109年台聯大轉學考-工程數學詳解

臺灣聯合大學系統109學年度學士班轉學生聯合招生考試

科目名稱：工程數學
類組別：A5

解答： $$(a)\;y'+xy=xy^{-1} \Rightarrow y{dy\over dx}+ xy^2 = x \Rightarrow y{dy\over dx}=x(1-y^2) \Rightarrow \int {y\over 1-y^2}dy = \int x\,dx \\ \Rightarrow -{1\over 2}\ln|1-y^2| ={1\over 2}x^2 +C_1 \Rightarrow \ln|1-y^2|=-x^2+C_2 \Rightarrow 1-y^2 = C_3e^{-x^2}\\ \Rightarrow y=\pm \sqrt{1-C_3e^{-x^2}},將y(0)=3代入 \Rightarrow C_3=-8 \Rightarrow \bbox[red, 2pt]{y=  \sqrt{1+8e^{-x^2}}} \\(b)\;令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-3xy'+3y=0\\ \Rightarrow m(m-1)x^m-3mx^m+3x^3=0 \Rightarrow (m^2-4m+3)x^m=0 \Rightarrow (m-3)(m-1)=0\\ \Rightarrow m=1,3 \Rightarrow y_h= C_1 x+ C_2x^3 \Rightarrow y_h'=C_1+3C_2x，將\cases{y(1)=0 \\ y'(0)=1}代入y_h \Rightarrow \cases{0=C_1+C_2\\ C_1=1}\\ \Rightarrow \cases{C_1=1\\ C_2=-1} \Rightarrow y_h=x-x^3\\ 令y_p=A\ln x \Rightarrow y_p'={A\over x} \Rightarrow y_p''=-{A\over x^2} \Rightarrow x^2y_p''-3xy_p'+3y_p= -A-3A+3A\ln x=3\ln x-4\\ \Rightarrow A=1 \Rightarrow y_p=\ln x \Rightarrow y=y_h +y_p \Rightarrow \bbox[red, 2pt]{y=x-x^3+\ln x}$$

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

解答： $$\cases{m=1\\ k=1\\ F(t)=1-t^2/\pi^2} 代入-ky+F(t) = my'' \Rightarrow y''+y=1-t^2/\pi^2 \\\Rightarrow y= C_1\cos t+C_2 \sin t-{t^2\over \pi^2} +1+{2\over \pi^2} \Rightarrow y'=-C_1\sin t+C_2 \cos t-{2t\over \pi^2}\\ 將初始值\cases{y(0)=0\\ y'(0)=0}代入\Rightarrow \cases{C_1+1+{2\over \pi^2}=0 \\C_2=0} \Rightarrow \cases{C_1=-1-2/\pi^2\\ C_2=0}\\ \Rightarrow \bbox[red,2pt]{y''+y=1-t^2/\pi^2}，其解為\bbox[red,2pt]{y(t)=(-1-2/\pi^2)\sin t-t^2/\pi^2+1 +2/\pi^2}$$

解答： $$A=\left[\begin{matrix} -2-\lambda & 2 & -3 \\2 & 1-\lambda & -6 \\-1 & -2 & 0-\lambda \end{matrix}\right] \Rightarrow \det(A-\lambda I)=\left|\begin{matrix} -2-\lambda & 2 & -3 \\2 & 1-\lambda & -6 \\-1 & -2 & 0-\lambda \end{matrix}\right| =-(\lambda-5)(\lambda+3)^2=0\\ \Rightarrow  {\lambda_1 = 5, \lambda_2=-3}\\ \lambda_1= 5 \Rightarrow (A-\lambda_1 I)\textbf{x} =\left[\begin{matrix}-7 & 2 & -3 \\2 & -4 & -6 \\-1 & -2 & -5\end{matrix}\right] \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_1+x_3=0 \\x_2+2x_3= 0}, 取v_1=\begin{bmatrix} 1\\ 2\\ -1\end{bmatrix}\\ \lambda_2=-3 \Rightarrow (A-\lambda_2 I)\textbf{x} =\begin{bmatrix} 1 & 2 & -3 \\2 & 4 & -6 \\-1 & -2 & 3\end{bmatrix}  \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow x_1+2x_2= 3x_3,取v_2=\begin{bmatrix} 2\\-1\\ 0\end{bmatrix},v_3=\begin{bmatrix} 3\\ 0\\ 1\end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{特徵值為5及-3，相對應的特徵向量為\begin{bmatrix} 1\\ 2\\ -1\end{bmatrix},\begin{bmatrix} 2\\-1\\ 0\end{bmatrix}及\begin{bmatrix} 3\\ 0\\ 1\end{bmatrix}}$$

解答： $$(a)\;\cases{x(t)= a\cos^3t\\ y(t)= a\sin^3 t} \Rightarrow \cases{x'(t)= -3a\cos^2 t \sin t \\ y'(t)= 3a\sin^2t \cos t} \Rightarrow \cases{(x'(t))^2 = 9a^2\cos^4 t\sin^2 t\\ (y'(t))^2 = 9a^2\sin^4t \cos^2 t}\\ \Rightarrow \int_0^{\pi/2} \sqrt{(x'(t))^2 +(y'(t))^2}\,dt =\int_0^{\pi/2} \sqrt{9a^2\cos^2 t\sin^2t(\cos^2t+\sin^2t)}\,dt \\=\int_0^{\pi/2} \sqrt{9a^2\cos^2 t\sin^2t }\,dt =\int_0^{\pi/2} 3a\cos t\sin t\,dt =\int_0^{\pi/2} {3\over 2}a\sin (2t)\,dt =\left. \left[ -{3\over 4}a\cos (2t)\right]\right|_0^{\pi/2} \\ ={3\over 4}a +{3\over 4}a= \bbox[red, 2pt]{{3\over 2}a}\\ (b)\;f(x,y)=16x^2-y^2=399 \Rightarrow \nabla f=(f_x,f_y)= (32x,-2y) \Rightarrow \nabla f(1/8,1)=(4,-2)\\ \Rightarrow {(4,-2)\over \sqrt{4^2+2^2}} = \bbox[red, 2pt]{({2\over \sqrt 5},-{1\over \sqrt 5})}\\註:P不在surface 上，題目有誤?!\\(c)\;f=x^2+y^2+z^2 \Rightarrow \nabla f=(f_x,f_y,f_z)= (2x,2y,2z) \Rightarrow \nabla f(2,2,-1)= (4,4,-2)\\ \Rightarrow \nabla f(2,2,-1)\cdot \vec a =(4,4,-2)\cdot (1,1,3) = 4+4-6=\bbox[red, 2pt]2$$

解答： $$\cases{A\vec x=\vec e_1\\ A\vec y=\vec e_n} \Rightarrow A\vec z=2\vec e_1+3\vec e_n =2A\vec x+ 3A\vec y \Rightarrow  \bbox[red, 2pt]{\vec z = 2\vec x+3\vec y}$$

解答： $$\int_S \vec u\cdot \vec n\,dA = \int_V \nabla \cdot \vec u\,dV = \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} ({\partial \over \partial x},{\partial \over \partial y},{\partial \over \partial z})\cdot (x,y,z)\,dxdydz\\ =  \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} 3\,dxdydz =\bbox[red, 2pt]3$$

解答： $$依\text{d'Alembert's solution}, u_{tt}=c^2u_{xx}的通解為 u(x,t)=F(x+ct)+G(x-ct)\\ =F(x+t)+ G(x-t) = {1\over 2}(f(x+t)+f(x-t)) +{1\over 2}\int_{x-t}^{x+t} g(s)\,ds \\ ={1\over 2}(\sin(\pi(x+t))+ 3\sin(3\pi(x+t)) +\sin(\pi(x-t))+ 3\sin(3\pi(x-t))+ 0\\ \Rightarrow \bbox[red, 2pt]{u(x,t)= {1\over 2}(\sin(\pi(x+t))+ 3\sin(3\pi(x+t)) +\sin(\pi(x-t))+ 3\sin(3\pi(x-t))}$$

解答： $$令u(x,t)= X(x)T(t) \Rightarrow \cases{u_t=XT'\\ u_{xx}=X''T} \Rightarrow u_t=u_{xx}-u \Rightarrow XT'=X''T-XT\\ \Rightarrow {T'\over T}={X''-X\over X} =k \Rightarrow k為一常數 \Rightarrow X''-(k+1)X=0\\ 原邊界條件\cases{u(0,t) = X(0)T(t)=0\\ u(1,t)= X(1)T(t)=0},若T(t)=0 \Rightarrow u(x,t)=0為明顯解，不討論;\\ 因此邊界條件變為X(1)=X(0)=0\\ \text{Case I: }k=-1 \Rightarrow  X''=0 \Rightarrow X= ax+b 代入邊介條件\Rightarrow\cases{X(1)=a+b=0\\ X(0)= b=0}\\ \qquad \Rightarrow X=0 \Rightarrow u(x,t)=0為明顯解，不討論\\ \text{Case II: }k>-1 \Rightarrow  X=C_1e^{\sqrt{k+1}x} +C_2e^{-\sqrt{k+1}x}代入邊介條件\\\qquad \Rightarrow\cases{X(1)=C_1e^{\sqrt{k+1}} +C_2e^{-\sqrt{k+1}}=0\\ X(0)= C_1+C_2=0} \Rightarrow C_1=C_2=0 \Rightarrow X=0 \Rightarrow u(x,t)=0為明顯解，不討論\\ \text{Case III: }k<-1，假設k+1= -\mu^2 \Rightarrow X=A \cos \mu x+ B\sin \mu x 代入邊界條件 \Rightarrow \cases{A\cos \mu +B\sin\mu =0\\ A=0}\\\qquad \Rightarrow \cases{A=B=0 \Rightarrow 明顯解，不討論\\ A=0且\sin \mu =0} \Rightarrow \sin u=0  \Rightarrow u=n\pi \Rightarrow X_n(x)= \sin n\pi x, n\in \mathbb{N}\\接著求{T'\over T}=k \Rightarrow T'-kT=0 \Rightarrow T'+(u^2+1)T=0 \Rightarrow T(t)= A e^{-(u^2+1)t} \Rightarrow T_n(t)= A_ne^{-(n^2\pi^2+1)t}\\ \Rightarrow u(x,t)= A_ne^{-(n^2\pi^2+1)t} \sin (n\pi x);再將初始條件u(x,0)=\sin(\pi x) \Rightarrow A_n\sin (n\pi x)= \sin(\pi x)\\ \Rightarrow \cases{A_1=1\\ A_m=0,m\ge 2} \Rightarrow \bbox[red, 2pt]{u(x,t)=e^{-( \pi^2+1)t} \sin (\pi x)}$$

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