 \Rightarrow \lambda=0,2,3 \\ \lambda_1=0 \Rightarrow (A-\lambda_1 I) \mathbf x=0 \Rightarrow \begin{bmatrix}1&0& -1\\0&3&0\\-1& 0&1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\ 0\\0 \end{bmatrix} \cases{x_1=x_3\\ x_2=0} \Rightarrow \mathbf x= x_3 \begin{bmatrix}1\\0\\1 \end{bmatrix} \\ \qquad \Rightarrow \text{Setting }x_3=1 \Rightarrow \mathbf v_1= \begin{bmatrix}1\\0\\1 \end{bmatrix} \\ \lambda_2= 2 \Rightarrow (A-\lambda_2 I) \mathbf x=0 \Rightarrow  \begin{bmatrix} -1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{x_1+x_3=0 \\x_2=0} \Rightarrow \mathbf x= x_3 \begin{bmatrix}1\\0 \\-1 \end{bmatrix} \\ \qquad \Rightarrow \text{Setting }x_3=1 \Rightarrow \mathbf v_2= \begin{bmatrix}1\\0\\-1 \end{bmatrix} \\\lambda_3=3  \Rightarrow (A-\lambda_3 I) =0 \Rightarrow \begin{bmatrix} -2 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{x_1=0\\x_3=0} \Rightarrow \mathbf x= x_2 \begin{bmatrix}0\\1\\0 \end{bmatrix} \\ \qquad \Rightarrow \text{Setting }x_2=1 \Rightarrow \mathbf v_3= \begin{bmatrix}0\\1\\0 \end{bmatrix} \\ \text{Applying the Gram-Schmidt Process,} \\\mathbf u_1= \mathbf v_1= \begin{bmatrix}1\\0\\1 \end{bmatrix} \\ \mathbf u_2= \mathbf v_2-{\mathbf v_2\cdot \mathbf u_1\over \mathbf u_1\cdot \mathbf u_1} \mathbf u_1 = \mathbf v_2= \begin{bmatrix}1\\0\\-1 \end{bmatrix} \\ \mathbf u_3= \mathbf v_3-{\mathbf v_3\cdot \mathbf u_1\over \mathbf u_1\cdot \mathbf u_1} \mathbf u_1 -{\mathbf v_3\cdot \mathbf u_2\over \mathbf u_2\cdot \mathbf u_2} \mathbf u_2  = \mathbf v_3 = \begin{bmatrix}0\\1\\0 \end{bmatrix} \\ \mathbf e_1={\mathbf u_1\over ||\mathbf u_1||} = \begin{bmatrix}1/\sqrt 2\\0\\ 1/\sqrt 2 \end{bmatrix} , \mathbf e_2= {\mathbf u_2\over ||\mathbf u_2||} = \begin{bmatrix}1/\sqrt 2\\0\\ -1/\sqrt 2 \end{bmatrix} , \mathbf e_3 ={\mathbf u_3\over ||\mathbf u_3||} = \begin{bmatrix}0\\1\\0 \end{bmatrix} \\ \Rightarrow P= \begin{bmatrix}\mathbf e_1 & \mathbf e_2 & \mathbf e_3 \end{bmatrix} \Rightarrow P=  \bbox[red, 2pt]{\begin{bmatrix}1/\sqrt 2& 1/\sqrt 2& 0\\ 0& 0& 1\\ 1/\sqrt 2& -1/\sqrt 2& 0 \end{bmatrix}}$$

解答:$$\cases{\det(A)=a_{11} a_{22}-a_{12}a_{21} \\ \det(B) =b_{11}b_{22} -b_{12}b_{21}} \Rightarrow \det(A) \cdot \det(B) =(a_{11} a_{22}-a_{12}a_{21}) (b_{11}b_{22} -b_{12}b_{21}) \\ =a_{11}a_{22}b_{11}b_{22} -a_{11}a_{22}b_{12}b_{21} -a_{12}a_{21}b_{11}b_{22}+ a_{12}a_{21} b_{12}b_{21} \\ AB= \begin{bmatrix}a_{11} & a_{12}\\a_{21}& a_{22} \end{bmatrix} \begin{bmatrix} b_{11} &b_{12} \\ b_{21} & b_{22}\end{bmatrix} = \begin{bmatrix}a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+ a_{12}b_{22} \\ a_{21}b_{11}+ a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} \end{bmatrix} \\\Rightarrow \det(AB) =(a_{11}b_{11}+a_{12}b_{21}) (a_{21}b_{12}+a_{22}b_{22}) -(a_{11}b_{12}+ a_{12}b_{22}) (a_{21}b_{11}+ a_{22}b_{21}) \\=a_{11}a_{21}b_{11}b_{12}+ a_{11} a_{22}b_{11}b_{22} + a_{12}a_{21}b_{12}b_{21}+a_{12}a_{22}b_{21}b_{22} \\\qquad -a_{11}a_{21}b_{11}b_{12}-a_{11} a_{22}b_{12} b_{21} -a_{12}a_{21}b_{11} b_{22}-a_{12}a_{22}b_{21}b_{22} \\ \qquad =a_{11}a_{22}b_{11}b_{22} -a_{11}a_{22}b_{12}b_{21} -a_{12}a_{21}b_{11}b_{22}+ a_{12}a_{21} b_{12}b_{21} = \det(A)\cdot \det(B) \\ \Rightarrow \bbox[red, 2pt]{\det(A)\cdot \det(B)=\det(AB)}$$

 解答:$$a_0={1\over 2\pi} \int_{-\pi}^\pi f(x)\,dx = {1\over 2\pi} \left[ \int_{-\pi}^0 -2\,dx +\int_0^\pi 1\,dx  \right] =-{1\over 2} \\ a_n = {1\over \pi}   \left[ \int_{-\pi}^0 -2\cos(nx)\,dx +\int_0^\pi \cos(nx)\,dx \right]  =0\\ b_n={1\over \pi} \left[ \int_{-\pi}^0 -2\sin(nx)\,dx +\int_0^\pi  \sin(nx)\,dx \right] = {3\over n\pi} \left( 1-(-1)^n \right) \\ \Rightarrow f(x) \sim \bbox[red, 2pt]{-{1\over 2}+ \sum_{n=1}^\infty {3\over n\pi} (1-(-1)^n) \sin(x)}$$

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''+ 2xy'+2y= (m^2+m-2)x^m=0\\ \Rightarrow m^2+m-2=(m+2)(m-1)=0 \Rightarrow m=1,-2 \Rightarrow y_h=c_1x+c_2 x^{-2} \\x^2y''+ 2xy'+2y=x^3\sin x \Rightarrow y''+{2\over x}y'+{2\over x^2}y=x\sin x\\ \text{Using variation of parameters, we have }\cases{y_1=x\\ y_2=x^{-2} \\ r(x)=x\sin x} \Rightarrow W = \begin{vmatrix} y_1& y_2\\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix} x& x^{-2}\\ 1& -2x^{-3} \end{vmatrix} =-3x^{-2}\\ \Rightarrow y_p =-x \int{x^{-2} \cdot x\sin x\over -3x^{-2}}\,dx + x^{-2}\int{x\cdot x\sin x\over -3x^{-2}}\,dx = -x\sin x-4\cos x+{8\sin x\over x}+{8\cos x\over x^2 } \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1x+c_2 x^{-2}-x\sin x-4\cos x+{8\sin x\over x}+{8\cos x\over x^2 }}$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解