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2022年10月2日 星期日

109年台綜大轉學考-工程數學D37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:D37


解答:x2y5xy+8y=0y=xmy=mxm1y=m(m1)xm2x2y5xy+8y=(m2m)xm5mxm+8xm=(m26m+8)xm=0m26m+8=0(m4)(m2)=0m=4,2yh=C1x4+C2x2yp=ax6yp=6ax5yp=30ax4x2y5xy+8y=30ax630ax6+8ax6=8ax6=8x6a=1y=yh+yp=C1x4+C2x2+x6y=4C1x3+2C2x+6x5{y(1)=2y(1)=6{C1+C2+1=24C1+2C2+6=6{C1=1C2=2y=x4+2x2+x6

解答y=n=0cnxny=n=1ncnxn1y=n=2n(n1)cnxn2(x1)y+y=n=2n(n1)cnxn1n=2n(n1)cnxn2+n=1ncnxn1=(c12c2)+(4c26c3)x+(9c312c4)x2+(16c420c5)x3+=0c1=2c2=3c3=4c4=ck=1kc1,kNy=c0+n=11nc1xn

解答F=6xyi+4yzj+xeykdiv F=x6xy+y4yz+zxey=6y+4z+0

\cases{S_1=\triangle OAB =\{(x,y,0)\mid x+y=1,0\le x,y\le 1\} \Rightarrow \vec n_1=(0,0,-1)\\ S_2=\triangle OAC =\{(x,0,z)\mid x+z=1,0\le x,z\le 1\} \Rightarrow \vec n_2= (0,-1,0)\\S_3=\triangle OBC =\{(0,y,z)\mid y+z=1,0\le  y,z\le 1\} \Rightarrow \vec n_3 =(-1,0,0) \\ S_4=\triangle ABC =\{(x,y,z)\mid x+y+z=1,0\le x,y,z\le 1\} \Rightarrow \vec n_4= (1/\sqrt 3,1/\sqrt 3,1/\sqrt 3)} \\ \Rightarrow \cases{\iint_{S_1} (\vec F\cdot \vec n_1)dS =\iint_{S_1} -xe^{-y}dS =\int_0^1 \int_0^{1-y} -xe^{-y}\,dxdy =1/e-1/2\\ \iint_{S_2} (\vec F\cdot \vec n_2)dS =\iint_{S_2} -4yzdS =0\\ \iint_{S_3} (\vec F\cdot \vec n_3)dS =\iint_{S_3} -6xydS = 0\\ \iint_{S_4} (\vec F\cdot \vec n_4)dS = {1\over \sqrt 3}\iint_{S_4} 6xy+4yz+xe^{-y}dS \\ \qquad= \int_0^1\int_0^{1-x} 6xy+4y(1-x-y)+ xe^{-y} dydx=11/12-1/e} \\\iint_S (\mathbf F\cdot \mathbf n)dS = {1\over e}-{1\over 2} +{11\over 12}-{1\over e}={5\over 12}, 因此 \bbox[red,2pt]{\iint_S (\mathbf F\cdot \mathbf n)dS = \iiint_D \text{div }\mathbf F\,dV={5\over 12}}

解答A=\left(\begin{matrix}1 & 0 & 1 \\0 & 1 & 0 \\ 1 & 0 & 1\end{matrix}\right) \Rightarrow \det(A-\lambda I)= -\lambda (\lambda-1) (\lambda-2)=0 \Rightarrow \lambda=0,1,2\\   \lambda_1=0 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}1 & 0 & 1 \\0 & 1 & 0 \\ 1 & 0 & 1\end{matrix}\right) \left(\begin{matrix}x_1\\x_2 \\ x_3 \end{matrix}\right) =0 \Rightarrow \cases{x_1=-x_3\\ x_2=0},取v_1=\left(\begin{matrix}-1\\0 \\ 1 \end{matrix}\right) \\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}0 & 0 & 1 \\0 & 0 & 0 \\ 1 & 0 & 0\end{matrix}\right) \left(\begin{matrix}x_1\\x_2 \\ x_3 \end{matrix}\right) =0 \Rightarrow \cases{x_1=0\\ x_3=0},取v_2=\left(\begin{matrix}0\\ 1 \\ 0 \end{matrix}\right) \\ \lambda_3=2 \Rightarrow (A-\lambda_3 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}-1 & 0 & 1 \\0 & -1 & 0 \\ 1 & 0 & -1\end{matrix}\right) \left(\begin{matrix}x_1 \\x_2 \\ x_3 \end{matrix}\right) =0 \Rightarrow \cases{x_1=x_3\\ x_2=0},取v_3=\left(\begin{matrix} 1\\0 \\ 1 \end{matrix} \right) \\ \Rightarrow  \left(\begin{matrix}x'(t) \\y'(t) \\ z'(t) \end{matrix}\right) =A \left( \begin{matrix}x(t) \\y(t) \\ z(t) \end{matrix}\right) \Rightarrow \left( \begin{matrix}x(t) \\y(t) \\ z(t) \end{matrix}\right) =(C_1v_1e^{\lambda_1 t}+C_2v_2e^{\lambda_2 t}+ C_3v_3e^{\lambda_3 t}) \\ =C_1\left(\begin{matrix}-1\\0 \\ 1 \end{matrix}\right)e^{0t} +C_2\left(\begin{matrix}0\\ 1 \\ 0 \end{matrix}\right) e^t+ C_3\left(\begin{matrix} 1\\0 \\ 1 \end{matrix} \right) e^{2t} \Rightarrow \bbox[red,2pt]{\cases{x(t)= -C_1+ C_3e^{2t} \\ y(t)= C_2e^t\\ z(t)=C_1 +C_3e^{2t}}}

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