115年成功大學航太碩士班-工程數學詳解

國立成功大學115學年度碩士班招生考試試題

系所:航空太空工程學系、能源工程國際碩士學位學程
科目:工程數學

 

解答:$$\textbf{(a) }y=\sum_{n=0}^\infty a_nx^{n+r} \Rightarrow y'=\sum_{n=0}^\infty (n+r)a_n x^{n+r-1} \Rightarrow y''=\sum_{n=0}^\infty (n+r)(n+r-1)a_n x^{n+r-2} \\ \Rightarrow x\sum_{n=0}^\infty (n+r)(n+r-1)a_n x^{n+r-2} +(1-2x) \sum_{n=0}^\infty (n+r)a_n x^{n+r-1} +(x-1) \sum_{n=0}^\infty a_nx^{n+r} =0 \\ \Rightarrow \sum_{k=0}^\infty (k+r)^2 a_k x^{k+r-1} -\sum_{k=1}^\infty (2k+2r-1) a_{k-1} x^{k+r-1} +\sum_{k=2}^\infty a_{k-2} x^{k+r-1} =0 \\ k=0 \Rightarrow r^2a_0=0 \Rightarrow r=0 \Rightarrow \cases{k=1 \Rightarrow a_1=a_0\\ k\ge 2 \Rightarrow k^2a_k-(2k-1)a_{k-1}+a_{k-2}=0} \\ \Rightarrow \cases{k=2 \Rightarrow a_2={1\over 2}a_0 \\k=3 \Rightarrow a_3={1\over 6}a_0} \Rightarrow a_k={a_0\over k!} \Rightarrow y_1(x)=a_0 \sum_{k=0}^\infty {x^k\over k!} = e^x, \text{ setting }a_0=1 \\ \Rightarrow \text{Let }y_2(x) = u(x)y_1(x)  =u(x)e^x \Rightarrow y_2'=u'e^x+ue^x \Rightarrow y_2''=u''e^x +2u'e^x +ue^x \\ \Rightarrow x(u''e^x +2u'e^x +ue^x) +(1-2x)(u'e^x+ue^x) +(x-1)ue^x=0 \Rightarrow xu''+u'=0 \\ \Rightarrow (xu')'=0 \Rightarrow xu'=c_1 \Rightarrow u'={c_1\over x} \Rightarrow u=\ln(x) \Rightarrow y_2=e^x\ln (x) \\ \Rightarrow y=y_1+y_2 \Rightarrow \bbox[red, 2pt]{y= Ae^x+Be^x\ln (x)}$$

$$\textbf{(b) } \cases{y_1'= 4y_1-8y_2 + 2\cosh t+ t\sinh t\\ y_2'=2y_1-6y_2+ 2\sinh t+ t\cosh t} \Rightarrow  \begin{bmatrix}y_1'\\ y_2' \end{bmatrix} = \begin{bmatrix}4&-8\\ 2& -6 \end{bmatrix} \begin{bmatrix}y_1\\ y_2 \end{bmatrix} + \begin{bmatrix}2\cosh t+ t\sinh t\\ 2\sinh t+ t\cosh t \end{bmatrix} \\ \text{Let }A= \begin{bmatrix}4&-8\\ 2& -6 \end{bmatrix} , g(t) =  \begin{bmatrix}2\cosh t+ t\sinh t\\ 2\sinh t+ t\cosh t \end{bmatrix} =\begin{bmatrix}e^t+e^{-t} +{t\over 2}(e^t-e^{{-t}}) \\ e^t-e^{-t}+{t\over 2}(e^t+e^{-t}) \end{bmatrix} \\ = e^t \begin{bmatrix}1+t/2\\ 1+t/2 \end{bmatrix} +e^{-t} \begin{bmatrix}1-t/2\\ -1+t/2 \end{bmatrix} =e^t \begin{bmatrix}1\\ 1 \end{bmatrix}+ te^t \begin{bmatrix}1/2\\ 1/2 \end{bmatrix}+ e^{-t} \begin{bmatrix}1\\ -1 \end{bmatrix}+ te^{-t} \begin{bmatrix}-1/2\\ 1/2 \end{bmatrix}\\ \det(A-\lambda I)=0 \Rightarrow \cases{\lambda_1=2\\ \lambda_2=-4} \Rightarrow \cases{v_1= \begin{bmatrix}4\\1 \end{bmatrix} \\v_2= \begin{bmatrix}1\\ 1 \end{bmatrix}} \Rightarrow y_h(t) = c_1e^{2t} \begin{bmatrix}4\\1 \end{bmatrix} +c_2 e^{-4t} \begin{bmatrix}1\\1 \end{bmatrix}\\\Rightarrow A=PDP^{-1} ,\text{ where }P= \begin{bmatrix}4& 1\\1& 1 \end{bmatrix} , D= \begin{bmatrix}2&0\\0&-4 \end{bmatrix} \\ y_p(t)= ae^t+bte^t +ce^{-t}+ dte^{-t} \Rightarrow y_p'=(a+b)e^t+bte^t+(-c+d)e^{-t}-dte^{-t} =Ay+g(t) \\ \Rightarrow \cases{\text{For }te^t: b=Ab+ \begin{bmatrix}1/2\\ 1/2 \end{bmatrix} \Rightarrow (A-I)b= \begin{bmatrix}-1/2\\ -1/2 \end{bmatrix} \Rightarrow b= (A-I)^{-1}\begin{bmatrix}-1/2\\ -1/2 \end{bmatrix} = \begin{bmatrix}1/10\\ 1/10 \end{bmatrix} \\ \text{For }e^t: a+b=Aa+ \begin{bmatrix}1\\ 1 \end{bmatrix} \Rightarrow (A-I)a=b- \begin{bmatrix}1\\1 \end{bmatrix} \Rightarrow a= \begin{bmatrix}9/50\\ 9/50 \end{bmatrix} \\ \text{For } te^{-t}: -d=Ad+ \begin{bmatrix}-1/2\\ 1/2 \end{bmatrix} \Rightarrow (A+I)d = \begin{bmatrix}1/2\\ -1/2 \end{bmatrix} \Rightarrow d= \begin{bmatrix}13/18\\ 7/18 \end{bmatrix} \\ \text{For }e^{-t}: -c+d=Ac+ \begin{bmatrix}1\\ -1 \end{bmatrix} \Rightarrow (A+I)c= d- \begin{bmatrix}1\\ -1 \end{bmatrix} \Rightarrow c= \begin{bmatrix}-25/18\\ -15/18 \end{bmatrix}} \\ \Rightarrow y=y_h+y_p \\\Rightarrow \bbox[red, 2pt]{\begin{bmatrix}y_1\\ y_2 \end{bmatrix} =c_1 e^{2t} \begin{bmatrix}4\\1 \end{bmatrix} +c_2 e^{-4t} \begin{bmatrix}1\\ 1 \end{bmatrix} +e^t \begin{bmatrix}9/50\\ 9/50 \end{bmatrix} +te^t \begin{bmatrix}1/10\\ 1/10 \end{bmatrix}+ e^{-t} \begin{bmatrix}-25/18\\ -15/18 \end{bmatrix} +te^{-t} \begin{bmatrix}13/18\\ 7/18 \end{bmatrix}}$$

解答:$$L\{y(t)\} + L\left\{ \int_0^t y(\tau) \sinh(t-\tau)\,d\tau\right\} = L\left\{2-{1\over 2}t^2 +(t-2) u(t-2) \right\} \\ \Rightarrow Y(s)+ Y(s)\cdot {1\over s^2-1}= {2\over s} -{1\over s^3}+ {e^{-2s} \over s^2} \Rightarrow Y(s)= \left( 1-{1\over s^2} \right) \left( {2\over s}-{1\over s^3}+{e^{-2s}\over s^2} \right) \\={2\over s}-{3\over s^3}+{1\over s^5}+e^{-2s} \left( {1\over s^2}-{1\over s^4} \right)\\ \Rightarrow y(t)= L^{-1}\{ Y(s)\} =L^{-1} \left\{{2\over s}-{3\over s^3}+{1\over s^5} \right\} +L^{-1} \left\{ e^{-2s} \left( {1\over s^2}-{1\over s^4} \right)\right\} \\ \Rightarrow \bbox[red, 2pt]{y(t)=2-{3\over 2}t^2+{1\over 24}t^4+ \left[ (t-2)-{1\over 6}(t-2)^3 \right] u(t-2)}$$

解答:$$\textbf{(a) }a_0= {1\over \pi} \int_{-\pi}^\pi f(x)\,dx = {1\over \pi} \int_{0}^\pi x\,dx={\pi\over 2} \\a_n={1\over \pi} \int_{-\pi}^\pi f(x)\cos(nx)\,dx = {1\over \pi} \int_{0}^\pi x\cos(nx)\,dx = {1\over n^2\pi}\left( (-1)^n-1 \right) \\b_n={1\over \pi} \int_{-\pi}^\pi f(x)\sin(x)\,dx = {1\over \pi} \int_0^\pi x\sin(nx)\,dx ={1\over n} (-1)^{n+1}   \\ \Rightarrow f(x)\sim \bbox[red, 2pt]{{\pi\over 4}+ \sum_{n=1}^\infty \left( {1\over n^2\pi}\left( (-1)^n-1 \right)\cos(nx)+{1\over n}(-1)^{n+1} \sin(nx) \right)} \\\textbf{(b) } F(\omega) =\int_{-\infty}^0 e^{at}\cdot e^{-i\omega t} \,dt+ \int_0^\infty e^{-at} \cdot e^{-i\omega t} \,dt =\int_{-\infty}^0  e^{(a-i\omega )t} \,dt+ \int_0^\infty   e^{-(a+i\omega) t} \,dt \\= \left. \left[ {e^{(a-i\omega)t} \over a-i\omega} \right] \right|_{-\infty}^0 + \left. \left[ {e^{-(a+i\omega)t}\over -(a+i\omega)} \right] \right|_0^\infty ={1\over a-i\omega}+{1\over a+i\omega} ={2a\over a^2+\omega^2} \Rightarrow \bbox[red, 2pt]{F(\omega)={2a\over a^2+\omega^2}} \\ F(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt  \Rightarrow f(t)={1\over 2\pi} \int_{-\infty}^\infty F(\omega)e^{i\omega t}\,d\omega ={1\over 2\pi} \int_{-\infty}^\infty {2a\over a^2+\omega^2}e^{i\omega t}\,d\omega \\ \Rightarrow \bbox[red, 2pt]{f(t)={a\over \pi} \int_{-\infty}^\infty{ e^{i\omega t}\over a^2+\omega^2} \,d\omega}$$

解答:$$f(x,y,z)= e^x \sin(4y)-z^3 \Rightarrow \nabla f=(f_x,f_y,f_z) =(e^x\sin(4y), 4e^x \cos(4y), -3z^2) \\ \Rightarrow \nabla f(0,\pi,1)=(0,4,-3) \Rightarrow \text{ the direction of steepest descent: }-\nabla f =(0,-4,3) \\ \Rightarrow \text{unit vector: }{(0,-4,3)\over ||(0,-4,3)||} =\bbox[red, 2pt]{\left(0,-{4\over 5},{3\over 5} \right)}$$

解答:$$\textbf{(a) }\cases{P(x,y)= 2x\sin(y^3)-2y\\ Q(x,y)= 3x^2y^2 \cos(y^3)+x} \Rightarrow \cases{P_y=6xy^2\cos (y^3)-2\\ Q_x=6xy^2\cos(y^3)+1}\\ \Rightarrow \oint_C \mathbf F \cdot d r = \iint_D \left( Q_x-P_y \right)dA = \iint_D 3\,dA =3\times (\text{ area of }\triangle PQR) =\bbox[red, 2pt]3 \\\textbf{(b) }\mathbf F=(3x+y^5z^2)\mathbf i+(2y+x^4z^3) \mathbf j+(z+x^3y^4) \mathbf k \Rightarrow \nabla \cdot \mathbf F=3+2+1=6 \\ \Rightarrow \iint_S \mathbf F\cdot ndA = \iiint_V \nabla \cdot \mathbf F\;dV =\iiint_V 6\,dV= 6\times (\text{volume of cube}) = \bbox[red, 2pt] 6$$

解答:$$\text{Using the given hint, } {\partial U\over \partial t} =k\nabla^2 U = k \left[{1\over r}{\partial \over \partial r} \left( r{\partial U\over \partial r} \right) +{\partial^2 U\over \partial z^2} \right]. \\\text{Then, } U(r,z,t)= R(r)Z(z)T(t) \Rightarrow RZ {dT\over dt}=k \left[ ZT {1\over r}{d\over dr} \left( r{dR\over dr} \right) +RT{d^2Z\over dz^2} \right] \\ \Rightarrow {1\over kT}{dT\over dt}={1\over R}\left( {1\over r}{d\over dr}\left( r{dR\over dr} \right) \right)+{1\over Z}{d^2Z\over dz^2}=-\lambda^2 \\\text{Solving for the Time Equation: } {dT\over dt}+k\lambda^2 T=0 \Rightarrow T(t)=e^{-k\lambda^2 t} \\{1\over R}\left( {1\over r}{d\over dr}\left( r{dR\over dr} \right) \right)=-\lambda^2-{1\over Z}{d^2Z\over dz^2}=-\mu^2 \Rightarrow {d^2Z\over dz^2}+(\lambda^2-\mu^2)Z=0 \\\text{Let }\nu^2=\lambda^2-\mu^2. \text{ Using BCs.:} \cases{Z(0)=0\\ Z(H)=0} \Rightarrow Z_n=\sin{n\pi z\over H}, \nu_n={n\pi\over H} \\ {1\over r}{d\over dr} \left( r{dR\over dr} \right)+ \mu^2R=0 \Rightarrow r^2{d^2R\over dr^2}+r{dR\over dr}+\mu^2r^2 R=0 \\ \text{It is Bessel's Equation, the solution bounded at }r=0 : R(r)=J_0(\mu r) \\ \text{Applyint the BC }U(a,z,t)=0 \Rightarrow J_0(\mu a)=0 \\\text{the total eigenvalue is }\lambda_{mn}^2 = \mu_m^2+ \nu_n^2 = \left( {\xi\over a} \right)^2+ \left( {n\pi\over H} \right)^2,  \text{ where }\mu_m={\xi_m\over a}, \xi_m \text{ is the }m\text{-th root of }J_0 \\ \Rightarrow \bbox[red, 2pt]{U(r,z,t)= \sum_{m=1}^\infty \sum_{n=1}^\infty A_{mn} J_0 \left( {\xi_m r\over a} \sin \left( {n\pi z\over H} \right)\right) e^{-k\lambda_{mn}^2 t},} \\ \bbox[red, 2pt]{\text{ where }A_{mn} = \left( {2\over a^2J_1^2(\xi_m)} \int_0^a r \alpha(r) J_0 \left( {\xi_m r\over a}dr \right)\right) \times \left( {2\over H} \int_0^H \beta(z) \sin {n\pi z\over H}\,dz \right)}$$

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解題僅供參考，碩士班歷年試題及詳解