113年中山材料碩士班-工程數學詳解

國立中山大學113學年度碩士班招生考試

科目名稱:工程數學【材料系碩士班】

解答: $$\cot x\,dy+y\,dx=0 \Rightarrow {1\over y}dy=-\tan x\,dx \Rightarrow \ln y=\ln(\cos x)+c_1\\ y(0)=1 \Rightarrow \ln 1=\ln 1+c_1 \Rightarrow c_1=0 \Rightarrow \bbox[red, 2pt]{y=\cos x}$$

解答: $$I=e^{\int P\,dx}  \Rightarrow I'=Pe^{\int P\,dx}  = PI \Rightarrow I'=PI\\y={\int IQ\,dx +c\over I} \Rightarrow I y=\int IQ\,dx +c \Rightarrow I'y+Iy'= IQ \Rightarrow PI y+Iy'=IQ \Rightarrow Py+y'=Q \\ \Rightarrow {dy\over dx}+ Py=Q. \bbox[red, 2pt]{QED}$$

解答: $$xy'+3y={\cos x\over x^2} \Rightarrow x^3y'+3x^2y=\cos x \Rightarrow (x^3y)'=\cos x \Rightarrow x^3y=\sin x+c_1\\  \Rightarrow \bbox[red, 2pt]{y={\sin x\over x^3}+ {c_1\over x^3}}, 初始值y(0)=0有疑義$$

解答: $$x^2y^4+y+3xy'=0 \Rightarrow y'+{y\over 3x}=-{x\over 3}y^4 \\ \text{Let }v={1\over y^3} \Rightarrow v'=-{3\over y^4}y' \Rightarrow y'=-{1\over 3}y^4v' \Rightarrow -{1\over 3}y^4v'+{y\over 3x}=-{x\over 3}y^4 \\ \Rightarrow v'-{v\over x}=x \Rightarrow {1\over x}v'-{v\over x^2}=1 \Rightarrow ({v\over x})'=1 \Rightarrow {v\over x}=x+c_1 \Rightarrow v={1\over y^3}=x^2+c_1x \\ \Rightarrow \bbox[red, 2pt]{y=\sqrt[3]{1\over x^2+c_1x}}, 初始值y(0)=0有疑義$$

解答: $$2y''+6y'-20y=0 \Rightarrow \lambda^2+3\lambda-10=0 \Rightarrow (\lambda+5)(\lambda-2)=0 \Rightarrow \lambda=2,-5\\ \Rightarrow y_h=c_1e^{2x}+ c_2e^{-5x}\\ y_p=Ae^{5x} \Rightarrow y_p'=5Ae^{5x} \Rightarrow y_p''=25Ae^{5x} \Rightarrow y_p''+3y_p'-10y_p=30Ae^{5x}=60e^{5x} \\ \Rightarrow A=2 \Rightarrow y_p=2e^{5x} \Rightarrow y=y_h+y_p =c_1e^{2x}+ c_2e^{-5x}+2e^{5x} \\ \Rightarrow y'=2c_1e^{2x}-5c_2e^{-5x}+10e^{5x} \Rightarrow \cases{y(0)=c_1+c_2 + 2=7\\ y'(0)=2c_1-5c_2+10= -1} \\ \Rightarrow \cases{c_1=2\\ c_2=3} \Rightarrow \bbox[red, 2pt]{y=2e^{2x}+3e^{-5x}+ 2e^{5x}}$$

解答: $$y''-4y'+8y=4e^{2x} \Rightarrow \lambda^2-4\lambda+8=0 \Rightarrow \lambda ={2\pm 2i  } \Rightarrow y_h=e^{2x}(c_1\cos(2x)+ c_2\sin(2x))\\ \text{Let }\cases{y_1=e^{2x} \cos(2x)\\ y_2=e^{2x}\sin (2x)}, \text{ then }W =\begin{vmatrix}y_1 & y_2 \\y_1' & y_2' \end{vmatrix} =2e^{4x}\\\text{ By variations of parameters,} \\y_p = -e^{2x}\cos(2x) \int{ e^{2x} \sin(2x) \cdot 4e^{2x} \sec(2x)\over 2e^{4x}} \,dx+e^{2x} \sin(2x) \int {e^{2x}\cos(2x) \cdot 4e^{2x} \sec(2x) \over 2e^{4x}}\,dx \\\qquad =-2e^{2x}\cos(2x) \int \tan(2x)\,dx +2e^{2x}\sin(2x) \int 1\,dx\\\qquad  =e^{2x}\cos(2x) \ln(\cos(2x)) +2xe^{2x} \sin(2x) \\ \Rightarrow y=y_h+y_p = e^{2x}(c_1\cos(2x)+ c_2\sin(2x))+e^{2x}\cos(2x) \ln(\cos(2x)) +2xe^{2x} \sin(2x) \\ \Rightarrow y'=e^{2x}((2c_1+2c_2)\cos(2x) +(2c_2-2c_1)\sin(2x))+ 2e^{2x}( \cos(2x)-\sin(2x))\ln(\cos(2x))\\\qquad \qquad  +4xe^{2x}( \sin(2x)+\cos(2x)) \\ \Rightarrow \cases{y(0)=c_1=0\\ y'(0)=2c_1+2c_2=0} \Rightarrow c_2=0 \\ \Rightarrow \bbox[red, 2pt]{y=e^{2x}\cos(2x) \ln(\cos(2x)) +2xe^{2x} \sin(2x)}$$

解答: $$\cases{x'=2x-4y\\ y'=x-3y} \Rightarrow \begin{bmatrix}x' \\ y' \end{bmatrix} = \begin{bmatrix}2& -4 \\ 1& -3 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} \\ A=\begin{bmatrix}2& -4 \\ 1& -3 \end{bmatrix} = \begin{bmatrix}4 & 1 \\1 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix}\frac{1}{3} & -\frac{1}{3} \\-\frac{1}{3} & \frac{4}{3} \end{bmatrix} \\ \Rightarrow e^{At} =\begin{bmatrix}2& -4 \\ 1& -3 \end{bmatrix} = \begin{bmatrix}4 & 1 \\1 & 1 \end{bmatrix} \begin{bmatrix}e^t & 0 \\ 0 & e^{-2t} \end{bmatrix} \begin{bmatrix} \frac{1}{3} & -\frac{1}{3} \\-\frac{1}{3} & \frac{4}{3} \end{bmatrix} = \begin{bmatrix}{4\over 3}e^t-{1\over 3}e^{-2t} & -{4\over 3}e^t+{4\over 3}e^{-2t} \\{1\over 3} e^t-{1\over 3}e^{-2t} & -{1\over 3}e^t+{4\over 3}e^{-2t} \end{bmatrix} \\ \Rightarrow  \begin{bmatrix}x \\ y \end{bmatrix} =e^{At}  \begin{bmatrix}c_1 \\ c_2 \end{bmatrix} \Rightarrow \begin{bmatrix}c_1 \\ c_2 \end{bmatrix} =\begin{bmatrix}x(0)=9 \\ y(0)=3 \end{bmatrix} \Rightarrow \begin{bmatrix}x \\ y \end{bmatrix} =e^{At}  \begin{bmatrix} 9 \\ 3 \end{bmatrix} =\begin{bmatrix} 8e^t+e^{-2t}  \\2e^t+e^{-2t} \end{bmatrix}\\ \Rightarrow  \bbox[red, 2pt]{\cases{x= 8e^t+e^{-2t}\\y =2e^t+e^{-2t}}}$$

解答: $$\textbf{(A)}\; u=t-x \Rightarrow du=-dx \Rightarrow \int_0^t f(x)g(t-x)\,dx = \int_t^0 f(t-u)g(u)(-du) \\\qquad =\int_0^t f(t-u)g(u)\,du = \int_0^t f(t-x)g(x)\,dx. \bbox[red, 2pt]{QED} \\\textbf{(B)}\;L\{f*g\} =\int_0^\infty \left( \int_0^t f(u)g(t-u)du\right)e^{-st}\,dt \\\qquad =\int_0^\infty \left( \int_0^\infty f(u)g(t-u)du\right)e^{-st}\,dt \text{ (assume that }f(t)=g(t)=0, t\lt 0) \\\qquad = \int_0^\infty f(u) \left( \int_0^\infty g(t-u)e^{-st}\,dt \right)\,du \\\qquad = \int_0^\infty f(u) \left( \int_0^\infty g(\tau)e^{-s(\tau+u)}\,d\tau \right)\,du \;(\tau=t-u) \\\qquad =\int_0^\infty f(u)e^{-su} \left( \int_0^\infty g(\tau)e^{-s\tau}\,d\tau \right)\,du \\\qquad =\left( \int_0^\infty f(u)e^{-su}\,du \right) \left( \int_0^\infty g(\tau)e^{-s\tau } \,d\tau \right) \\\qquad =F(s)G(s). \bbox[red, 2pt]{\text{QED}} \\\textbf{(C)}\; \cases{f(x)=e^{-\alpha x} \\g(x)=\sin(\omega x)} \Rightarrow L\left\{ \int_0^t e^{-\alpha x} \sin[\omega(t-x)]\,dx \right\} =L\{f(x)\} L\{g(x)\} ={1\over s+\omega}\cdot {\omega\over s^2+\omega^2} \\\qquad = \bbox[red, 2pt]{{\omega \over (s+\omega)(s^2+\omega^2)}}$$

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