114年台科機械碩士班甲組-工程數學詳解

國立臺灣科技大學114學年度碩士班招生試題

系所組別:機械工程系碩士班甲組
科目:工程數學

解答: $$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-3xy'+3y=m(m-1)x^m-3mx^m+3x^m=0 \\ \Rightarrow (m^2-4m+3)x^m=0 \Rightarrow (m-3)(m-1)=0 \Rightarrow m=1,3 \Rightarrow y_h= c_1x+ c_2x^3 \\  \cases{y_1=x\\ y_2=x^3} \Rightarrow W=\begin{vmatrix} y_1& y_2\\ y_1'& y_2'\end{vmatrix} =2x^3  \Rightarrow \text{By variation of parameters,}\\ y_p=-x\int{x^3\cdot 2x^2e^x \over 2x^3}\,dx + x^3\int {x\cdot 2x^2e^x \over 2x^3}\,dx = -x \int x^2e^x\,dx +x^3 \int e^x\,dx \\=-x(x^2-2x+2)e^x +x^3e^x = (2x^2-2x)e^x \\ \Rightarrow y= y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1x+ c_2x^3+ (2x^2-2x)e^x}$$

解答: $$\cases{P(x,y)=x+2y\\ Q(x,y)= 2x} \Rightarrow P_y =2=Q_x \Rightarrow \text{ Exact} \\ \Rightarrow \Phi(x,y) =\int (x+2y)\,dx = \int 2x\,dy \Rightarrow \Phi(x,y)={1\over 2}x^2+ 2xy+ \phi(y)= 2xy+ \rho(x) \\ \Rightarrow \Phi(x,y)={1\over 2}x^2+2xy =c_1 \Rightarrow y= {c_1\over 2x}-{1\over 4}x \Rightarrow y(1)={c_1\over 2}-{1\over 4}=0 \Rightarrow c_1={1\over 2} \\ \Rightarrow \bbox[red, 2pt]{y= {1\over 4x}-{x\over 4}}$$

解答: $$\textbf{(1) }L^{-1}\left\{{6s+3\over s^2+4} \right\} = L^{-1}\left\{6\cdot {s\over s^2+ 2^2} +{3\over 2} \cdot {2\over s^2+2^2}\right\} = \bbox[red, 2pt]{6\cos(2t)+{3\over 2}\sin(2t)} \\\textbf{(2) }L\{e^{2t} * \sin(t)\} =L\{e^{2t}\} \cdot L\{ \sin(t)\} ={1\over s-2}\cdot {1\over s^2+1^2} = \bbox[red, 2pt]{1\over (s-2)(s^2+1)}$$

解答: $$\textbf{(1) }\vec u=2\vec i+\vec k \Rightarrow \vec e={\vec u\over |\vec u|} ={2\over \sqrt 5}\vec i+{1\over \sqrt 5} \vec k \\ D_\vec ef(x,y,z)= \nabla f\cdot \vec e =(3x^2,2yz, y^2) \cdot ({2\over \sqrt 5}, 0,{1\over \sqrt 5}) ={1\over \sqrt 5}(6x^2+y^2) \\ \Rightarrow D_{\vec e}f(P) =D_{\vec e}f(1,1,1) ={7\over \sqrt 5} = \bbox[red, 2pt] {{7\over 5}\sqrt 5} \\\textbf{(2) }\cases{x(t)=t\\ y(t) =1\\ z(t)=2+5t},0\le t\le 1 \Rightarrow \cases{x'(t)=1\\ y'(t)=0\\ z'(t)=5} \\\quad \Rightarrow I= \int_C (3x^2dx +2yzdy+ y^2dz) =\int_0^1 (3t^2,4+10t,1)\cdot (1,0,5)\,dt = \int_0^1(3t^2+5)\,dt \\\qquad =\left. \left[ t^3+5t \right] \right|_0^1 =\bbox[red, 2pt]6$$

解答: $$\int_{-\infty}^\infty {x^2\over (x^2+a^2)(x^2+b^2)}\,dx =\int_{-\infty}^\infty {z^2\over (z^2+a^2)(z^2+b^2)}\,dz  =\int_{-\infty}^\infty {z^2\over (z+ai)(z-ai)(z+bi)(z-bi)}\,dz \\=2\pi i\left(\left. {z^2\over (z+ai)(z^2+b^2)} \right|_{z=ai}+ \left. {z^2\over (z^2+a^2)(z+bi)} \right|_{z=bi} \right) =2\pi i\left( {-a^2\over 2ai(b^2-a^2)} +{-b^2\over (a^2-b^2)(2bi)}\right) \\= \pi\left({-a\over b^2-a^2} +{-b\over a^2-b^2}\right) =\pi \cdot {a-b\over a^2-b^2} =\bbox[red, 2pt]{\pi\over a+b}$$

解答: $$\textbf{(1) }\phi(x) \text{ is a solution} \Rightarrow 0=\phi''+2 \Rightarrow \phi''(x)=-2 \Rightarrow \phi'(x)=-2x+c_1 \Rightarrow \phi(x)=-x^2+c_1x+c_2\\ \cases{\theta(1,t)=0\\ {\partial \theta(0,t) \over \partial x} =0} \Rightarrow \cases{\phi(1)=0\\ \phi'(0)=0} \Rightarrow \cases{-1+ c_1+c_2=0\\ c_1=0} \Rightarrow \cases{c_1=0\\ c_2=1} \Rightarrow \bbox[red, 2pt]{\phi(x)=-x^2+1}\\\textbf{(2) } \varphi(x,t)=X(x)T(t) \Rightarrow XT'=X''T \Rightarrow {T'\over T}={X'' \over X}=\lambda= -k^2 \lt 0\\ \Rightarrow X''+k^2X=0 \Rightarrow X=c_1 \cos(kx)+c_2 \sin(kx) \Rightarrow X'=-c_1k \sin(kx) +c_2k \cos(kx) \\\Rightarrow \cases{\theta(1,t)=0\\ {\partial \theta(0,t) \over \partial x} =0} \Rightarrow \cases{X(1)=0\\ X'(0)=0}  \Rightarrow \cases{c_1 \cos(k)+ c_2\sin(k) =0\\c_2k=0} \Rightarrow \cases{c_2 = 0\\ \cos(k)=0} \\\Rightarrow k={(2n-1)\pi\over 2 }\Rightarrow X_n=A_n \cos{(2n-1)\pi x\over 2},n=1,2,\dots\\ T'=-k^2T \Rightarrow   T_n= B_ne^{-k^2 t} \Rightarrow \varphi(x,t)= \sum_{n=1}^\infty C_ne^{-(2n-1)^2\pi^2 t/4} \cos{(2n-1)\pi x\over 2} \\ \Rightarrow \theta(x,t) =\varphi(x,t)+ \phi(x)= \sum_{n=1}^\infty C_ne^{-(2n-1)^2\pi^2 t/4} \cos{(2n-1)\pi x\over 2}-x^2+1 \\ \Rightarrow \text{Initial condition: } \theta(x,0) =0=\sum_{n=1}^\infty C_n  \cos {(2n-1)\pi x\over 2}-x^2+1 \Rightarrow x^2-1= \sum_{n=1}^\infty C_n  \cos{(2n-1)\pi x\over 2} \\\Rightarrow C_n= 2\int_0^1(x^2-1) \cos{(2n-1)\pi x\over 2}\,dx =2\cdot {16(-1)^n \over (2n-1)^3 \pi^3} ={32 \over (2n-1)^3\pi^3}(-1)^n \\ \Rightarrow \bbox[red, 2pt]{\varphi(x,t)= \sum_{n=1}^\infty {32\over (2n-1)^3 \pi^3}(-1)^ne^{-(2n-1)^2\pi^2t/4} \cos{(2n-1)\pi x\over 2}}$$

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解題僅供參考，碩士班歷年試題及詳解