112年台師大機電碩士班-工程數學詳解

國立臺灣師範大學112學年度碩士班招生考試試題

科目:工程數學
適用系所:機電工程學系

解答:$$\textbf{(a) }y''+y'-2y=0 \Rightarrow \lambda^2+\lambda-2=0 \Rightarrow (\lambda+2)(\lambda-1)=0 \Rightarrow \lambda=1,-2 \Rightarrow y_h=c_1e^x+ c_2e^{-2x}\\\qquad y_p=Axe^x \Rightarrow y_p'=Ae^x +Axe^x \Rightarrow y_p''=2Ae^x +Axe^x \Rightarrow y_p''+y_p'-2y_p= 3Ae^x=e^x\\ \qquad \Rightarrow A={1\over 3} \Rightarrow y_p={1\over 3}xe^x \Rightarrow  y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^x+ c_2e^{-2x}+ {1\over 3}xe^x} \\ \textbf{(b) }y''+y'-2y=0 \text{ is unstable, because }-2\lt 0 \Rightarrow y''+y'-2y=e^x \bbox[red, 2pt]{\text{ is unstable}}$$

解答:$$\textbf{(a) }\det(-4A) =(-4)^3 \det(A)=-64\times5=\bbox[red, 2pt]{-320} \\ \textbf{(b)} \det(AA^{-1}) =\det(A)\cdot \det(A^{-1})= \det(I)=1 \Rightarrow \det(A^{-1}) ={1\over  \det(A)} =\bbox[red, 2pt]{1\over 5} \\\textbf{(c) } \det(A^2) =\det(AA) =\det(A)\cdot \det(A)=5^2= \bbox[red, 2pt]{25} \\\textbf{(d) }\det((3A^{-1})^T) =\det(3A^{-1}) =3^3 \det(A^{-1}) = \bbox[red, 2pt] {27\over 5} \\\textbf{(e)} \det \begin{bmatrix}t & r & s \\w & u & v \\z &  x& y\end{bmatrix} =-\det\begin{bmatrix}s & r & t \\v & u & w \\y &  x& z\end{bmatrix} =\det\begin{bmatrix}r & s & t \\u & v & w \\x &  y& z\end{bmatrix} = \bbox[red, 2pt]5$$

解答:$$L\{y''\}+3L\{y'\}+ 2L\{y\} =L\{u(t-1)\}+  L\{\delta(t-2)\} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0) +3(sY(s)-y(0)) +2Y(s) ={e^{-s}\over s}+ e^{-2s} \\ \Rightarrow (s^2+3s+2)Y(s)-1 ={e^{-s}\over s}+ e^{-2s} \Rightarrow Y(s)= {e^{-s}\over s(s^2+3s+2)}+ {e^{-2s}\over s^2+3s+2}+{1\over s^2+3s+2} \\ \Rightarrow y(t) =L^{-1}\{Y(s)\} =L^{-1}\left\{ {e^{-s}\over s(s^2+3s+2)}+ {e^{-2s}\over s^2+3s+2}+{1\over s^2+3s+2}\right\}\\= L^{-1} \left\{e^{-s}{\left( -{1\over s+1}+ {1\over 2(s+2)}+ {1\over 2s}\right)} \right\}+ L^{-1} \left\{ e^{-2s}\left({1\over s+1}-{1\over s+2} \right) \right\}+ L^{-1} \left\{ {1\over s+1}-{1\over s+2}\right\} \\\Rightarrow \bbox[red, 2pt]{y(t)=u(t-1) \left(-e^{-(t-1)} +{1\over 2}e^{-2(t-1)} +{1\over 2}\right) +u(t-2)\left( e^{-(t-2)}{-e^{-2(t-2)}}\right)+ e^{-t}-e^{-2t}}$$

解答:$$\textbf{(a) }L\{f(t)\} =L\{ \cos t\}+ L\left\{ \int_0^t f(\tau) e^{-2(t-\tau)} \,d\tau\right\} \Rightarrow F(s)= {s\over s^2+1}+ L\{f(t)\}L\{e^{-2t}\} \\ \Rightarrow F(s)= {s\over s^2+1}+F(s)\cdot {1\over s+2} \Rightarrow F(s)={s(s+2) \over (s+1)(s^2+1)} ={3s\over 2(s^2+1)} +{1\over 2(s^2+1)}-{1\over 2(s+1)} \\ \Rightarrow f(t)=L^{-1}\{F(s)\} \Rightarrow \bbox[red, 2pt]{f(t)={3\over 2}\cos t+{1\over 2} \sin t -{1\over 2}e^{-t}} \\\textbf{(b) }h(t) =L^{-1}\{H(s)\} =L^{-1} \left\{ {2\over (s+1)(s+2)(s+4)} \right\} \\ =L^{-1} \left\{ {2\over 3(s+1)}-{1\over s+2}+{1\over 3(s+4)} \right\}  \Rightarrow \bbox[red, 2pt]{h(t)={2\over 3}e^{-t}-e^{-2t}+{1\over 3}e^{-4t}}$$

解答:$$\textbf{(a) }\cases{x(t)=2 \cos t\\ y(t)=2\sin t\\ z(t)=6t} \Rightarrow \cases{x'(t)=-2\sin t\\ y'(t)= 2\cos t\\ z'(t)= 6}\\ \Rightarrow \int_C \vec F \cdot d\vec r = \int_0^{\pi/2} (4\cos^2 t, 4\sin^2 t, 8\cos t\sin^2 t) \cdot (-2\sin t,2\cos t, 6)\,dt\\= \int_0^{\pi/2} \left( -8\cos^2 t\sin t+56 \cos t\sin^2 t\right)\,dt =\left. \left[ {8\over 3}\cos^3 t+{56\over 3}\sin^3 t\right] \right|_0^{\pi/2} ={56\over 3}-{8\over 3}= \bbox[red, 2pt]{16} \\\textbf{(b) }r(t)=(\cos t,\sin t,t) \Rightarrow r'(t)=(-\sin t, \cos t, 1) \Rightarrow |r'(t)|= \sqrt{\sin^2t +\cos^2 t+1} =\sqrt 2 \\ \Rightarrow \int_C (xy+z^2)\,ds =\int_0^\pi (\cos t\sin t+t^2) \sqrt 2\,dt =\int_0^\pi \left( {\sqrt 2\over 2}\sin(2t) +\sqrt 2t^2 \right)\,dt \\=\left. \left[ -{\sqrt 2\over 4} \cos(2t) +{\sqrt 2\over 3}t^3\right] \right|_0^\pi = \bbox[red, 2pt]{{\sqrt 2\over 3}\pi^3}$$

解答:$$\textbf{(a) }f(x)=|x| \Rightarrow f(-x)=f(x) \Rightarrow f(x) \text{ is even} \Rightarrow b_n=0\\\qquad a_0={1\over 2\pi} \int_{-\pi}^\pi |x|\,dx ={1\over \pi} \int_0^\pi x\,dx={1\over 2}\pi \\ a_n={1\over \pi} \int_{-\pi}^\pi |x| \cos(nx)\,dx ={2\over \pi} \int_0^\pi x\cos(nx) \,dx ={2\over n^2\pi} \left((-1)^n-1 \right) \\ \Rightarrow \bbox[red, 2pt] {f(x)\sim {1\over 2}\pi + {2\over \pi} \sum_{n=1}^\infty {1\over n^2}\left((-1)^n-1 \right) \cos(nx)} \\\textbf{(b) } f(x)=|x|=  {1\over 2}\pi +{2\over \pi}\sum_{n=1}^\infty {1\over n^2}\left((-1)^n-1 \right)\cos(nx) \\ \Rightarrow f(0)=0={1\over 2}\pi + {2\over \pi}\sum_{n=1}^\infty {1\over n^2}\left((-1)^n-1 \right) ={1\over 2 } \pi-{4\over \pi}\left( 1+{1\over 3^2} +{1\over 5^2} +\cdots\right) \\ \Rightarrow 1+{1\over 3^2} +{1\over 5^2} +\cdots= \bbox[red, 2pt]{\pi^2 \over 8}$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解