110年台科大材料碩士班-工程數學詳解

 國立臺灣科技大學110學年度碩士班招生考試

系所組別:材料科學與工程系碩士班乙組
科目名稱:工程數學 

解答: $$xy'+(1+x)y=e^{-x}\sin 2x \Rightarrow y'+(1+{1\over x})y= {1\over x}e^{-x}\sin 2x\\ \text{integration factor }I(x)=e^{\int (1+1/x)dx} =xe^x \Rightarrow I(x)y'+ I(x)(1+{1\over x})y= I(x){1\over x}e^{-x}\sin 2x \\ \Rightarrow xe^x y'+(1+x)e^x y=\sin 2x \Rightarrow (xe^x y)'=\sin 2x \Rightarrow xe^x y=-{1\over 2}\cos 2x+c_1 \\ \Rightarrow \bbox[red, 2pt]{y= -{1\over 2xe^x}\cos 2x+ {c_1\over xe^x}}$$

解答: $$z=2x^{1/3} \Rightarrow {dz\over dx} ={2\over 3}x^{-2/3} \Rightarrow {dy\over dx} ={dy\over dz}\cdot {dz\over dx} ={2\over 3}x^{-2/3} {dy\over dz} \\ \Rightarrow {d^2y \over dx^2} =-{4\over 9}x^{-5/3}{dy \over dz} +{2\over 3}x^{-2/3}{d^2y\over dz^2}\cdot {2\over 3}x^{-2/3}=-{4\over 9}x^{-5/3}{dy \over dz} +{4\over 9}x^{-4/3} {d^2y\over dz^2} \\ \Rightarrow 9x^2y''+ 9xy'+(4x^{2/3}-16)y= 9x^2 \left( -{4\over 9}x^{-5/3}{dy \over dz} +{4\over 9}x^{-4/3} {d^2y\over dz^2} \right) +9x \left( {2\over 3}x^{-2/3} {dy\over dz}\right) +(4x^{2/3}-16)y \\=  4x^{2/3} {d^2y\over dz^2}+ 2x^{1/3}{dy\over dz} +(4x^{2/3}-16)y =z^2{d^2y\over dz^2}+ z{dy\over dz}+(z^2-4^2)y=0\\ z^2{d^2y \over dz^2}+z{dy\over dz}(z^2-4^2)y=0 \text{ is a Bessel function of the second kind} \\ \Rightarrow y=c_1J_v(z)+c_2Y_v(z) \Rightarrow \bbox[red, 2pt]{y =c_1J_4(2x^{1/3}) +c_2Y_{4}(2x^{1/3}) }$$

解答: $$3y''-6y'+30y=15\sin x+{e^x \over \cos 3x} \Rightarrow y''-2y'+10y= 5\sin x+ {e^x\over 3\cos 3x} \\ \Rightarrow y''-2y'+10y=0 \Rightarrow \lambda^2-2\lambda+10 =0 \Rightarrow \lambda = 1\pm 3i \Rightarrow y_h=e^x(c_1 \cos 3x +c_2 \sin 3x) \\ \text{Let }\cases{y_1=e^x \cos 3x\\ y_2= e^x \sin 3x}, \text{then }W=\begin{vmatrix} y_1& y_2\\ y_1'& y_2'\end{vmatrix} =\begin{vmatrix} e^x\cos 3x&  e^x \sin 3x\\ e^x\cos 3x-3e^x\sin 3x& e^x \sin 3x+3e^x \cos 3x\end{vmatrix} =3e^{2x} \\ \text{Using variation of parameters, }\\y_p= -e^x\cos 3x \int{ e^x\sin 3x\cdot (5\sin x+{e^x \over 3\cos 3x})\over 3e^{2x}} \,dx +e^x \sin 3x \int{e^x \cos 3x \cdot (5\sin x+{e^x\over 3\cos 3x}) \over 3e^{2x}}\,dx\\\bbox[red, 2pt] {y=c_1e^x \sin(3x)+ c_2e^x \cos(3x)+{9\over 17}\sin x+ {1\over 9}xe^x \sin(3x)+{2\over 17}\cos x+ {1\over 27} e^x \cos(3x) \ln (\cos 3x)}$$

解答: $$f(t)=\begin{cases} 2& 0\le t\lt 3\\ 4& t\ge 3\end{cases} \Rightarrow L\{f(t)\} =\int_0^\infty f(t)e^{-st}\,dt = \int_0^3 2e^{-st}\,dt +\int_3^\infty 4e^{-st}\,dt\\ =\left. \left[ -{2\over s}e^{-st} \right] \right|_0^3 +\left. \left[ -{4\over s}e^{-st} \right] \right|_3^\infty ={2\over s}(1-e^{-3s}) +{4\over s}e^{-3s} ={2\over s}(1+e^{-3s}) \\ \Rightarrow L\{y''\}-4L\{ y'\}+ 4L\{y\}=L\{f(t)\} \Rightarrow s^2Y(s)-sy(0)-y'(0)-4(sY(s)-y(0)) +4Y(s) ={2\over s}(1+e^{-3s}) \\ \Rightarrow (s-2)^2Y(s)+2s-9= {2\over s}(1+e^{-3s}) \Rightarrow Y(s) ={2\over s(s-2)^2}(1+e^{-3s}) +{9-2s \over (s-2)^2} \\ \Rightarrow y(t) =L^{-1}\{Y(s)\} = 2L^{-1}\left\{{1\over s(s-2^2)} \right\} +2L^{-1} \left\{ {e^{-3s} \over s(s-2)^2)} \right\}-2L^{-1}\left\{ {1\over s-2}\right\}+ 5L^{-1}\left\{ {1\over (s-2)^2} \right\} \\={1\over 2}H(t)-{1\over 2}e^{2t}+te^{2t} +H(t-3) \left({1\over 2}-{1\over 2}e^{2(t-3)}+ (t-3)e^{2(t-3)} \right) -2e^{2t} +5te^{2t} \\ \Rightarrow \bbox[red, 2pt]{y(t)={1\over 2}-{5\over 2}e^{2t}+ 6te^{2t} +{1\over 2}H(t-3)(1+2te^{2(t-3)}-7e^{2(t-3)})}$$

解答: $$f(x)=\begin{cases} x,& 0\lt x\lt \pi/2\\ \pi-x,& \pi/2\le x\lt \pi\end{cases} \Rightarrow f(x)=f(-x) \Rightarrow f(x)\text{ is even} \Rightarrow f(x)=a_0 +\sum_{n=1}^\infty a_n \cos (nx),\\ \text{where }a_0={1\over \pi}\int_0^\pi f(x)\,dx ={1\over \pi} \cdot {\pi^2 \over 4} ={\pi \over 4}, \\\text{and }a_n={2 \over \pi }\int_0^\pi f(x)\cos(nx)\,dx ={2\over \pi} \left( \int_0^{\pi/2} x\cos(nx)\,dx + \int_{\pi/2}^\pi (\pi-x) \cos(nx) \,dx\right) \\={2\over \pi } \left. \left[ {x\over n} \sin(nx) +{1\over n^2}\cos(nx)\right] \right|_0^{\pi/2} + {2\over \pi}\left. \left[ {\pi-x\over n}\sin(nx)-{1\over n^2} \cos(nx) \right] \right|_{\pi/2}^\pi \\={2\over  n^2\pi}(2\cos(n\pi/2)-1-(-1)^n) \\ \Rightarrow \bbox[red, 2pt]{f(x)={\pi\over 4}+ \sum_{n=1}^\infty {2\over  n^2\pi}(2\cos(n\pi/2)-1-(-1)^n)  \cos(nx)}$$


解答: $$A=\begin{bmatrix}4 & \frac{1}{3} \\9 & 6\end{bmatrix} \Rightarrow \det(A-\lambda I) =(\lambda-3)(\lambda-7) =0 \Rightarrow \lambda=3,7\\ \lambda_1=3 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1 & \frac{1}{3} \\9 & 3\end{bmatrix} \begin{bmatrix}x_1 \\x_2\end{bmatrix} =0 \Rightarrow 3x_1+x_2=0 \Rightarrow v=x_2 \begin{pmatrix}-1/3 \\1 \end{pmatrix}, \\\qquad\text{we choose }v_1= \begin{pmatrix}-1/3 \\1 \end{pmatrix} \\ \lambda_2=7 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-3 & \frac{1}{3} \\9 & -1\end{bmatrix} \begin{bmatrix}x_1 \\x_2\end{bmatrix} =0 \Rightarrow 9x_1=x_2 \Rightarrow v=x_2  \begin{pmatrix} 1/9 \\1 \end{pmatrix}, \\ \qquad \text{we choose }v_2= \begin{pmatrix} 1/9 \\1 \end{pmatrix}\\ \Rightarrow X_c(t) =c_1 e^{3t} \begin{pmatrix}-1/3 \\1 \end{pmatrix}+c_2e^{7t} \begin{pmatrix} 1/9 \\1 \end{pmatrix} \Rightarrow X=\begin{bmatrix} -{1\over 3}e^{3t} & {1\over 9}e^{7t} \\ e^{3t} &e^{7t} \end{bmatrix} \Rightarrow X^{-1} =\begin{bmatrix} -{9\over 4}e^{-3t} & {1\over 4}e^{-3t} \\ {9\over 4}e^{-7t} & {3\over 4}e^{-7t} \end{bmatrix} \\ \Rightarrow X^{-1}g=\begin{bmatrix} -{9\over 4}e^{-3t} & {1\over 4}e^{-3t} \\ {9\over 4}e^{-7t} & {3\over 4}e^{-7t} \end{bmatrix} \begin{bmatrix} -3e^{t}  \\10e^{ t} \end{bmatrix} = \begin{bmatrix} {37\over 4}e^{-2t}  \\ {3\over 4} e^{ -6t} \end{bmatrix} \\ \Rightarrow X_p(t)= X \int X^{-1}g\,dt =\begin{bmatrix} -{1\over 3}e^{3t} & {1\over 9}e^{7t} \\ e^{3t} &e^{7t} \end{bmatrix} \int \begin{bmatrix} {37\over 4}e^{-2t}  \\ {3\over 4} e^{ -6t} \end{bmatrix}\,dt= \begin{bmatrix} -{1\over 3}e^{3t} & {1\over 9}e^{7t} \\ e^{3t} &e^{7t} \end{bmatrix} \begin{bmatrix} -{37\over 8}e^{-2t}  \\ -{1\over 8} e^{ -6t} \end{bmatrix} \\= \begin{bmatrix} {55\over 36}e^{t}  \\ -{19\over 4} e^{ t} \end{bmatrix} \Rightarrow X(t)=X_c(t)+ X_p(t) \\ \Rightarrow \bbox[red, 2pt]{X(t)= c_1 e^{3t} \begin{pmatrix}-1/3 \\1 \end{pmatrix} +c_2e^{7t} \begin{pmatrix} 1/9 \\1 \end{pmatrix} +\begin{pmatrix} 55/36 \\-19/4 \end{pmatrix}e^t}$$

解答: $$\cases{x(t)=6\cos t\\ y(t)= 6\sin t} \Rightarrow \cases{x'(t)=-6\sin t\\ y'(t) =6\cos t} \Rightarrow ds= \sqrt{(x'(t))^2+ (y'(t))^2} dt =6\,dt\\ \Rightarrow \oint_C (x^2-y^2)\,ds = \int_0^{2\pi} (36\cos^2t -36\sin^2 t) 6\,dt =216 \int_0^{2\pi} \cos (2t)\,dt \\=216 \left. \left[ {1\over 2}\sin 2t \right] \right|_0^{2\pi} = \bbox[red, 2pt]0$$

解答: $$AA=I \Rightarrow \begin{bmatrix}4 & -3 \\x & -4 \end{bmatrix}\begin{bmatrix}4 & -3 \\x & -4 \end{bmatrix} =\begin{bmatrix}16-3x & 0 \\0 & 16-3x \end{bmatrix} =\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} \Rightarrow 16-3x=1 \Rightarrow x=\bbox[red, 2pt]5$$

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解題僅供參考，碩士班歷年試題及詳解