112年成大航太碩士班-工程數學詳解

國立成功大學112學年度碩士班招生考試

系所:　航空太空工程學系、民航研究所、能源工程國際碩士學位學程

解答: $$\oint_C \vec F \cdot d \vec r =\int_0^{2\pi} (\sin\theta,\cos \theta, \cos\theta+\sin \theta)\cdot (-\sin \theta, \cos \theta,0)dt =\int_0^{2\pi} -\sin^2\theta + \cos^2 \theta \,d\theta\\= \int_0^{2\pi } \cos (2\theta)\,d\theta =\bbox[red, 2pt]0$$

解答: $$\mathbf{(a)}\; \oint_C {z^2-3z+4i\over z+2i} =2\pi i \times \left[\left. z^2-3z+4i \right|_{z=-2i} \right] =2\pi i \times (-4+10i) = \bbox[red, 2pt]{-20\pi -8\pi i } \\ \mathbf{(b)}\; 2\pi i \times \left. \frac{d^2 }{dz^2}( e^{-z} \sin(z))\right|_{z=0} =2\pi i \times \left. -2e^{-z} \cos z \right|_{z=0} =\bbox[red, 2pt]{-4\pi i}$$

解答: $$\mathbf{(a)}\; L\{y''\}+ 5L\{y'\}+ 6L\{y\}=L\{\delta(t-\pi/2\}+ L\{u(t-\pi)\sin(t)\} \\ \Rightarrow s^2Y(s)+5sY(s)+6Y(s)=e^{-\pi s/2} +e^{-\pi s}\cdot {-1\over s^2+1}\\ \Rightarrow Y(s)={1\over s^2+5s+6}\cdot \left( e^{-\pi s/2} -e^{-\pi s}\cdot {1\over s^2+1}\right)\\=\left({1\over s+2}-{1\over s+3} \right)e^{-\pi s/2} -\left({1\over s+2}-{1\over s+3} \right)\cdot {1\over s^2+1}\cdot e^{-\pi s} \\\Rightarrow \bbox[red,2pt]{y(t)=u(t-\pi/2)\left(e^{-2(t-\pi/2)} -e^{-3(t-\pi/2)} \right)\\-u(t-\pi)\left( {1\over 5} e^{-2(t-\pi} -{1\over 10}e^{-3(t-\pi)}-{1\over 10} \cos(t-\pi)+ {1\over 10}\sin(t-\pi)\right)}\\ \mathbf{(b)}\; L\{y(t)\}+ L\{\int_0^t y(\tau)\sinh(t-\tau)\,d\tau\}= 3L\{t\}+ L\{e^t\} \\ \Rightarrow Y(s)+Y(s)\cdot {1\over s^2-1}={3\over s^2}+{1\over s-1}\\ \Rightarrow Y(s)={3\over s^2}+{1\over s-1}-{3\over s^4}-{1\over s^2(s-1)} \\ \Rightarrow y(t)=L^{-1}\{ {3\over s^2}\}+L^{-1}\{ {1\over s-1}\}-L^{-1}\{ {3\over s^4}\}-L^{-1}\{ {1\over s^2(s-1)}\}\\ =3t+e^t-{1\over 2}t^3-e^t+t+u(t) \\ \Rightarrow \bbox[red, 2pt]{y(t)=-{1\over 2}t^3+4t+u(t)}$$

解答: $$先求齊次解,y''+2y'+3y=0 \Rightarrow \lambda^2+2\lambda+0 \Rightarrow \lambda =-1\pm \sqrt 2i  \\\Rightarrow y_h=c_1e^{-x}\cos (\sqrt 2x) +c_2 e^{-x} \sin(\sqrt 2 x)\\ 令y_p=a\sin x+b \cos x +cx+d \Rightarrow y_p'=a\cos x-b\sin x+c \Rightarrow y_p''= -a\sin x-b \cos x\\ \Rightarrow y_p''+2y_p'+3y_p = (2a+2b)\cos x+(2a-2b)\sin x +3cx+(3d+2c) =\sin x-x\\ \Rightarrow \cases{2a+2b=0\\ 2a-2b=1\\ 3c=-1\\ 3d+2c=0} \Rightarrow \cases{a=1/4\\ b=-1/4\\c=-1/3 \\ d=2/9} \Rightarrow y_p={1\over 4}\sin x-{1\over 4}\cos x-{1\over 3}x+{2\over 9} \\ \Rightarrow y=y_h+y_p =c_1e^{-x}\cos (\sqrt 2x) +c_2 e^{-x} \sin(\sqrt 2 x)+{1\over 4}\sin x-{1\over 4}\cos x-{1\over 3}x+{2\over 9} \\ \Rightarrow y'=-e^{-x}(c_1\cos (\sqrt 2x)+ c_2 \sin(\sqrt 2x)) +e^{-x}(-\sqrt 2c_2\sin (\sqrt 2x)+ \sqrt 2c_2 \cos(\sqrt 2x)) \\\qquad +{1\over 4} \cos x+{1\over 4}\sin x-{1\over 3} \\ \cases{y(0)=c_1-{1\over 4}+{2\over 9}=0\\ y'(0)=-c_1+\sqrt 2c_2 +{1\over 4}-{1\over 3}=0} \Rightarrow \cases{c_1=1/36\\ c_2= 1/9\sqrt 2} \\ \Rightarrow \bbox[red, 2pt]{ y={1\over 36}e^{-x}\cos(\sqrt 2x) +{1\over 9\sqrt 2}e^{-x}\sin(\sqrt 2 x)+{1\over 4}\sin x-{1\over 4}\cos x-{1\over 3}x+{2\over 9}}$$

解答: $$先求齊次解,即x^3y'''-3x^2y''+6xy'-6y=0, 令y=x^m \Rightarrow \cases{y'=mx^{m-1}\\ y''= m(m-1)x^{m-2} \\ y'''=m(m-1)(m-2) x^{m-3}} \\ \Rightarrow x^3y'''-3x^2y''+6xy'-6y=(m-1)(m-2)(m-3)x^m=0 \Rightarrow m=1,2,3\\ \Rightarrow \bbox[red, 2pt]{y_h=c_1x +c_2x^2 +c_3x^3}\\ 接著利用\text{ variations of parameter }求y_p, \\令\cases{y_1=x\\ y_2=x^2 \\ y_3=x^3} \Rightarrow W= \begin{vmatrix} y_1 & y_2 & y_3\\ y_1'& y_2' & y_3'\\ y_1'' & y_2'' & y_3''\end{vmatrix} = \begin{vmatrix} x & x^2 & x^3\\ 1& 2x & 3x^2\\ 0 & 2 & 6x\end{vmatrix} =2x^3 \Rightarrow \cases{W_1= \begin{vmatrix} 0 & x^2 & x^3\\ 0& 2x & 3x^2\\ 1 & 2 & 6x\end{vmatrix} =x^4 \\ W_2= \begin{vmatrix} x & 0 & x^3\\ 1& 0 & 3x^2\\ 0 & 1 & 6x\end{vmatrix} = -2x^3 \\ W_3= \begin{vmatrix} x & x^2 & 0\\ 1& 2x & 0\\ 0 & 2 & 1 \end{vmatrix} =x^2} \\ \Rightarrow y_p = x\int {x^4 \ln x\over 2x^3}\,dx + x^2 \int{(-2x^3) \ln x\over 2x^3}+ x^3 \int{ x^2\ln x\over 2x^3}\,dx \\={x\over 2}\left( {1\over 2}x^2\ln x-{1\over 4}x^2\right)-x^2(x\ln x-x)+{x^3\over 2}({1\over 2}\ln^2(x)) \\ \Rightarrow\bbox[red, 2pt]{ y_p={1\over 4}x^3\ln^2(x)-{3\over 4} x^3\ln (x)+{7\over 8}x^3}$$
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