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2024年5月25日 星期六

113年台科大材料碩士班-工程數學詳解

 國立臺灣科技大學113學年度碩士班招生試題

系所組別:材料科學與工程系碩士班乙組
科目:工程數學

解答(1)No,it is nonlinear. It cannot be written in the form of a(x)y+b(x)y=c(x)(2)(y4x)dx+(yx)dy=0{P(x,y)=y4xQ(x,y)=yx{Py=1Qx=1PyQxNot Exact(3){PyQxQ=2yx is dependent on both of x,yPyQxP=2y4x is dependent on both of x,yNo,the ODE have no integrating factor I(x) or I(y)(4)xM+yN=4x2+y2μ(x,y)=14x2+y2(μM)y=4x2+8xyy2(4x2y2)2=(μN)x exactΦ(x,y)=4xy4x2y2dx=xy4x2y2dyΦ=14ln(2xy)(2x+y)3+ϕ(y)=14ln(y2x)(2x+y)3+ρ(x)ln(2xy)+3ln(2x+y)=c1



解答{y(0)=c1=2y(2π)=c1=2y=2cos(2x)+c2sin(2x)There are infinitely many solutions with different c2

解答y=vexy=vex+vexy

解答\textbf{(1)}\;f(t)=t(u(t)-u(t-2))+u(t-2) \Rightarrow \bbox[red, 2pt]{f(t)= tu(t)+(1-t)u(t-2)} \\ \textbf{(2)}\; L\{f(t)\} =\int_0^\infty f(t) e^{-st}\,dt  = \int_0^2 te^{-st}\,dt +\int_2^\infty e^{-st}\,dt \\\quad = \left.\left[ -{1\over s}te^{-st}-{1\over s^2}e^{-st}\right] \right|_0^2 + \left. \left[ -{1\over s}e^{-st} \right] \right|_2^\infty ={1\over s^2}-{2\over s}e^{-2s}-{1\over s^2}e^{-2s}+{1 \over s}e^{-2s} \\ \quad \Rightarrow \bbox[red, 2pt]{ L\{f(t)\} ={1\over s^2}-{1\over s}e^{-2s}-{1\over s^2}e^{-2s}} \\\textbf{(3)}\;L\{y'\}+ L\{y\}=L\{f(t)\}  \Rightarrow sY(s)-1+Y(s)= {1\over s^2}-{1\over s}e^{-2s}-{1\over s^2}e^{-2s} \\\quad \Rightarrow Y(s)={1\over (s+1)s^2}-{1\over s(s+1)}e^{-2s}-{1\over (s+1)s^2}e^{-2s}+ {1\over s+1}\\  \Rightarrow y(t)=L^{-1}\{ Y(s)\} =-1+t+e^{-t}-u(t-2)(1-e^{2-t})-u(t-2)(t-3+e^{2-t})+e^{-t} \\ \Rightarrow \bbox[red, 2pt]{y(t)=-1+t+2e^{-t}-u(t-2)\left( t-2\right)}

解答\begin{cases}x_1 &    & +x_3 &-x_4  & = & 1 \\  & 2x_2  & + x_3 & +x_4  & = & 3 \\x_1 & -x_2 &   &+x_4  & = & -1\\ x_1& +x_2& +x_3& +x_4 &=& 2\end{cases} \Rightarrow \begin{bmatrix}1& 0  & 1&-1 \\0 & 2 & 1& 1\\ 1& -1 & 0 & 1\\ 1& 1& 1& 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\\x_4\end{bmatrix}=\begin{bmatrix}1 \\3 \\-1\\2 \end{bmatrix} \\ \Rightarrow \text{augmented matrix} \left[ \begin{array}{rrrr|r}1 & 0 & 1 & -1 & 1 \\ 0 & 2 & 1 & 1 & 3 \\1 & -1 & 0 & 1 & -1 \\1 & 1 & 1 & 1 & 2\end{array} \right] \xrightarrow{R_3-R_1\to R_3,R_4-R_1\to R_4} \\ \left[ \begin{matrix} 1 & 0 & 1 & -1 & 1 \\0 & 2 & 1 & 1 & 3 \\0 & -1 & -1 & 2 & -2 \\0 & 1 & 0 & 2 & 1\end{matrix} \right] \xrightarrow{R_3+(1/2)R_2\to R_3, R_4-(1/2)R_2\to R_4} \left[ \begin{matrix}1 & 0 & 1 & -1 & 1 \\0 & 2 & 1 & 1 & 3 \\0 & 0 & \frac{-1}{2} & \frac{5}{2} & \frac{-1}{2} \\0 & 0 & \frac{-1}{2} & \frac{3}{2} & \frac{-1}{2} \end{matrix} \right] \\ \xrightarrow{R_4-R_3\to R_4} \left[ \begin{matrix}1 & 0 & 1 & -1 & 1 \\0 & 2 & 1 & 1 & 3 \\0 & 0 & \frac{-1}{2} & \frac{5}{2} & \frac{-1}{2} \\0 & 0 & 0 & -1 & 0 \end{matrix} \right] \Rightarrow \begin{cases} x_1 &  & +x_3 & -x_4 & = & 1 \\ &2x_2 & +x_3 & +x_4 & = & 3 \\ & &\frac{-1}{2}x_3 & +\frac{5}{2}x_4 & = & \frac{-1}{2} \\ & & &-x_4 & = & 0\end{cases} \\ \Rightarrow x_4=0 \Rightarrow -{1\over 2}x_3=-{1\over 2} \Rightarrow x_3=1 \Rightarrow 2x_2+1=3 \Rightarrow x_2=1 \Rightarrow x_1+1=1 \Rightarrow x_1=0 \\ \Rightarrow \bbox[red, 2pt]{\cases{x_1=0\\ x_2=1\\ x_3=1\\ x_4=0}}

解答\begin{cases}x_1 & +2x_2 & +x_3 &  & = & 8 \\2x_1 & -2x_2 & +2x_3 &  & = & 7 \\x_1 & -4x_2 & +3x_3 &  & = & 1\end{cases} \Rightarrow \begin{bmatrix}1 & 2 & 1 \\2 & -2 & 2 \\1 & -4 & 3 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\x_3\end{bmatrix} =\begin{bmatrix}8 \\ 7 \\ 1\end{bmatrix} \\ \Rightarrow \cases{\triangle =\begin{vmatrix}1 & 2 & 1 \\2 & -2 & 2 \\1 & -4 & 3 \end{vmatrix} =-12 \\[1ex] \triangle_1= \left|\begin{matrix} 8 & 2 & 1 \\7 & -2 & 2 \\ 1 & -4 & 3 \end{matrix}\right| = -48\\[1ex] \triangle_2 =\left|\begin{matrix} 1 & 8 & 1 \\ 2 & 7 & 2 \\1 & 1 & 3\end{matrix}\right| = -18 \\[1ex] \triangle_3= \left|\begin{matrix} 1 & 2 & 8 \\2 & -2 & 7 \\1 & -4 & 1 \end{matrix}\right| =-12} \Rightarrow \cases{x_1=\triangle_1/\triangle =-48/-12=4 \\x_2= \triangle_2/\triangle = -18/-12= 3/2 \\ x_3=\triangle_3/ \triangle = -12/-12 =1}  \Rightarrow \bbox[red, 2pt]{ \cases{x_1= 4 \\x_2 = 3/2 \\ x_3  =1}}

解答A=\begin{bmatrix}1 & 6 &0 \\0 & 2& 1\\ 0 & 1& 2 \end{bmatrix} \Rightarrow \det(A-\lambda I)= -(\lambda-1)^2(\lambda-3)\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}0 & 6 & 0 \\0 & 1 & 1 \\0 & 1 & 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_2=x_3=0\\\qquad \Rightarrow v=x_1 \begin{pmatrix}1 \\0\\0 \end{pmatrix}, \text{ choose }v_1= \begin{pmatrix}1 \\0\\0 \end{pmatrix} \\ \lambda_2=3  \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-2 & 6 & 0 \\0 & -1 & 1 \\ 0 & 1 & -1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=3x_3\\ x_2=x_3} \\\qquad \Rightarrow v=x_3 \begin{pmatrix}3 \\1\\ 1 \end{pmatrix}, \text{ choose }v_2= \begin{pmatrix}3 \\1\\ 1 \end{pmatrix}\\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{1,3}, \text{ eigenvectors: } \bbox[red, 2pt]{\begin{pmatrix}1 \\0\\0 \end{pmatrix} ,\begin{pmatrix}3 \\1\\ 1 \end{pmatrix}}

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