114年台大碩士班-工程數學E詳解

國立臺灣大學114學年度碩士班招生考試

題號:174
科目:工程數學(E)

解答: $$\textbf{(a) }y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x'''=m(m-1) (m-2)x^{m-3}\\ \Rightarrow x^3y'''+x^2y''-2xy'+2y= (m(m-1)(m-2)+ m(m-1)-2m+2)x^m=0 \\ \Rightarrow (m-2) (m-1)(m+1)=0 \Rightarrow m=2,1,-1 \Rightarrow y_h= c_1x^2+ c_2x+c_3x^{-1}\\ \text{Using variation of parameters, let} \cases{y_1 =x^2\\ y_2=x\\ y_3=1/x} \Rightarrow W=\begin{vmatrix} y_1& y_2& y_3\\ y_1'& y_2'& y_3'\\ y_1''& y_2''& y_3''\end{vmatrix} =-{6\over x} \\ \Rightarrow W_1= \begin{vmatrix} y_2& y_3\\ y_2'& y_3' \end{vmatrix} =-{2\over x},W_2= \begin{vmatrix} y_1& y_3\\ y_1'& y_3' \end{vmatrix} =-3, W_3= \begin{vmatrix} y_1& y_2\\ y_1'& y_2' \end{vmatrix} = -x^2\\ \Rightarrow R(x)={3\ln(x) \over W} =-{1\over 2}x\ln(x) \Rightarrow \cases{u_1'= RW_1=\ln x \\ u_2'=-RW_2= -{3\over 2}  x\ln x \\u_3'=RW_3= {1\over 2}x^3\ln x} \Rightarrow \cases{u_1=x\ln x-x \\u_2={3 \over 8}x^2-{3\over 4}x^2\ln x \\u_3= {1\over 8}x^4 \ln x-{1\over 32}x^4} \\ \Rightarrow y_p= u_1y_1+ u_2y_2 +u_3y_3 =x^3\ln x-x^3+{3\over 8}x^3-{3\over 4}x^3\ln x+{1\over 8}x^3\ln x-{1\over 32}x^3 \\\Rightarrow y_p =-{21\over 32}x^3 +{3\over 8}x^3\ln x \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{c_1x^2+ c_2x+c_3x^{-1}-{21\over 32}x^3 +{3\over 8}x^3\ln x} \\\textbf{(b) }y''-5y'+6y=0 \Rightarrow \lambda^2-5\lambda+6=0 \Rightarrow (\lambda-2)(\lambda-3)=0 \Rightarrow \lambda=2,3 \Rightarrow y_h=c_1e^{2x} +c_2e^{3x} \\ y_p=A\cos(3x)+ B\sin(3x) \Rightarrow y_p'=-3A\sin(3x)+ 3B\cos(3x) \Rightarrow y_p''=-9A\cos(3x)-9B\sin(3x) \\ \Rightarrow y_p''-5y_p'+6y_p= (-3A-15B) \cos(3x)+(15A-3B)\sin(3x)= 9\cos(3x) \\ \Rightarrow \cases{A+5B=-3\\ 5A-B=0} \Rightarrow \cases{A=-3/26\\ B=-15/26} \Rightarrow y_p=-{3\over 26}\cos(3x)-{15\over 26}\sin(3x) \Rightarrow y=y_h+ y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1e^{2x} +c_2e^{3x}-{3\over 26}\cos(3x)-{15\over 26}\sin(3x)} \\\textbf{(c) }4y''-12y'+9y=0 \Rightarrow 4\lambda^2-12\lambda+9=0 \Rightarrow  (2\lambda-3)^2=0 \Rightarrow \lambda={3\over 2} \Rightarrow y_h =c_1e^{1.5x} +c_2xe^{1.5x} \\ y_p=Ax^2e^{1.5x} \Rightarrow y_p'=2Axe^{1.5x}+{3\over 2}Ax^2e^{1.5x} \Rightarrow y_p''=2Ae^{1.5x}+ 6Axe^{1.5x}  +{9\over 4}Ax^2e^{1.5x} \\ \Rightarrow 4y_p''-12y_p'+9y_p =8Ae^{1.5x}= 2e^{1.5x} \Rightarrow A={1\over 4} \Rightarrow y_p={1\over 4}x^2e^{1.5x} \Rightarrow y=y_h+ y_p\\ \Rightarrow \bbox[red, 2pt]{y= c_1e^{1.5x} +c_2xe^{1.5x} +{1\over 4}x^2e^{1.5x}}$$

解答: $$\textbf{(a) }L\left\{ e^{-t}( 8\cosh(2t)-3\sinh(4t))\right\} =L\left\{ e^{-t}( 8\cdot {e^{2t}+e^{-2t}\over 2}-3\cdot {e^{4t}-e^{-4t}\over 2})\right\} \\= L\left\{ 4e^{t} +4e^{-3t}-{3\over 2}e^{3t}+ {3\over 2}e^{-5t}\right\} = \bbox[red, 2pt]{{4\over s-1}+{4\over s+4}-{3\over 2(s-3)} +{3\over 2(s+5)}} \\\textbf{(b) }L^{-1}\left\{  {2s-5\over s^2-6s+25} \right\} =L^{-1}\left\{  2\cdot {s-3\over (s-3)^2+4^2} +{1\over 4}\cdot {4\over (s-3)^2+4^2}\right\} \\\qquad = \bbox[red, 2pt]{e^{3t}(2\cos(4t)+{1\over 4}\sin(4t))} \\\textbf{(c) }L\{y''\}-2L\{y'\}-3L\{y\} =L\{u(t-1)\} \\\qquad \Rightarrow s^2Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))-3Y(s) ={e^{-s}\over s} \\\qquad \Rightarrow (s^2-2s-3)Y(s)+1={e^{-s}\over s} \Rightarrow Y(s)= {e^{-s}\over s(s^2-2s-3)} -{1\over s^2-2s-3} \\\qquad \Rightarrow y(t)= L^{-1}\{Y(s)\} =L^{-1}\left\{ {e^{-s}\over s(s^2-2s-3)}\right\} -L^{-1}\left\{ {1\over s^2-2s-3}\right\} \\\qquad =L^{-1} \left\{ -{e^{-s} \over 3s}+{e^{-s}\over 4(s+1)} +{e^{-s} \over 12(s-3)}\right\}-L^{-1}\left\{ {1\over 4(s-3)} -{1\over 4s(s+1)} \right\}\\\qquad \Rightarrow \bbox[red, 2pt]{y(t)=u(t-1)\left( -{1\over 3}+{1\over 4}e^{-(t-1)}+ {1\over 12}e^{3(t-1)}\right) -\left({1\over 4}e^{3t}-{1\over 4}e^{-t} \right)}$$  

$$\textbf{(d) }L\left\{ {\partial^2 u \over \partial x^2}\right\} =L\left\{ {\partial u \over \partial t}\right\} \Rightarrow {\partial^2  \over \partial x^2}U(x,s)=sU(x,s)-u(x,0) \Rightarrow {\partial^2  \over \partial x^2}U(x,s)-sU(x,s)=-25 \\\qquad \Rightarrow \cases{U_h(x,s)= c_1e^{\sqrt sx} +c_2e^{-\sqrt s x} \\ U_p(x,s)=25/s} \Rightarrow U(x,s)=U_h+U_p \Rightarrow U(x,s)=c_1e^{\sqrt sx} +c_2e^{-\sqrt s x}+{25\over s}\\ U(\infty,s) =\int_0^\infty e^{-st} u(\infty,t)\,dt =\int_0^\infty e^{-st} \cdot 25\,dt={25\over s} \Rightarrow c_1=0 \Rightarrow U(x,s) =c_2e^{-\sqrt s x}+{25\over s} \\ u(0,t)=0 \Rightarrow U(0,s) =0 \Rightarrow  c_2+{25\over s}=0 \Rightarrow c_2=-{25\over s} \Rightarrow U(x,s)= -{25\over s}e^{-\sqrt sx} +{25\over s}\\ u(x,t)= L^{-1} \left\{ -{25\over s}e^{-\sqrt sx} +{25\over s} \right\} \Rightarrow \bbox[red, 2pt]{u(x,t)=-25 \text{erfc}({x\over 2\sqrt t})+25}$$

解答: $$3xy''+y'-y=0 \Rightarrow y''+{1\over 3x}y'-{1\over 3x}y=0 \\\Rightarrow \cases{P(x)={1\over 3x} \\Q(x) =-{1\over 3x}} \Rightarrow \cases{xP(x) ={1\over 3} \text{ is analytic at }x=0\\ x^2 Q(x)=-{x\over 3}  \text{ is analytic at }x=0} \\ \text{Set }y=x^r \sum_{n=0}^\infty c_nx^n = \sum_{n=0}^\infty c_nx^{n+r} \Rightarrow y'=\sum_{n=0}^\infty c_n(n+r)x^{n+r-1}  \\\Rightarrow y'' =\sum_{n=0}^\infty c_n(n+r)(n+r-1)x^{n+r-2} \Rightarrow y''+{1\over 3x}y'-{1\over 3x}y=0\\\Rightarrow  \sum_{n=0}^\infty c_n(n+r)(n+r-1)x^{n+r-2}+{1\over 3x} \sum_{n=0}^\infty c_n(n+r)x^{n+r-1}-{1\over 3x}\sum_{n=0}^\infty c_nx^{n+r} =0 \\ \Rightarrow x^r \left[ \sum_{n=0}^\infty 3c_n(n+r) (n+r-1)x^{n-1} +\sum_{n=0}^\infty c_n (n+r)x^{n-1} -\sum_{n=0}^\infty  c_n x^n\right]=0 \\ \Rightarrow x^r\left[c_0r(3r-2)+ \sum_{n=0}^\infty (c_{n+1}(n+r+1)(3n+3r+1)+c_n)x^n \right]=0 \\ \Rightarrow c_0r(3r-2)=0 \Rightarrow r=0,{2\over 3}\\ \textbf{Case I }r=0 \Rightarrow c_{n+1}(n+0+1)(3n+0+1) +c_n=0 \Rightarrow c_{n+1} =-{c_n\over (n+1)(3n+1)} \\\qquad \Rightarrow c_1=-c_0 \Rightarrow c_2=-{c_1\over 2\cdot 4}={c_0\over 8} \Rightarrow c_3= -{c_2\over 3\cdot 7} =-{c_0\over 168} \Rightarrow c_4=-{c_3\over 4\cdot 10}={1\over 6720}c_0 \\ \Rightarrow y=c_0(1- x+{1\over 8}x^2-{1\over 168}x^3+{1\over 6720}x^4+\cdots) \\ \textbf{Case II }r={2\over 3} \Rightarrow c_{n+1}(n+{5\over 3}) (3n+3) +c_n=0 \Rightarrow c_{n+1} =-{c_n\over (n+1)(3n+5)} \\ \qquad \Rightarrow c_1=-{c_0\over 5} \Rightarrow c_2=-{c_1\over 16}={c_0\over 80} \Rightarrow c_3=-{c_2\over 33}= -{c_0\over 2640} \Rightarrow c_4=-{c_3\over 56} ={c_0\over 147840} \\ \qquad \Rightarrow y=c_0x^{2/3}(1-{1\over 5}x+{1\over 80}x^2-{1\over 2640}x^3+ {1\over 147840}x^4+ \cdots) \\\qquad \Rightarrow y=c_0(x^{2/3}-{1\over 5}x^{5/3}+{1\over 80}x^{8/3}-{1\over 2640}x^{11 /3}+ {1\over 147840}x^{14/3}+ \cdots) \\ \Rightarrow \text{General solution is }\\\bbox[red, 2pt]{y=A(1- x+{1\over 8}x^2-{1\over 168}x^3+{1\over 6720}x^4+\cdots) +}\\\qquad \qquad \bbox[red, 2pt]{B(x^{2/3}-{1\over 5}x^{5/3}+{1\over 80}x^{8/3}-{1\over 2640}x^{11/3}+ {1\over 147840}x^{14/3}+ \cdots)}$$

解答: $$\langle1,\cos(mx) \rangle =\int_{-\pi}^\pi \cos(mx)\,dx= \left. \left[ {1\over m}\sin(mx)\right] \right|_{-\pi}^\pi =0, m=1,2,3\\ \langle1,\sin(mx) \rangle =\int_{-\pi}^\pi \sin(mx)\,dx= \left. \left[ -{1\over m}\cos(mx)\right] \right|_{-\pi}^\pi =0, m=1,2,3 \\ \langle \cos(mx), \sin(nx) \rangle =\int_{-\pi}^\pi \cos(mx) \sin(nx)\,dx = \left. \left[ {\cos((m-n)x) \over 2(m-n)}-{ \cos((m+n))x\over 2(m+n)} \right] \right|_{-\pi}^\pi=0, m\ne n  \\ \langle \cos(mx), \sin(mx) \rangle =\int_{-\pi}^\pi \cos(mx)\sin(mx)\,dx ={1\over 2} \int_{-\pi}^\pi \sin(2mx)\,dx =0, m=n\\\Rightarrow \bbox[red, 2pt]{\text{Yes, it forms an orthogonal set}} \\ \langle 1,1\rangle =\int_{-\pi}^\pi 1\,dx =2\pi \Rightarrow 1\to {1\over \sqrt{2\pi}} \\ \langle \cos(mx), \cos(mx) \rangle =\int_{-\pi}^\pi \cos^2(mx)\,dx ={1\over 2}\int_{-\pi}^\pi (\cos(2mx)+1)\,dx =\pi \Rightarrow \cos(mx) \to {1\over \sqrt \pi} \cos(mx) \\ \langle \sin(mx), \sin(mx)\rangle =\int_{-\pi}^\pi \sin^2(mx) \,dx ={1\over 2}\int_{-\pi}^\pi (1-\cos(2mx))\,dx =\pi \Rightarrow \sin(mx) \to {1\over \sqrt \pi}\sin(mx) \\ \Rightarrow \text{ orthonomal set =} \bbox[red, 2pt]{\{{1\over \sqrt{2\pi}}, {1\over \sqrt \pi} \cos(mx), {1\over \sqrt{\pi}} \sin(mx)\}},m=1,2,3$$

解答: $$f(x)=x^3 \Rightarrow f(-x)=-f(x) \Rightarrow f(x)\text{ is odd }\Rightarrow a_n=0\\ b_n={1\over \pi} \int_{-\pi}^\pi x^3\sin(nx)\,dx =(-1)^n({12\over n^3}-{2\pi^2\over n}) \\ \Rightarrow f(x)= \bbox[red, 2pt]{\sum_{n=1}^\infty (-1)^n({12\over n^3}-{2\pi^2\over n}) \sin(nx)}$$

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解題僅供參考，碩士班歷年試題及詳解