109年台綜大轉學考-工程數學D04詳解

臺灣綜合大學系統109學年度學士班轉學生聯合招生考試

科目名稱：工程數學
類組代碼：D04

解答：$$\cases{3x+2y+2z =1\\ x-3y+3z = -3\\ y-2z=1} \Rightarrow A\mathbf x=\mathbf b,其中A=\begin{bmatrix} 3 & 2 & 2\\ 1 & -3 & 3\\ 0 & 1 & -2\end{bmatrix},\mathbf x=\begin{bmatrix} x\\ y\\ z\end{bmatrix},\mathbf b=\begin{bmatrix}1\\-3\\ 1 \end{bmatrix} \\ \Rightarrow \cases{A_x= \begin{bmatrix} 1 & 2 & 2\\ -3 & -3 & 3\\ 1 & 1 & -2\end{bmatrix} \\[1ex] A_y=\begin{bmatrix} 3 & 1 & 2\\ 1 & -3 & 3\\ 0 & 1 & -2\end{bmatrix} \\[1ex] A_z =\begin{bmatrix} 3 & 2 & 1\\ 1 & -3 & -3\\ 0 & 1 & 1\end{bmatrix}} \Rightarrow \cases{\det(A)= 15\\ \det(A_x)=-3\\ \det(A_y) = 13 \\ \det(A_z)= -1} \Rightarrow \cases{x= \det(A_x)/\det(A) = -1/5\\ y=\det(A_y)/\det(A) = 13/15\\ z= \det(A_z)/\det(A) = -1/15} \\ \Rightarrow \bbox[red, 2pt]{\cases{x=-1/5\\ y=13/15\\ z=-1/15}}$$

解答：$$\cases{x(t)=\cos t\\ y(t)= \sin t\\ z(t)=t} \Rightarrow \cases{x'(t)=-\sin t\\ y'(t)=\cos t\\ z'(t)=1} \Rightarrow 線積分\int \mathbf F\cdot dr =\int_0^{2\pi} (t,\cos  t,\sin t)\cdot (-\sin t,\cos t,1)\;dt\\ =\int_0^{2\pi } -t\sin t+\cos^2 t+\sin t\,dt =\left. \left[ t\cos t-\sin t +{1\over 2}(t+ \sin t\cos t)-\cos t \right]\right|_0^{2\pi} =\bbox[red, 2pt]{3 \pi}$$

解答：$$f(x,y,z)=4x^2 +2y^2+ z^2-31 \Rightarrow \cases{f_x= 8x\\ f_y=4y\\ f_z =2z} \Rightarrow \vec u=(f_x,f_y,f_z)|_{(1,1,5)} =(8,4,10) \\ \Rightarrow {\vec u\over |\vec u|} ={1\over 6\sqrt{5 }}(8,4,10) = ({4\over 3\sqrt 5},{2\over 3\sqrt 5}, {5\over 3\sqrt 5}) =\bbox[red, 2pt]{({4\over 15}\sqrt 5,{2\over 15}\sqrt 5,{1\over 3}\sqrt 5)} $$

解答：$$v_1=[2e^x \cos y,e^x\sin y] \Rightarrow \cases{v_{1x}=[2e^x \cos y,e^x\sin y]\\ v_{1y}= [-2e^x\sin y,e^x\cos y]} \\ v_2=[\cos x\cosh y,-\sin x \sinh y] \Rightarrow \cases{v_{2x}= [-\sin x\cosh y,-\cos x\sinh y]\\ v_{2y}= [\cos x \sinh y,-\sin x\cosh y]}$$

解答：$$\left(\begin{array}{rrr|rrr}1 & 8 & -7 & 1 & 0 & 0 \\0 & 1 & 3 & 0 & 1 & 0 \\0 & 0 & 1 & 0 & 0 & 1\end{array}\right) \underrightarrow{7r_3+r_1\to r_1\\ (-3)r_3+r_2\to r_2} \left(\begin{array}{rrr|rrr}1 & 8 & 0 & 1 & 0 & 7 \\0 & 1 & 0 & 0 & 1 & -3 \\0 & 0 & 1 & 0 & 0 & 1\end{array}\right) \\\underrightarrow{  (-8)r_2+r_1\to r_1} \left(\begin{array}{rrr|rrr}1 & 0 & 0 & 1 & -8 & 31 \\0 & 1 & 0 & 0 & 1 & -3 \\0 & 0 & 1 & 0 & 0 & 1\end{array}\right) \Rightarrow A^{-1}= \bbox[red, 2pt]{\left(\begin{array}{rrr }  1 & -8 & 31 \\  0 & 1 & -3 \\  0 & 0 & 1\end{array}\right)}$$

解答：$$y(t)+ \int_0^t y(\tau)\,d\tau=3 \Rightarrow y'(t)+ y(t)=0 \Rightarrow y(t)=Ce^{-t} \Rightarrow Ce^{-t} +\int_0^t Ce^{-\tau}\,d\tau\\ =Ce^{-t}-Ce^{-t}+C=3 \Rightarrow C=3 \Rightarrow \bbox[red,2pt]{y(t)=3e^{-t}}$$

解答：$$\lim_{n\to \infty} \left|{a_{n+1} \over a_n} \right| = \lim_{n\to \infty}  \left|{ (n+2)(n+1)x^{n+1} \over (n+1)n x^n} \right| = \lim_{n\to \infty} {n+1\over n}|x| =|x| \lt 1\\ \Rightarrow 收斂半徑r=\bbox[red, 2pt]1$$

解答：$$先求齊次解:y'''+3y''-3y'-y=0 \Rightarrow \lambda^3+3\lambda^2-3\lambda-1=0\\ \Rightarrow (\lambda-1)(\lambda^2+4\lambda +1)=0 \Rightarrow \lambda=1, -2\pm \sqrt 3 \Rightarrow y_h =C_1 e^{-(2+\sqrt 3)x} +C_2e^{(-2+\sqrt 3)x} +C_3e^x\\ 接著令y_p = Axe^x +Bx+C \Rightarrow y'_p= Axe^x+ Ae^x+B \Rightarrow y_p''= Axe^x+ 2Ae^x \Rightarrow y_p'''=A xe^x+ 3Ae^x\\ \Rightarrow y_p'''+ 3y_p''-3y_p'-y_p =6Ae^x-Bx-3B-C= 2e^x-x-1 \Rightarrow \cases{A=1/3\\ B=1\\ C=-2} \\ \Rightarrow \bbox[red, 2pt]{y=C_1 e^{-(2+\sqrt 3)x} +C_2e^{(-2+\sqrt 3)x} +C_3e^x +{1\over 3}xe^x+x-2}$$

解答：$$貝色方程式x^2y''+ xy'+(x^2-v^2)y=0之通解為 y=C_1J_v(x)+ C_2Y_v(x);\\ 本題v^2=9 \Rightarrow v=\pm 3為整數解，因此y=C_1J_v(x)+ C_2J_{-v}(x) \\ \Rightarrow \bbox[red,2pt]{ y=C_1\sum_{n=0}^\infty {(-1)^n\over n!(n+3)! 2^{2n+3}}x^{2n+3} +C_2\sum_{n=0}^\infty {(-1)^n\over n!(n-3)! 2^{2n-3}}x^{2n-3} }$$

============================== END =================================

解題僅供參考，其他轉學考歷屆試題及詳解