國立臺灣科技大學112學年度碩士班招生試題
系所組別:材料科學與工程系碩士班乙組
科目:工程數學
解答:$$\textbf{(1) }\cases{P(x,y)= y^4+2y\\ Q(x,y)=xy^3+2y^4-4x} \Rightarrow \cases{P_y=4y^3+2\\ Q_x=y^3-4} \Rightarrow P_y \ne Q_x \Rightarrow \bbox[red, 2pt]{\text{No, it is not exact.}} \\\textbf{(2) } u'=-{P_y-Q_x\over P}u =-{3y^3+6\over y^4+2y}u = -{3 \over y }u \Rightarrow u'+{3\over y}u=0 \Rightarrow y^3u'+3y^2 u=0 \\\qquad \Rightarrow \left( y^3 u\right)'=0 \Rightarrow y^3u=c_1 \Rightarrow \text{integration factor }u={1\over y^3} \Rightarrow \cases{uP=y+2/y^2\\ uQ=x+2y-4x/y^3} \\\qquad \Rightarrow \cases{(uP)_y=1-4/y^3\\ (uQ)_x=1-4/y^3} \Rightarrow (uP)_y= (uQ)_x \Rightarrow \Phi(x,y) = \int \left( y+{2\over y^2} \right)dx =\int \left( x+2y-{4x\over y^3} \right)dy \\ \qquad \Rightarrow \Phi(x,y)= xy+{2x\over y^2} +\phi(y) =xy+y^2+{2x\over y^2} +\rho(x) \Rightarrow \bbox[red, 2pt] {xy+y^2+{2x\over y^2}=c_1}$$
解答:$$\textbf{(1) }y''+4y=0 \Rightarrow \lambda^2+4=0 \Rightarrow \lambda=\pm 2i \Rightarrow \bbox[red, 2pt]{ y_h= c_1\cos(2x) +c_2\sin(2x)}\\ \textbf{(2) }y_p=A+Bx\cos(2x)+ Cx\sin(2x) \Rightarrow y_p'=B\cos(2x)-2Bx\sin(2x) +C\sin(2x)+ 2Cx\cos(2x) \\\qquad \Rightarrow y_p''=-4B\sin(2x) - 4Bx\cos(2x)+ 4C\cos(2x) -4Cx\sin(2x) \\ \qquad \Rightarrow y_p''+4y_p=4A-4B\sin(2x)+4C\cos(2x)={1\over 2} -{\cos(2x) \over 2} \Rightarrow \cases{A=1/8\\ B=0\\ C=-1/8} \\\qquad \Rightarrow y_p={1\over 8}-{1\over 8}x\sin(2x) \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos(2x) +c_2\sin(2x) +{1\over 8}-{1\over 8}x\sin(2x)}$$
解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-2xy'+2y=0 \\ \Rightarrow m(m-1)x^m-2mx^m+2x^m=(m^2-3m+ 2)x^m=0 \Rightarrow m^2-3m+2=0\\ \Rightarrow (m-2)(m-1)=0 \Rightarrow y_h=c_1x^2+ c_2 x\\ \text{By variations of parameters, let }\cases{y_1=x^2\\ y_2=x \\r(x)=1+2/x^2}, \text{ then }\cases{y_1'=2x\\ y_2' =1} \Rightarrow W=\begin{vmatrix} y_1 & y_2\\ y_1' & y_2'\end{vmatrix} =-x^2 \\ \Rightarrow y_p= -y_1 \int{ y_2 r(x) \over W}\,dx +y_2 \int {y_1 r(x)\over W}\,dx =x^2 \int \left({1\over x}+{2\over x^3} \right) \,dx -x \int \left(1 +{2 \over x^2} \right)\,dx \\ \Rightarrow y_p=x^2 \ln x-x^2 +1 \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_3x^2+c_2x+ x^2\ln x+1}$$
解答:$$\textbf{(1) }f(t)= \begin{cases} t-1 & 1\le t\le 2\\ 0& \text{otherwise}\end{cases} \Rightarrow L\{f(t)\} =\int_0^\infty f(t)e^{-st}\,dt = \int_1^2 (t-1)e^{-st}\,dt \\ \qquad= \left. \left[ {e^{-st} \over s^2} (-st+s-1)\right] \right|_1^2= {e^{-2s} \over s^2}(-s-1) +{e^{-s} \over s^2} \Rightarrow \bbox[red, 2pt]{L\{f(t)\} ={e^{-s} \over s^2}\left( 1-e^{-s}(s+1)\right)} \\\textbf{(2) }L\{y'+4y\} =L\{f(t) \} \Rightarrow sY(s)-y(0)+4Y(s)= {e^{-s} \over s^2}\left( 1-e^{-s}(s+1)\right) \\ \Rightarrow Y(s)= {e^{-s} \over s^2(s+4)}\left( 1-e^{-s}(s+1)\right) \Rightarrow y(t) =L^{-1}\{ Y(s)\} =L^{-1}\{ {e^{-s} \over s^2(s+4)}\left( 1-e^{-s}(s+1)\right)\} \\= L^{-1}\{ {e^{-s} \over s^2(s+4)}\} - L^{-1}\{ {se^{-2s} \over s^2(s+4)}\}- L^{-1}\{ {e^{-2s} \over s^2(s+4)}\} \\=u(t-1)\left( -{1\over 16}+{t-1\over 4} +{1\over 16}e^{-4(t-1)}\right) -u(t-2) \left( {1\over 4}-{1\over 4} e^{-4(t-2)}\right)\\\qquad \qquad -u(t-2)\left(-{1\over 16}+{t-2\over 4}+{1\over 16} e^{-4(t-2)} \right) \\ \Rightarrow \bbox[red, 2pt]{y(t)=u(t-1)\left( -{1\over 16}+{t-1\over 4} +{1\over 16}e^{-4(t-1)}\right) -u(t-2) \left( {3\over 16}+{1\over 4}(t-2)-{3\over 16} e^{-4(t-2)}\right)}$$
解答:$$\cases{\mathbf a=4i+6j\\ \mathbf b=-2i+6j-6k\\ \mathbf c=5/2 i+3j+1/2k} \Rightarrow \begin{vmatrix} 4& 6& 0\\ -2 & 6&-6\\ {5\over 2}& 3& {1\over 2}\end{vmatrix}=12+0-90-0+6+72=0 \\ \Rightarrow \text{They are linear dependent.} \Rightarrow \bbox[red, 2pt]{\text{They are coplanar.}}$$
解答:$$\textbf{(1)} A^TA =\begin{bmatrix}4& 8 & -10 \end{bmatrix}\begin{bmatrix} 4\\ 8\\ -10\end{bmatrix} =16+64+100= \bbox[red, 2pt]{180} \\\textbf{(2) }B^TB= \begin{bmatrix} 2\\4\\5\end{bmatrix} \begin{bmatrix} 2& 4 & 5\end{bmatrix}= \bbox[red, 2pt]{\begin{bmatrix} 4& 8& 10\\ 8& 16& 20\\ 10& 20& 25\end{bmatrix}} \\\textbf{(3) }A+B^T =\begin{bmatrix} 4\\ 8\\ -10\end{bmatrix} +\begin{bmatrix} 2\\4\\ 5\end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix} 6\\ 12\\ -5\end{bmatrix}}$$
解答:$$|a\cdot (b\times c)| =|(3,1,1) \cdot ((1,4,1)\times (1,1,5))|= |(3,1,1) \cdot (19,-4,-3)|= \bbox[red, 2pt]{50}$$
解答:$$A= \begin{bmatrix}2 & -1 & 0 \\5 & 2 & 4 \\0 & 1 & 2 \end{bmatrix} \Rightarrow \det(A-\lambda I) =0 \Rightarrow -(\lambda-2)(\lambda^2-4\lambda+ 5)=0 \Rightarrow \lambda=2, 2\pm i\\ \lambda_1=2 \Rightarrow (A-\lambda_1 I) v=0 \Rightarrow \begin{bmatrix} 0 & -1 & 0 \\5 & 0 & 4 \\0 & 1 & 0\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{5x_1+4x_3=0\\ x_2=0} \\\qquad \Rightarrow v=x_3\begin{pmatrix}\frac{-4}{5} \\0 \\1 \end{pmatrix}, \text{choose }v_1= \begin{pmatrix}\frac{-4}{5} \\0 \\1 \end{pmatrix} \\\lambda_2=2-i \Rightarrow (A-\lambda_2 I) v=0 \Rightarrow \begin{bmatrix} i & -1 & 0 \\5 & i & 4 \\0 & 1 & i\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+x_3=0\\ x_2+ix_3=0} \\\qquad \Rightarrow v=x_3 \begin{pmatrix} -1 \\-i \\1 \end{pmatrix}, \text{choose }v_2= \begin{pmatrix}-1 \\-i \\1 \end{pmatrix} \\\lambda_3=2+i \Rightarrow (A-\lambda_3 I) v=0 \Rightarrow \begin{bmatrix} -i & -1 & 0 \\5 & -i & 4 \\0 & 1 & -i\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+ x_3=0\\ x_2=ix_3} \\\qquad \Rightarrow v=x_3\begin{pmatrix}-1 \\i \\1 \end{pmatrix}, \text{choose }v_3= \begin{pmatrix}-1 \\i \\1 \end{pmatrix} \\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{2,2\pm i}, \text{ eigenvectors: } \bbox[red, 2pt]{\begin{pmatrix}\frac{-4}{5} \\0 \\1 \end{pmatrix},\begin{pmatrix}-1 \\-i \\1 \end{pmatrix},\begin{pmatrix}-1 \\i \\1 \end{pmatrix}}$$
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解題僅供參考，碩士班歷年試題及詳解
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