108年聯合大學轉學考-工程數學詳解

國立聯合大學108學年度暑假轉學生招生考試

科目：工程數學

一、非選擇題(50%)

解答：$$\mathbf{(1)}\;y'(x)-1=y^2 \Rightarrow {dy\over dx}=y^2+1 \Rightarrow \int {1\over y^2+1}dy =\int 1dx \Rightarrow \tan^{-1}y=x+C\\ 將y(1)=0代入上式\Rightarrow 0=1+C \Rightarrow C=-1 \Rightarrow \bbox[red,2pt]{y=\tan(x-1)}\\ \mathbf{(2)}\;令\cases{M(x,y)=\cos(xy)+x/y\\ N(x,y)=1+(x/y)\cos(xy)} \Rightarrow M_y \ne N_x \Rightarrow 非恰當\\ 令I(y)為積分因子\Rightarrow \cases{{\partial \over\partial y}(MI) = I'(\cos(xy)+x/y)+I(-x\sin(xy)-{x\over y^2}) \\{\partial \over\partial x}(NI)= I({1\over y}\cos(xy)-x\sin(xy))}\\ 因此{\partial \over\partial y}(MI) ={\partial \over\partial x}(NI) \Rightarrow I'(\cos(xy)+x/y)= I({1\over y}\cos(xy)-x\sin(xy)) \Rightarrow {I'\over I}={1\over y}\\ \Rightarrow I(y)=y \Rightarrow (y\cos (xy)+x)dx +(y+x\cos(xy))dy=0 \Rightarrow \int y\cos(xy)+x \,dx = \int y+x\cos(xy)\,dy\\ \Rightarrow \sin(xy)+{1\over 2}x^2+g(y)={1\over 2}y^2+\sin(xy)+h(x) \Rightarrow \cases{h(x)=x^2/2 \\ g(y)=y^2/2} \\ \Rightarrow \sin(xy)+{1\over 2}(x^2+y^2)+C=0，再將y(0)=1代入\Rightarrow C=-{1\over 2}\\ \Rightarrow \bbox[red,2pt]{\sin(xy)+{1\over 2}(x^2+y^2-1)=0}\\\mathbf{(3)}\;令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-xy'-3y\\ = (m(m-1)-m-3)x^m= (m^2-2m-3)x^m=0 \Rightarrow (m-3)(m+1)x^m=0 \Rightarrow m=3,-1\\ \Rightarrow y=C_1x^3+{C_2\over x} \Rightarrow y'=3C_1x^2-{C_2\over x^2};\\將初始值\cases{y(1)= 1\\ y'(1)=2 }代入\Rightarrow \cases{C_1+C_2= 1\\ 3C_1-C_2=2} \Rightarrow \cases{ C_1=3/4\\ C_2= 1/4} \Rightarrow \bbox[red, 2pt]{y={3\over 4}x^3+ {1\over 4x}}$$

解答：$$\mathbf{(1)}\;y(t)+ \int_0^t y(\tau) \cos(t-\tau)\,d\tau= e^{-t} \Rightarrow \mathcal L\{ y(t)\}+ \mathcal L\{ \int_0^t y(\tau) \cos(t-\tau)\,d\tau\} =\mathcal L\{  e^{-t}\} \\ \Rightarrow Y(s)+ Y(s)\cdot {s\over s^2+1} ={1\over s+1} \Rightarrow Y(s)={1\over s+1} \cdot {s^2+1 \over s^2+s+1} ={2\over s+1}-{s+1\over  s^2+s+1}\\={2\over s+1}-{s+1/2\over  (s+1/2)^2+3/4}-{2\over 3}\cdot {3/4\over  (s+1/2)^2+3/4}\\ \Rightarrow \mathcal L^{-1}\{ Y(s)\} = \bbox[red,2pt]{y(t) =2e^{-t}-e^{-t/2}\cos(\sqrt 3t/2)-{2\over 3}e^{-t/2}\sin(\sqrt 3t/2)} \\\mathbf{(2)}\;y''(t)+ 3y'(t)+2y(t)= \delta(t-1)-\delta(t-2) \\\Rightarrow \mathcal L\{ y''(t)\}+ 3\mathcal L\{ y'(t)\}+2\mathcal L\{ y(t)\} =\mathcal L\{  \delta(t-1)\}-\mathcal L\{ \delta(t-2)\} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0) +3(sY(s)-y(0))+2Y(s)= e^{-s}-e^{-2s} \\ \Rightarrow (s^2+3s+2)Y(s)=e^{-s}-e^{-2s} \Rightarrow Y(s)=(e^{-s}-e^{-2s}){1 \over s^2+3s+2} \\=(e^{-s}-e^{-2s})({1 \over s+1}-{1\over s+2})  \\\Rightarrow \mathcal L^{-1}\{Y(s)\}= \bbox[red, 2pt]{y(t)= H(t-1)(e^{-(t-1)}-e^{-2(t-1)}) -H(t-2)(e^{-(t-2)}+ e^{-2(t-2)})}$$

二、選擇題(50%)

解答：$$\cases{y_1'(t)= -4y_1+ 4y_2-e^{-t}\\ y_2'(t)= -1.6y_1+1.2 y_2+e^{-t}} \Rightarrow A=\left[\begin{matrix} -4  & 4 \\\frac{-8}{5} & \frac{6}{5} \end{matrix}\right],\mathbf g=\begin{bmatrix}-e^{-t}\\ e^{-t} \end{bmatrix} \\ \det(A-\lambda I)=\left|\begin{matrix}-4-\lambda & 4 \\\frac{-8}{5} & \frac{6}{5}-\lambda\end{matrix}\right|= (\lambda+2)(\lambda +{4\over 5})=0 \Rightarrow 特徵值\lambda_1=-2,\lambda_2=-0.8，故選\bbox[red, 2pt]{(D)}$$

解答：$$(A-\lambda_1 I)\mathbf x=0 \Rightarrow \left[\begin{matrix}-2 & 4 \\\frac{-8}{5} & \frac{16}{5}\end{matrix} \right]\begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_1=2x_2，取v_1=\begin{bmatrix}2\\ 1 \end{bmatrix}\\ (A-\lambda_2 I)\mathbf x=0 \Rightarrow \left[\begin{matrix}\frac{-16}{5} & 4 \\\frac{-8}{5} & 2\end{matrix}\right] \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow 4x_1=5x_2,取v_2=\begin{bmatrix} 1\\ 0.8\end{bmatrix}，故選\bbox[red,2pt]{(A)}$$

解答：$$由(1)及(2)可知\mathbf Y= [v_1e^{\lambda_1t}\; v_2e^{\lambda_2 t} ]=\begin{bmatrix} 2 e^{-2t} & e^{-0.8t}\\ e^{-2t} & 0.8 e^{-0.8t}\end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$

解答：$$\mathbf Y^{-1}={1\over \det(Y)}\begin{bmatrix}0.8 e^{-0.8t}  & -e^{-0.8t}\\ -e^{-2t} & 2 e^{-2t}\end{bmatrix}={5\over 3} e^{2.8t}\begin{bmatrix}0.8 e^{-0.8t}  & -e^{-0.8t}\\ -e^{-2t} & 2 e^{-2t}\end{bmatrix} ={1\over 3}\begin{bmatrix}4 e^{2t}  & -5e^{2t}\\ -5e^{0.8t} & 10 e^{0.8t}\end{bmatrix}\\，故選\bbox[red, 2pt]{(C)}$$

解答：$$y_h=C_1v_1 e^{\lambda_1 t} +C_2v_2e^{\lambda_2t} =C_1 \begin{bmatrix}2\\ 1 \end{bmatrix}e^{-2t} + C_2\begin{bmatrix} 1 \\ 0.8\end{bmatrix}e^{-0.8t}，故選\bbox[red, 2pt]{(A)}$$

解答：$$\mathbf g=\begin{bmatrix}-e^{-t}\\ e^{-t} \end{bmatrix} \Rightarrow y_p=  \begin{bmatrix} ae^{-t}\\ be^{-t} \end{bmatrix} \Rightarrow y_p'= \begin{bmatrix} -ae^{-t}\\ -be^{-t} \end{bmatrix} \\ 將y_p及y_p'代回原式y'=Ay+g \Rightarrow   \begin{bmatrix} -ae^{-t}\\ -be^{-t} \end{bmatrix} = \begin{bmatrix} -4& 4\\ -1.6 & 1.2\end{bmatrix}  \begin{bmatrix} ae^{-t}\\ be^{-t} \end{bmatrix}+\begin{bmatrix}-e^{-t}\\ e^{-t} \end{bmatrix}\\ \Rightarrow \cases{-a=-4a+4b-1\\ -b=-1.6a+1.2b+1} \Rightarrow \cases{3a-4b=-1\\ 1.6a-2.2b=1} \Rightarrow \cases{a=-31\\ b=-23} \\ \Rightarrow y_p=  \begin{bmatrix} -31e^{-t}\\ -23e^{-t} \end{bmatrix}= \begin{bmatrix} -31\\ -23\end{bmatrix} e^{-t}，故選\bbox[red, 2pt]{(C)}$$

解答：$$y=y_h+ y_p=C_1  \begin{bmatrix} 2\\ 1 \end{bmatrix} e^{-2t} +C_2 \begin{bmatrix} 1\\ 0.8 \end{bmatrix} e^{-2t}+ \begin{bmatrix} -31\\ -23\end{bmatrix} e^{-t}\\ 將\cases{y_1(0)=0\\ y_2(0)=1} 代入\Rightarrow \begin{bmatrix} 0\\ 1 \end{bmatrix}=C_1  \begin{bmatrix} 2\\ 1 \end{bmatrix}+C_2 \begin{bmatrix} 1\\ 0.8 \end{bmatrix} + \begin{bmatrix} -31\\ -23\end{bmatrix} \Rightarrow \cases{2C_1+C_2 =31\\ C_1+0.8C_2 =24} \\ \Rightarrow \cases{C_1=4/3\\ C_2= 85/3} \Rightarrow y =\begin{bmatrix} y_1\\ y_2 \end{bmatrix}={4\over 3}\begin{bmatrix} 2\\ 1 \end{bmatrix} e^{-2t} +{85\over 3} \begin{bmatrix} 1\\ 0.8 \end{bmatrix} e^{-2t} -\begin{bmatrix} 31\\ 23\end{bmatrix} e^{-t}，故選\bbox[red, 2pt]{(B)}$$

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