114年中山大學材光碩士班-工程數學詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：工程數學【材光系碩士班選考、材料前瞻應材碩士班選考、材光聯合碩士班選考】

解答:$$y'={1\over x}y^2+{1\over x}y-{2\over x} \Rightarrow xy'= y^2+y-2 \Rightarrow {1\over y^2+y-2}\, dy={1\over x}\,dx \\ \Rightarrow \int {1\over 3}\left({1\over y-1}-{1\over y+2} \right) \,dy = \int {1\over x}\,dx \Rightarrow {1\over 3}\ln{y-1\over y+2}= \ln x+c_1 =\ln c_2x \\ \Rightarrow {y-1\over y-2}=c_3x^3 \Rightarrow {1\over y-2}=c_3x^3-1 \Rightarrow \bbox[red, 2pt]{y={1\over c_3x^3-1}+2}$$

解答:$$y''+2y'+5y=0 \Rightarrow \lambda^2+2\lambda+5=0 \Rightarrow \lambda =-1\pm 2i \Rightarrow y_h= e^{-x} (c_1 \cos 2x +c_2 \sin 2x) \\ y_p= Ae^{0.5x} +B \cos 4x+ C\sin 4x \Rightarrow y_p'=0.5Ae^{0.5x}-4B\sin 4x +4C\cos 4x \\ \Rightarrow y_p'' =0.25 Ae^{0.5x}-16B\cos 4x-16C\sin 4x  \Rightarrow y_p''+ 2y_p'+5y_p \\= 6.25A e^{0.5x} +(-11B+8C) \cos 4x+(-8B-11C) \sin 4x = 1.25 e^{0.5x} +40 \cos 4x-55 \sin 4x \\ \Rightarrow \cases{6.25A=1.25\\ -11B+8C= 40\\ -8B-11C=-55} \Rightarrow \cases{A=0.2\\ B=0\\ C=5} \Rightarrow y_p= 0.2 e^{0.5x}+ 5\sin 4x \Rightarrow y=y_h+ y_p \\ \Rightarrow \bbox[red, 2pt]{y= e^{-x} (c_1 \cos 2x +c_2 \sin 2x) + 0.2e^{0.5x}+ 5\sin 4x }$$

解答:$$y=\sum_{n=0}^\infty a_n x^n \Rightarrow \cases{xy= \sum_{n=0}^\infty a_n x^{n+1} \\ y'=\sum_{n=0}^\infty na_n x^{n-1}} \Rightarrow \cases{(x+4)y= \sum_{n=0}^\infty (a_nx^{n+1} +4a_n x^n) \\ 5xy'=\sum_{n=0}^\infty 5na_n x^n \\ y''=\sum_{n=0}^\infty n(n-1)a_n x^{n-2}} \\ \Rightarrow \cases{5xy'+(x+4)y = \sum_{n=0}^\infty a_n x^{n+1}+ (4a_n+ 5na_n)x^n \\ xy''= \sum_{n=0}^\infty n(n-1)a_n x^{n-1}} \\ \Rightarrow xy''+5xy'+(x+ 4)y = \sum_{n=0}^\infty (a_n x^{n+1}+ (4a_n+ 5na_n)x^n +n(n-1)a_nx^{n-1}) \\= \sum_{n=0}^\infty (a_{n-1} +(5n+4)a_n +(n^2+n) a_{n+1})x^n =0, a_{-1}=0 \\ \Rightarrow \cases{4a_0=0\\ a_0+9a_1+ 2a_2 =0 \\ a_1+14a_2+6a_3=0\\ \cdots \\ a_{n-1} +(5n+4)a_n +(n^2+n) a_{n+1}=0} \Rightarrow \cases{a_0=0\\ a_2=-{9\over 2}a_1\\ a_3= {31\over 3}a_1\\ \cdots} \\ \Rightarrow \bbox[red, 2pt]{y =a_1x-{9\over 2}a_1x^2 +{31\over 3}a_1x^3+ \cdots}$$

解答:$$z=x^\alpha \Rightarrow{dz\over dx} =\alpha x^{\alpha-1} \Rightarrow {d^2 z\over dx^2} =\alpha(\alpha-1)x^{\alpha-2} \\ \Rightarrow {dy\over dx}={dy\over dz}\cdot {dz\over dx} =\alpha x^{\alpha-1} y_z \Rightarrow {d^2 y\over dx^2}=\alpha(\alpha-1)x^{\alpha-2} y_z+ \alpha^2x^{2\alpha-2} y_{zz} \\ 原式: x^2y''+xy'+(\tau^2x^{2\alpha}-\omega^2)y=0 \Rightarrow \alpha(\alpha-1)x^{\alpha } y_z+ \alpha^2x^{2\alpha } y_{zz}+ \alpha x^\alpha y_z +(\tau^2x^{2\alpha}-\omega^2)y=0 \\ \Rightarrow \alpha^2 x^{2\alpha}y_{zz}+ \alpha^2x^\alpha y_z+(\tau^2x^{2\alpha}-\omega^2)y=0 \Rightarrow x^{2\alpha} y_{zz}+ x^\alpha y_z+ \left({\tau^2\over \alpha^2}x^{2\alpha}-{\omega^2\over \alpha^2} \right) y=0 \\ \Rightarrow z^2y_{zz}+ zy_z+\left({\tau^2\over \alpha^2}z^{2 }-{\omega^2\over \alpha^2} \right) y=0 \\ \Rightarrow \bbox[red, 2pt]{y=AJ_{\omega/\alpha} \left({\tau\over \alpha} x^{\alpha} \right) +BY_{\omega/\alpha} \left({\tau\over \alpha}x^\alpha \right)}, \\\quad \text{ where }J \text{ and }Y \text{ are Bessel function of the first and second kind, respectively.}$$

解答:$$L\{f(t) \} =\int_3^\infty te^{-st}\,dt =\left. \left[ -{t\over s}e^{-st}-{1\over s^2}e^{-st}  \right] \right|_3^\infty ={1\over s^2}e^{-3s}+ {3\over s}e^{-3s}\\ \Rightarrow L\{y'' \}+ 4L\{y\} =L\{f(t)\} \Rightarrow s^2Y(s)+4Y(s)= {1\over s^2}e^{-3s}+{3\over s}e^{-3s} \\ \Rightarrow Y(s)={1\over s^2(s^2+4)}e^{-3s} + {3\over s(s^2+4)}e^{-3s} \Rightarrow y(t)= L^{-1}\{Y(s) \} \\=L^{-1} \left\{ {1\over s^2(s^2+4)}e^{-3s}+ {3\over s(s^2+4)}e^{-3s}\right\} \\=L^{-1} \left\{ \left({1\over 4s^2 }-{1\over 4(s^2+1)} \right)e^{-3s}+ \left({3\over 4s }-{3s\over 4(s^2+4)} \right)e^{-3s}\right\}\\ =u(t-3)({t-3\over 4} -{1\over 8} \sin(2(t-3))+ {3\over 4}-{3\over 4} \cos(2(t-3))) \\ \Rightarrow \bbox[red, 2pt]{y(t) =u(t-3)\left( {t\over 4}-{1\over 8}\sin(2t-6)-{3\over 4}\cos(2t-6)\right)}$$

解答:$$C_n= {1\over 2\pi} \int_{-\pi}^\pi e^x e^{-in x}\,dx = {1\over 2\pi} \int_{-\pi}^\pi e^{(1-in) x}\,dx = {1\over 2\pi} \left. \left[ {1\over 1-in}e^{(1-in)x} \right] \right|_{-\pi}^\pi  ={e^\pi-e^{-\pi}\over 2\pi(1-in)} (-1)^n  \\ \Rightarrow \bbox[red, 2pt]{f(x)= \sum_{n=-\infty}^\infty {e^\pi-e^{-\pi}\over 2\pi(1-in)} (-1)^n e^{inx}}$$

解答:$$u(x,t) =X(x)T(t) \Rightarrow XT'=c^2X''T \Rightarrow {T'\over c^2T} ={X'' \over X} =k\\ \text{Ends are insulated, we can suppose that }\cases{X'(0)T(t) =0\\ X'(L)T(t)=0} \Rightarrow \cases{X'(0)=0\\ X'(L)=0 } \\ \textbf{Case I }k=0 \Rightarrow X=c_1x +c2 \Rightarrow X'=c_1 =0 \Rightarrow X=c_2\\ \textbf{Case II }k=\rho^2 \gt 0 \Rightarrow X''-\rho^2X=9 \Rightarrow X=c_3e^{\rho x}+c_4e^{-\rho x} \Rightarrow X'=c_3\rho e^{\rho x}-c_4\rho e^{-\rho x} \\\qquad \cases{X'(0) =\rho(c_3-c_4)=0\\ X'(L)=c_3\rho e^{\rho L}-c_4\rho e^{-\rho L} =0} \Rightarrow c_3\rho(e^{2\rho L}-1)=0 \Rightarrow c_3=0 \Rightarrow c_4=0 \Rightarrow X=0\\ \text{Case III }k=-\rho^2 \lt 0 \Rightarrow X''+\rho^2 X= 0 \Rightarrow X=c_3 \cos(\rho x)+ c_4\sin (\rho x)\\\qquad \Rightarrow X'=-c_3\rho \sin(\rho x)+ c_4 \rho \cos(\rho x) \Rightarrow \cases{X'(0)= c_4\rho =0\\ X'(L)= -c_3\rho \sin(\rho L)+c_4 \rho \cos(\rho L)=0} \\\qquad \Rightarrow c_4=0 \Rightarrow \sin(\rho L)=0 \Rightarrow \rho L=n\pi \Rightarrow \rho ={n\pi\over L }\Rightarrow X_n =c_3 \cos{n \pi x\over L},n=1,2,\dots \\ k=-\rho^2 \Rightarrow T'=-c^2\rho^2 T \Rightarrow T=c_5e^{-c^2\rho^2 t} \Rightarrow T_n=c_5e^{-n^2 c^2\pi^2t /L^2}  \\ \Rightarrow u_n(x,t)= X_nT_n \Rightarrow \bbox[red, 2pt]{u(x,t)= a_0+ \sum_{n=1}^\infty a_n\cos{n\pi x\over L}e^{-n^2 c^2 \pi^2t/L^2}}$$

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