115年台大碩士班-工程數學E詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 工程數學(E)

解答:$$\textbf{(a) }y''+y'-2y=0 \Rightarrow r^2+r-2=0 \Rightarrow (r+2)(r-1)=0 \Rightarrow r=-2,1 \Rightarrow y_h= c_1e^{-2x}+c_2e^x \\ y_p=Axe^x +B \cos x+C\sin x \Rightarrow y_p'=Ae^x+Axe^x-B\sin x+C\cos x \\ \Rightarrow y_p''=2Ae^x +Axe^x-B\cos x-C\sin x \Rightarrow y_p''+y_p'-2y_p=3Ae^x+ (C-3B)\cos x-(B+3C)\sin x \\=3e^x+\sin x \Rightarrow \cases{A=1\\ C=3B\\ B+ 3C=-1} \Rightarrow \cases{A=1\\B=-1/10\\ C=-3/10} \Rightarrow y_p=xe^x-{1\over 10}\cos x-{3\over 10} \sin x \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{-2x}+c_2e^x +xe^x-{1\over 10}\cos x-{3\over 10} \sin x} \\\textbf{(b) }(y')^2 =2yy'' \Rightarrow {(y')^2\over yy'} ={2yy'' \over yy'} \Rightarrow {y'\over y} ={2y''\over y'} \Rightarrow \int{y'\over y} \,dx= \int {2y''\over y'}\,dx \Rightarrow \ln|y|= 2\ln |y'|+c_1 \\ \Rightarrow \ln |y|=\ln (y')^2 +c_1 =\ln c_2(y')^2\Rightarrow y = c_2(y')^2 \Rightarrow (y')^2= c_3y \Rightarrow y'=c_4\sqrt y \\ \Rightarrow \int {1\over \sqrt y} dy=\int c_4\,dx \Rightarrow 2\sqrt y= c_4x+ c_5 \Rightarrow \sqrt y=c_6x+c_7 \Rightarrow \bbox[red, 2pt]{y=(c_6x+c_7)^2}$$

解答:$$\textbf{(a) } L\{tf(t)\} =-{d\over ds}F(s) \Rightarrow f(t)=-{1\over t} L^{-1}\left\{{d\over ds}F(s) \right\} \\ \quad \text{Now, }F(s)= \ln{s+1\over s-1} =\ln(s+1) -\ln(s-1) \Rightarrow {d\over ds}F(s)={1\over s+1}-{1\over s-1}\\\quad  \Rightarrow L^{-1}\left\{{d\over ds}F(s) \right\} = e^{-t}-e^t \Rightarrow \bbox[red, 2pt]{f(t) ={e^t-e^{-t}\over t}} \\\textbf{(b) } L\{ y''\}  -2 L\{y'\}-3L\{y\} = L\{t^2 \delta(t-2)\} \Rightarrow [s^2Y(s)-s]-2[sY(s)-1]-3Y(s) =4e^{-2s} \\ \quad \Rightarrow Y(s)={s-2\over (s-3)(s+1)} +{4e^{-2s} \over (s-3)(s+1)} = \left( {1/4\over s-3}+{3/4\over s+1} \right)+e^{-2s} \left( {1\over s-3}-{1\over s+1} \right) \\ \Rightarrow y(t)=L^{-1} \{Y(s)\} = L^{-1}\left\{ {1/4\over s-3}+{3/4\over s+1}\right\} +L^{-1}\left\{ e^{-2s} \left( {1\over s-3}-{1\over s+1} \right)\right\} \\ \quad \Rightarrow \bbox[red, 2pt]{y(t)={1\over 4}e^{3t}+{3\over 4}e^{-t} +u(t-2) \left( e^{3(t-2)}-e^{-(t-2)} \right)}$$

解答:$$y= \sum_{n=0}^\infty a_n x^n \Rightarrow y'=\sum_{n=1}^\infty na_n x^{n-1} \Rightarrow y''= \sum_{n=2}^\infty n(n-1)a_n x^{n-2}\\\Rightarrow(x^2-1)\sum_{n =2}^\infty n(n-1)a_n x^{n-2} +2x \sum_{n=1}^\infty na_n x^{n-1}-12\sum_{n=0}^\infty a_n x^n=0 \\ \Rightarrow \sum_{n=2}^\infty n(n-1)a_n x^{n}-\sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=1}^\infty 2na_n x^{n}-12 \sum_{n=0}^\infty a_n x^n=0 \\ \Rightarrow \sum_{n=2}^\infty n(n-1)a_n x^{n}-\sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^{n} + \sum_{n=1}^\infty 2na_n x^{n}-12 \sum_{n=0}^\infty a_n x^n=0  \\ \Rightarrow \text{coefficients of }x^n: n(n-1)a_n-(n+2)(n+1)a_{n+2}+2na_n-12a_n=0 \\ \Rightarrow a_{n+2}={(n+4)(n-3) \over (n+2)(n+1)} a_n \\ \text{Choosing }\cases{a_0=1\\ a_1=0} \Rightarrow \cases{n=0 \Rightarrow a_2=-6a_0=-6 \\n=2 \Rightarrow a_4=-{1\over 2}a_2=3\\ n=4 \Rightarrow a_6={8\over 30}a_2={4\over 5}} \Rightarrow \bbox[red, 2pt]{y_1(x)=1-6x^2+3x^4+{4\over 5}x^6+\cdots} \\ \text{singular points at }x^2-1=0 \Rightarrow x=\pm 1\Rightarrow \text{ radius of convergence: }\bbox[red, 2pt]{R_1=1} \\\text{Choosing }\cases{a_0=0 \\ a_1=1} \Rightarrow \cases{n=1 \Rightarrow a_3=-{5\over 3}a_1=-{5\over 3} \\ n=3 \Rightarrow a_5=0\\ a_7=a_9=\cdots =0} \Rightarrow \bbox[red, 2pt]{y_2(x)=x-{5\over 3}x^3} \\\text{Since }y_2(x) \text{ is a polynomial, radius of convergence }\bbox[red, 2pt]{R_2=\infty}  \\ \Rightarrow \text{The general solution is: } \bbox[red, 2pt]{y(x)=c_1 \left( 1-6x^2+3x^4+{4\over 5}x^6+ \cdots \right) +c_2 \left( x-{5\over 3}x^3 \right)}$$

解答:$$\text{The double Fourier sin series for }f(x,y) \text{ on }0\le x\le a\text{ and }0\le y\le b \text{ is given by}\\ f(x,y)= \sum_{m=1}^\infty \sum_{n=1}^\infty B_{mn} \sin {m\pi x\over a} \sin {n\pi y\over b} \\ \Rightarrow \bbox[red, 2pt]{3x+y^2 =\sum_{m=1}^\infty \sum_{n=1}^\infty B_{mn} \sin(m\pi x) \sin{n\pi y\over 2}}\\ \Rightarrow B_{mn} =2 \int_0^1\int_0^2 (3x+y^2) \sin(m\pi x) \sin {n\pi y\over 2} \,dydx \\=2 \left( \underbrace{ \int_0^1\int_0^2 3x \sin(m\pi x) \sin {n\pi y\over 2} \,dydx}_{I_1} + \underbrace{\int_0^1\int_0^2 y^2\sin(m\pi x) \sin {n\pi y\over 2} \,dydx}_{I_2}\right) \\ \Rightarrow I_1= 3 \left[ \int_0^1 x\sin(m\pi x)\,dx \right]\left[ \int_0^2 \sin{n\pi y\over 2}\,dy \right] =3 \left( -{(-1)^m\over m\pi} \right) \left( {2\over n\pi}(1-(-1)^n) \right) \\\qquad =-{6(-1)^m (1-(-1)^n) \over mn\pi^2} \\ \Rightarrow I_2=   \left[ \int_0^1  \sin(m\pi x)\,dx \right]\left[ \int_0^2 y^2\sin{n\pi y\over 2}\,dy \right] = \left( {1-(-1)^m\over m\pi} \right) \left( -{8(-1)^n\over n\pi}+{16\over n^3 \pi^3}((-1)^n-1) \right) \\ \Rightarrow \bbox[red, 2pt]{B_{mn} = {28(-1)^{m+n}-12(-1)^m-16(-1)^n \over mn\pi^2} -{32(1-(-1)^m)(1-(-1)^n) \over mn^3\pi^4}}$$

解答:$$f(t) = \sum_{-\infty}^\infty c_n e^{in\omega_0t}, \text{ where} \cases{T=2\\ \omega_0=2\pi/T=\pi \\ \text{interval of integration: }[-1,1]} \\ \Rightarrow c_n={1\over 2} \int_{-1}^1 e^{-t} e^{-in\pi t} \,dt ={1\over 2} \int_{-1}^1  e^{-(1+in\pi) t} \,dt = {-1\over 2(1+in\pi)} \left[ e^{-(1+in\pi)}  -e^{1+in\pi} \right] \\= {-1\over 2(1+in\pi)} \left[ e^{-1}(-1)^n  -e(-1)^n \right]= {(-1)^n\over 2(1+in\pi)} \left[e-e^{-1} \right]= {(-1)^n\over  (1+in\pi)}\sinh(1) \\ \Rightarrow \bbox[red, 2pt]{f(t)= \sum_{n=-\infty}^\infty \left(  {(-1)^n\over  (1+in\pi)}\sinh(1)  \right)e^{in\pi t}}$$

解答:$$u(x,y,t) =X(x)Y(y)T(t) \Rightarrow XYT''=4(X''YT+ XY''T) \Rightarrow {T''\over 4T}={X''\over X}+{Y''\over Y}=-\lambda \\ B.C.: \cases{X(0)Y(y)T(t) =0\\ X(2\pi)Y(y) T(t)=0\\ X(x)Y(0)T(t) =0\\ X(x)Y(2\pi)T(t)=0} \Rightarrow \cases{X(0)=0\\X(2\pi) =0\\ Y(0)=0\\ Y(2\pi) =0} \\\Rightarrow \cases{\text{For }X(x): \lambda_x= ({m\pi\over L})^2 =({m\pi\over 2\pi})^2 ={m^2\over 4}, m=1,2,\dots \\ \text{For }Y(y): \lambda_y=({n\pi\over W})^2 =({n\pi\over 2\pi})^2 ={n^2\over 4},n=1,2,\dots} \Rightarrow \cases{X_m(x) = \sin (mx/2) \\ Y_n(y)= \sin(ny/2)} \\ \text{Solving for }T(t): \lambda= \lambda_x+ \lambda_y= {1\over 4}(m^2+ n^2) \Rightarrow T''+4\lambda T= T''+(m^2+n^2)T=0 \\ \qquad \Rightarrow T_{mn}(t)= A_{mn} \cos \sqrt{m^2+n^2}t +B_{mn} \sin \sqrt{m^2+n^2}t \\ \Rightarrow u(x,y,0)= 0\Rightarrow A_{mn}=0 \Rightarrow u(x,y,t) = \sum_{m=1}^\infty \sum_{n=1}^\infty B_{mn} \sin{mx\over 2}\sin {ny\over 2} \sin \sqrt{m^2+n^2}t \\ \Rightarrow {\partial u(x,y,0) \over \partial t}=1 \Rightarrow \sum_{m=1}^\infty \sum_{n=1}^\infty B_{mn} \sqrt{m^2+n^2}\sin{mx\over 2}\sin {ny\over 2} =1 \\ \Rightarrow B_{mn} \sqrt{m^2+n^2} ={4\over (2\pi)^2} \int_0^{2\pi} \int_0^{2\pi} 1\cdot \sin{mx\over 2}\sin {ny\over 2} \,dxdy ={1\over \pi^2} \cdot {4\over m} \cdot {4 \over n} \\ \Rightarrow B_{mn} ={16\over \pi^2 mn\sqrt{m^2+n^2}}, \text{ where }m \text{ and }n \text{ are odd numbers.} \\ \Rightarrow \bbox[red, 2pt]{u(x,y,t) = \sum_{m\in odd} \sum_{n\in odd} {16\over \pi^2 mn\sqrt{m^2+n^2}} \sin{mx\over 2}\sin {ny\over 2} \sin (\sqrt{m^2+n^2}t)}$$

解答:$$\text{Let }U(x,s) = L\{u(x,t)\} \Rightarrow L\{u_t\} =L\{a^2 u_{xx}\} \Rightarrow sU(x,s)-u(x,0)=a^2{d^2U\over dx^2} \\ \Rightarrow {d^2U\over dx^2} -{s\over a^2}U=0 \Rightarrow U(x,s)=c_1 e^{-\sqrt sx/a} +c_2 e^{\sqrt sx/a} \\ \lim_{x\to \infty} u(x,t)\lt \infty \Rightarrow c_2=0 \text{ and } u(0,t)=A \Rightarrow L\{u(0,t)\} =L\{A\} \Rightarrow U(0,s)={A\over s} =c_1 \\ \Rightarrow U(x,s)={A\over s}e^{-\sqrt sx/a} \Rightarrow u(x,t)= L^{-1}\{U(x,s)\}= L^{-1}\left\{ {A\over s}e^{-\sqrt sx/a} \right\} =A\cdot \text{erfc} \left( {x\over 2a\sqrt t} \right) \\ \Rightarrow \bbox[red, 2pt]{u(x,t)= A \left[ 1-\text{erf}\left( {x\over 2a\sqrt t} \right)  \right]}$$

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解題僅供參考，碩士班歷年試題及詳解