115年台科大機械碩士班甲組-工程數學詳解

國立立臺灣科技大學115學年度碩士班招生

系所組別:0310機械工程系碩士班甲組
科目:工程數學

解答:$$\textbf{(1) }  y'-2y=x^2+5 \Rightarrow e^{-2x}y'-2ye^{-2x} =e^{-2x}(x^2+5) \Rightarrow (ye^{-2x })'=e^{-2x}(x^2+5) \\ \qquad \Rightarrow ye^{-2x} = \int e^{-2x}(x^2+5)\,dx =-{1\over 4}e^{-2x} (2x^2+2x+11)+c_1 \\ \Rightarrow y=-{1\over 4}(2x^2+2x+11)+c_1 e^{-2x} \Rightarrow y(0)=-{11\over 4}+c_1= 3 \Rightarrow c_1={23\over 4}\\\qquad  \Rightarrow  \bbox[red, 2pt]{y=-{1\over 4}(2x^2+2x+11)+{23\over 4} e^{-2x}}\\ \textbf{(2) }  y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda= \pm i \Rightarrow y_h= c_1\cos x+ c_2 \sin x\\ \quad \cases{y_1= \cos x\\ y_2=\sin x} \Rightarrow W= \begin{vmatrix} \cos x& \sin x\\ -\sin x& \cos x\end{vmatrix}=1 \Rightarrow y_p =-\cos x \int{\sin x \cdot \sec x}\, dx+ \sin x \int\cos x\cdot \sec x \,dx \\\quad \Rightarrow y_p =\cos x \ln |\cos x| +x\sin x \Rightarrow y= y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1\cos x+ c_2 \sin x+\cos x \ln |\cos x| +x\sin x}$$

解答:$$L\{f(t)\} = L\{ \cos(t)\} + L\left\{ \int_0^t e^{-\tau} f(t-\tau)\,d\tau \right\} = L\{ \cos(t)\} + L\left\{  e^{-t}  \right\}\cdot L\{f(t)\} \\ \Rightarrow F(s)={s\over s^2+1}+{1\over s+1}\cdot F(s) \Rightarrow F(s)= {s+1\over s^2+1} ={s\over s^2+1} +{1\over s^2+1} \\ \Rightarrow f(t) =L^{-1}\{F(s)\} = L^{-1} \left\{ {s\over s^2+1}\right\}+L^{-1} \left\{ {1\over s^2+1}\right\} \Rightarrow \bbox[red, 2pt] {f(t)=\cos t+\sin t}$$

解答:$$[A \mid I] = \left[ \begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0\\1 & 2 & 1 & 0 & 1 & 0\\0 & 0 & 1 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_2-R_1  \to R_2} \left[ \begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0\\0 & -1 & 0 & -1 & 1 & 0\\0 & 0 & 1 & 0 & 0 & 1\end{array} \right] \xrightarrow{-R_2\to R_2} \\\left[ \begin{array}{ccc|ccc}1 & 3 & 1 & 1 & 0 & 0\\0 & 1 & 0 & 1 & -1 & 0\\0 & 0 & 1 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_1-3R_2\to R_1} \left[ \begin{array}{ccc|ccc}1 & 0 & 1 & -2 & 3 & 0\\0 & 1 & 0 & 1 & -1 & 0\\0 & 0 & 1 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_1-R_3\to R_1}\\ \left[ \begin{array}{ccc|ccc}1 & 0 & 0 & -2 & 3 & -1\\0 & 1 & 0 & 1 & -1 & 0\\0 & 0 & 1 & 0 & 0 & 1\end{array} \right]  \Rightarrow A^{-1} = \bbox[red, 2pt]{\begin{bmatrix}  -2 & 3 & -1\\  1 & -1 & 0\\  0 & 0 & 1 \end{bmatrix}}$$

解答:

$$w(x,y) =\cosh x \Rightarrow \nabla^2 w = {\partial^2 w\over \partial x^2} +{\partial^2 w\over \partial y^2} =\cosh x \Rightarrow \oint_C {\partial w\over \partial n}\,dS = \iint_R \nabla^2 w\,dA= \iint_R \cosh x\,dA \\= \int_0^2 \int_0^{2y} \cosh \,dxdy =\int_0^2 \sinh(2y)\,dy = \bbox[red, 2pt]{{1\over 2}(\cosh 4-1)}$$

解答:$$\mathbf F=[y, xz^3,-zy^3] \Rightarrow \nabla \times \mathbf F = \begin{vmatrix} \mathbf i& \mathbf j& \mathbf k\\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z}  \\ y& xz^3& -zy^3\end{vmatrix} =[-3zy^2-3xz^2,0,z^3-1] \\ \text{Using Stokes' Theorem, } \oint_C \mathbf F\cdot d\mathbf r = \iint_S (\nabla \times \mathbf F)\cdot \mathbf n\,dS = \iint_S [-3zy^2-3xz^2,0,z^3-1] \cdot [0,0,1]dS \\=\iint_S (z^3-1)d S =((-3)^3-1) \times Area(S) =-28\times 2^2 \pi = \bbox[red, 2pt]{-112\pi}$$

  解答:$$u(x,y) =X(x) Y(y) \Rightarrow \nabla^2 u = {\partial^2 u\over \partial x^2} +{\partial^2 u\over \partial y^2} = X''Y+ XY''=0 \Rightarrow {X''\over X} =-{Y''\over Y}=-\lambda \\ B.C.: \cases{u(0,y)=0 \\u(1,y)=0} \Rightarrow \cases{X(0)=0\\ X(1)=0} \\ \text{Solving for }X(x)\\ \textbf{Case I }\lambda=0 \Rightarrow \text{trivial solution }X=0\\ \textbf{Case II }\lambda \lt 0 \Rightarrow \text{trivial solution }X=0 \\\textbf{Case III }\lambda =k^2 \gt 0 \Rightarrow X=A\cos(kx)+B\sin(kx) \Rightarrow \cases{X(0)=A=0\\ X(1)=A\cos k+B\sin k=0} \\\qquad \Rightarrow \sin k=0 \Rightarrow k=n\pi \Rightarrow \lambda_n=n^2\pi^2 \Rightarrow X_n(x)= \sin (n\pi x), n=1,2,\dots\\ \text{Solving for }Y(y) \\ \lambda_n=n^2\pi^2 \Rightarrow Y''-n^2\pi^2 Y=0 \Rightarrow Y_n(y)= C_n\cosh(n\pi y)+D_n \sinh(n\pi y)\\ u(x,1)=0 \Rightarrow Y_n(1)=0 \Rightarrow C_n=0 \Rightarrow Y_n(y)= \sinh(n\pi(1-y)) \\ \Rightarrow u(x,y) = \sum_{n=1}^\infty B_n \sin(n\pi x) \sinh(n\pi(1-y)) \Rightarrow u(x,0)=2=\sum_{n=1}^\infty B_n \sin(n\pi x) \sinh(n\pi ) \\ \Rightarrow B_n \sinh(n \pi ) =2\int_0^1 2\sin(n\pi x)\,dx ={4\over n\pi}(1-(-1)^n) \Rightarrow B_n={4\over n\pi \sinh(n\pi)}(1-(-1)^n) \\ \Rightarrow \bbox[red, 2pt]{u(x,y) = \sum_{n=1}^\infty {4\over n\pi \sinh(n\pi)}(1-(-1)^n) \sin(n\pi x) \sinh(n\pi(1-y))}$$

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解題僅供參考，碩士班歷年試題及詳解