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：$$先求齊次解:x^2y''-5xy'+8y=0，取y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow x^2y''-5xy'+8y = (m^2-m)x^m -5mx^m+8x^m =(m^2-6m+8)x^m=0 \\ \Rightarrow m^2-6m+8=0 \Rightarrow (m-4)(m-2)=0 \Rightarrow m=4,2 \Rightarrow y_h=C_1 x^4+C_2 x^2 \\ 取y_p =ax^6 \Rightarrow y_p'=6ax^5 \Rightarrow y_p''=30ax^4 \Rightarrow x^2y''-5xy'+8y =30ax^6 -30ax^6+ 8ax^6\\ =8ax^6 =8x^6 \Rightarrow a=1 \Rightarrow y=y_h+y_p = C_1 x^4+C_2 x^2+x^6 \Rightarrow y'= 4C_1 x^3 +2C_2 x +6x^5\\ 將\cases{y(1)=2 \\ y'(1)=6} 代入\Rightarrow \cases{C_1+C_2 +1=2\\ 4C_1+ 2C_2 +6=6} \Rightarrow \cases{C_1=-1\\ C_2= 2} \Rightarrow \bbox[red, 2pt]{y=-x^4+ 2x^2 +x^6}$$

解答：$$y= \sum_{n=0}^\infty c_nx^n \Rightarrow y'=\sum_{n=1}^\infty nc_nx^{n-1} \Rightarrow y''=\sum_{n=2}^\infty n(n-1)c_nx^{n-2} \\ \Rightarrow (x-1)y''+y'=\sum_{n=2}^\infty n(n-1)c_nx^{n-1}-\sum_{n=2}^\infty n(n-1)c_nx^{n-2}+\sum_{n=1}^\infty nc_nx^{n-1} \\ =(c_1 -2c_2) +(4c_2-6c_3 )x +(9c_3-12c_4)x^2 +(16c_4-20c_5)x^3+\cdots =0\\ \Rightarrow c_1=2c_2 =3c_3 =4c_4=\cdots  \Rightarrow  c_k={1\over k}c_1,k\in \mathbb{N}\\ \Rightarrow \bbox[red, 2pt]{y=c_0 +\sum_{n=1}^\infty {1\over n}c_1 x^n}$$

解答：$$\mathbf F=6xy\mathbf i+4yz\mathbf j+xe^{-y}\mathbf k \Rightarrow \text{div }\mathbf F={\partial \over \partial x}6xy + {\partial \over \partial y}4yz + {\partial \over \partial z} xe^{-y} =6y+4z +0\\ \Rightarrow \iiint_D \text{div }\mathbf F\,dV = \iiint_D 6y+4z\,dV =\int_0^1 \int_0^{1-z} \int_0^{1-y-z} 6y+4z\,dxdydz \\=\int_0^1 \int_0^{1-z}  (1-y-z) (6y+4z) \,dxdydz=\int_0^1 \int_0^{1-z} 6y-6y^2-10yz+4z -4z^2\,dydz \\ =\int_0^1 z^3-z^2-z+1\,dz ={5\over 12}$$

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

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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