113年中正大學地科碩士班-工程數學詳解

國立中正大學113學年度碩士班招生考試試題

科目名稱:工程數學
系所組別:地球與環境科學系地震學、地球與環境科學系

10% for each question.Solve the following questions

解答: $$y''-16y'+64y=0 \Rightarrow \lambda^2-16\lambda+64=0 \Rightarrow (\lambda-8)^2=0 \Rightarrow \lambda=8\\ \Rightarrow y=c_1e^{8x}+ c_2xe^{8x} \Rightarrow y'=(8c_1+c_2)e^{8x}+   8c_2xe^{8x} \\ \Rightarrow \cases{y(0)=c_1=1\\ y'(0)=8c_1+c_2=7} \Rightarrow \cases{c_1=1\\ c_2=-1} \Rightarrow \bbox[red, 2pt]{y=e^{8x}-xe^{8x}}$$

解答: $$y''-8y'+12y=0 \Rightarrow \lambda^2-8\lambda+12=0 \Rightarrow (\lambda-6)(\lambda-2)=0 \Rightarrow \lambda=2,6\\ \Rightarrow y=c_1e^{2x}+ c_2e^{6x} \Rightarrow y'=2c_1e^{2x}+ 6c_2e^{6x} \Rightarrow \cases{y(0) =c_1+c_2=-4\\ y'(0) =2c_1+6c_2 =0} \Rightarrow \cases{c_1=-6\\ c_2=2} \\ \Rightarrow \bbox[red, 2pt]{ y=-6e^{2x} +2e^{6x}}$$

解答: $$y''-14y'+170y=0 \Rightarrow \lambda^2-14\lambda+170 =0 \Rightarrow \lambda=7\pm 11i \Rightarrow y=e^{7x}(c_1\cos(11x)+ c_2\sin(11x)) \\ \Rightarrow y'= 7e^{7x}( c_1\cos(11x)+ c_2\sin(11x)) +e^{7x}(-11c_1\sin(11x)+ 11c_2\cos(11x)) \\ \Rightarrow \cases{y(0)=c_1=7 \\y'(0) =7c_1+11c_2=126} \Rightarrow \cases{c_1=7\\ c_2=7} \Rightarrow \bbox[red, 2pt]{ y=7e^{7x}( \cos(11x) +\sin (11x))}$$

解答: $$xy'=x-y \Rightarrow xy'+y=x \Rightarrow (xy)'=x \Rightarrow xy=\int x\,dx ={1\over 2}x^2 +c_1\\ \Rightarrow \bbox[red, 2pt]{y={1\over 2}x+ {c_1\over x}}$$

解答: $$y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda=\pm i \Rightarrow y_c=c_1\cos x+c_2\sin x\\ y_p=Ax^2+Bx+C \Rightarrow y_p'=2Ax+B \Rightarrow y_p''=2A \Rightarrow y_p''+y_p=Ax^2+Bx+2A+ C =x^2\\ \Rightarrow \cases{A=1\\ B=0\\ 2A+C=0} \Rightarrow \cases{A=1\\ B=0\\ C=-2} \Rightarrow y_p=x^2-2 \Rightarrow y=y_c+y_p \Rightarrow y=c_1\cos x+c_2\sin x+x^2-2 \\ \Rightarrow y'=-c_1\sin x+c_2\cos x+2x \Rightarrow \cases{y(0) =c_1-2=2\\ y'(\pi) =-c_2+2\pi=0} \Rightarrow \cases{c_1=4\\ c_2=2\pi} \\ \Rightarrow \bbox[red, 2pt]{y=4\cos x+2\pi\sin x+x^2-2}$$

解答: $$A=\begin{bmatrix} -2 & 2\\ 2 & -2\end{bmatrix} = \begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & -4\end{bmatrix} \begin{bmatrix} 1/2 & 1/2\\ -1/2 & 1/2 \end{bmatrix} \Rightarrow e^A =\begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix} \begin{bmatrix} e^0 & 0\\ 0 & e^{-4}\end{bmatrix} \begin{bmatrix} 1/2 & 1/2\\ -1/2 & 1/2 \end{bmatrix} \\= \begin{bmatrix} {1\over 2}(1+e^{-4}) & {1\over 2}(1-e^{-4})\\ {1\over 2}(1-e^{-4}) & {1\over 2}(1+e^{-4}) \end{bmatrix} \\ \mathbf y'=A\mathbf y \Rightarrow y=e^{At} \mathbf c=  \begin{bmatrix} {1\over 2}(1+e^{-4t}) & {1\over 2}(1-e^{-4t})\\ {1\over 2}(1-e^{-4t}) & {1\over 2}(1+e^{-4t}) \end{bmatrix} \begin{bmatrix}c_1\\ c_2 \end{bmatrix}= \begin{bmatrix}{1\over 2}c_1(1+e^{-4t})+ {1\over 2}c_2(1-e^{-4t})\\ {1\over 2}c_1(1-e^{-4t}) +{1\over 2}c_2(1+e^{-4t})\end{bmatrix}  \\ \Rightarrow \bbox[red, 2pt]{\cases{y_1(t)=c_3+c_4e^{-4t}\\ y_2(t) = c_3-c_4e^{-4t}}}$$

解答: $$(1-x^2)y''-2xy'+6y=0 \Rightarrow (x^2-1)y''+2xy'=6y \Rightarrow ((x^2-1)y')'=6y \\ \Rightarrow (x^2-1)y'=\int 6y\,dy=3y^2+c_1 \Rightarrow (x^2-1)dy=(3y^2+c_1)dx \\ \Rightarrow \int{1\over 3y^2+c_1} \,dy = \int {1\over x^2-1}\,dx \Rightarrow {1\over \sqrt{3c_1}} \tan^{-1} {\sqrt 3y\over \sqrt c_1}={1\over 2}  \ln{x-1\over x+1}  +c_2 \\ \Rightarrow {\sqrt 3 y\over \sqrt{c_1}} =\tan\left({\sqrt {3c_1} \over 2} \ln{x-1\over x+1} +c_2\sqrt{3c_1}\right) \Rightarrow y=\sqrt{c_1\over 3} \tan\left({\sqrt {3c_1} \over 2} \ln{x-1\over x+1} +c_2\sqrt{3c_1}\right) \\ \Rightarrow y={c_3\over 3}\tan\left({c_3 \over 2} \ln{x-1\over x+1} +c_2c_3\right) \Rightarrow \bbox[red, 2pt]{ y={c_3\over 3}\tan\left({c_3 \over 2} \ln{x-1\over x+1} +c_4\right) }$$

解答: $$A= \left[ \begin{matrix}3 & 0 & 2 & 2\\-6 & 42 & 24 & 54\\21 & -21 & 0 & -15\end{matrix} \right] \xrightarrow{R_2+2R_1\to R_2,R_3-7R_1\to R_3} \left[\begin{matrix}3 & 0 & 2 & 2\\0 & 42 & 28 & 58\\0 & -21 & -14 & -29\end{matrix}\right] \\ \xrightarrow{R_2+2R_3\to R_2 } \left[\begin{matrix}3 & 0 & 2 & 2\\0 & 0 & 0 & 0\\0 & -21 & -14 & -29\end{matrix}\right] \Rightarrow rank(A)=\bbox[red, 2pt]2$$

解答: $$B=\left[\begin{matrix}1 & 0 & -1 & -3\\3 & 3 & -9 & -18\\4 & \frac{1}{2} & 6 & \frac{127}{2}\end{matrix}\right] \xrightarrow{R_2-3R_1\to R_2, R_3-4R_1\to R_3} \left[\begin{matrix}1 & 0 & -1 & -3\\0 & 3 & -6 & -9\\0 & \frac{1}{2} & 10 & \frac{151}{2}\end{matrix}\right] \xrightarrow {R_2/3 \to R_2, 2R_3 \to R3} \\\left[ \begin{matrix}1 & 0 & -1 & -3\\0 & 1 & -2 & -3\\0 & 1 & 20 & 151\end{matrix}\right] \xrightarrow{R_3-R_2 \to R_3} \left[\begin{matrix}1 & 0 & -1 & -3\\0 & 1 & -2 & -3\\0 & 0 & 22 & 154\end{matrix}\right] \xrightarrow{R_3/22 \to R_3} \left[ \begin{matrix}1 & 0 & -1 & -3\\0 & 1 & -2 & -3\\0 & 0 & 1 & 7\end{matrix}\right] \\ \xrightarrow{R_1 +R_3\to R_1, R_2+2R_3\to R_2} \left[\begin{matrix}1 & 0 & 0 & 4\\0 & 1 & 0 & 11\\0 & 0 & 1 & 7\end{matrix} \right] \Rightarrow rref(B)= \bbox[red, 2pt]{\left[\begin{matrix}1 & 0 & 0 & 4\\0 & 1 & 0 & 11\\0 & 0 & 1 & 7\end{matrix}\right]}$$

解答: $$f(x)=a_0+ \sum_{n=1}^\infty \left( a_n \cos{n\pi x\over L} +b_n\sin{n\pi x\over L}\right)\\\qquad =a_0+a_1\cos{\pi x\over L} +a_2\cos{2\pi x\over L}+\cdots+a_n\cos{n\pi x\over L}+\cdots\\\qquad \qquad + b_1\sin{\pi x\over L} +b_2\sin{2\pi x\over L}+\cdots+ b_n\sin{n\pi x\over L}+ \cdots \\ \Rightarrow \int_0^{L} f(x) \cos{n\pi x\over L}\,dx =\int_0^{L} a_0 \cos{n\pi x\over L}\,dx+ \int_0^La_1 \cos{\pi x\over L} \cos{n\pi x\over L}\,dx+ \cdots+ \int_0^L a_n \cos^2{n \pi x\over L}\,dx +\cdots\\\qquad \qquad + \int_0^L b_1 \sin{\pi x\over L} \cos{n\pi x\over L}\,dx+ \int_0^L b_2\sin{2\pi x\over L} \cos {n\pi x\over L}\,dx +\cdots+ \int_0^Lb_n \sin{n\pi x\over L} \cos{n\pi x\over L}\,dx+ \cdots \\\qquad =0+0+\cdots +{L\over 2} a_n +0+\cdots+0+ \cdots \\ \Rightarrow a_n={2\over L} \int_0^L f(x) \cos{n\pi x\over L}\,dx \quad \bbox[red, 2pt]{QED}$$

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解題僅供參考，碩士班歷年試題及詳解