國立臺灣師範大學115學年度碩士班招生考試
科目: 工程數學 適用系所:電機工程學系
解答:$$ \mathcal{L}\{f'(t)\} = \int_0^\infty e^{-st} f'(t) dt = \left[ e^{-st} f(t) \right]_0^\infty - \int_0^\infty (-s e^{-st}) f(t) dt =sF(s)-f(0)\\ \mathcal{L}\{f''(t)\} = \mathcal{L}\{(f'(t))'\} = s\mathcal{L}\{f'(t)\} - f'(0)=s(sF(s)-f(0))-f'(0)= s^2F(s)-sf(0)-f'(0)\\ \text{Assume the formula holds for the $k$-th derivative:}\\ \mathcal{L}\{f^{(k)}(t)\} = s^k F(s) - s^{k-1}f(0) - s^{k-2}f'(0) - \dots - f^{(k-1)}(0) \\ \Rightarrow \mathcal{L}\{f^{(k+1)}(t)\} = \mathcal{L}\{(f^{(k)}(t))'\} = s\mathcal{L}\{f^{(k)}(t)\} - f^{(k)}(0) \\ \qquad = s\left[ s^k F(s) - s^{k-1}f(0) - \dots - f^{(k-1)}(0) \right] - f^{(k)}(0) \\\qquad = s^{k+1} F(s) - s^k f(0) - \dots - sf^{(k-1)}(0) - f^{(k)}(0) \\ \Rightarrow \mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \dots - sf^{(n-2)}(0) - f^{(n-1)}(0) \\ \Rightarrow \bbox[red, 2pt]{ \mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - \sum_{k=1}^{n} s^{n-k} f^{(k-1)}(0) }$$
解答:$$y''-3y'+2y=0 \Rightarrow r^2-3r+2r=0 \Rightarrow (r-1)(r-2)=0 \Rightarrow r=1,2 \\ \Rightarrow y_h= c_1e^x+ c_2e^{2x} \Rightarrow y_p=(Ax^2 +Bx)e^{2x} \Rightarrow y_p'= [2Ax^2+(2A+2B)x+ B]e^{2x} \\ \Rightarrow y_p''= \left[4Ax^2 + (8A + 4B)x + (2A + 4B)\right]e^{2x} \Rightarrow y_p''-3y_p'+2y_p= xe^{2x} \\ \Rightarrow \left[4Ax^2 + (8A + 4B)x + (2A + 4B)\right]e^{2x} - 3\left[2Ax^2 + (2A + 2B)x + B\right]e^{2x} + 2\left[Ax^2 + Bx\right]e^{2x} = xe^{2x} \\ \Rightarrow 2Ax+2A+B=x \Rightarrow \cases{2A=1\\ 2A+B=0} \Rightarrow \cases{A=1/2\\ B=-1} \Rightarrow y_p= \left( {1\over 2}x^2-x \right)e^{2x} \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{ y = c_1 e^x + c_2 e^{2x} + \left(\frac{1}{2}x^2 - x\right)e^{2x} }$$
解答:$$ \frac{dy}{dx} = \frac{y^2 + 2y}{y^4 + 2xy + 4x} \Rightarrow \frac{dx}{dy} = \frac{y^4 + 2xy + 4x}{y^2 + 2y} = \frac{y^4}{y^2 + 2y} + \frac{2y + 4}{y^2 + 2y}x = \frac{y^4}{y(y + 2)} + \frac{2(y + 2)}{y(y + 2)}x \\ \Rightarrow \frac{dx}{dy} = \frac{y^3}{y + 2} + \frac{2}{y}x \Rightarrow \frac{dx}{dy} - \frac{2}{y}x = \frac{y^3}{y + 2} \Rightarrow \text{ integration factor } I(y)= e^{\int -(2/y)dy}={1\over y^2} \\ \Rightarrow \frac{1}{y^2}\frac{dx}{dy} - \frac{2}{y^3}x = \frac{1}{y^2}\left(\frac{y^3}{y + 2}\right) \Rightarrow \frac{d}{dy}\left( \frac{x}{y^2} \right) = \frac{y}{y + 2} \\ \Rightarrow \frac{x}{y^2} = \int \frac{y}{y + 2} \,dy = \int \left( 1-{2\over y+2} \right)\,dy =y-2\ln|y+2|+C \Rightarrow \bbox[red, 2pt]{x=y^3-2y^2 \ln|y+2|+Cy^2}$$
解答:$$\textbf{1) }\bbox[red, 2pt]{True} :\text{By definition, an $n \times n$ matrix $A$ is invertible if and only if it is row equivalent to }\\ \text{the $n \times n$ identity matrix $I$. Therefore, performing Gauss-Jordan elimination on $A$ will}\\ \text{ always result in the identity matrix as its reduced row echelon form (RREF).} \\ \textbf{2) }\bbox[red, 2pt]{False} :\cases{A=0\\ B= \begin{bmatrix}1& 2\\3& 4 \end{bmatrix}} \Rightarrow AB=BA=0 \\\textbf{3) }\bbox[red, 2pt]{False}: \text{The standard basis vectors $(1, 0)$ and $(0, 1)$ form an orthogonal basis. However,}\\ \text{ rotating these vectors by any angle $\theta$ (e.g., $45^\circ$) produces a new pair of orthogonal vectors}\\ \text{ that also serve as a completely valid orthogonal basis for the same space.} \\\textbf{4) }\bbox[red, 2pt]{True}: \text{A rotation matrix $R$ is an orthogonal matrix with a determinant of $+1$. Because} \\\text{ it is orthogonal, its inverse is equal to its transpose ($R^{-1} = R^T$). The transpose }\\ \text{of a matrix has the same determinant as the original matrix ($\det(R^T) = \det(R) = 1$).}\\ \text{ Since $R^{-1}$ is orthogonal and has a determinant of $+1$, it is also a valid rotation matrix.} \\ \textbf{5) }\bbox[red, 2pt]{False}: A^TA= I \Rightarrow \det(A^TA)= \det(A^T)\det(A) =(\det(A))^2=1 \Rightarrow \det(A)=\pm 1 \\ \textbf{6) }\bbox[red, 2pt]{False}: \text{This statement is only true for square matrices. For an $m \times n$ matrix $A$} \\\text{, $\text{rank}(A) + \text{nullity}(A) = n$ and $\text{rank}(A^T) + \text{nullity}(A^T) = m$} \\ \text{If $m \neq n$ (a non-square matrix), then $\text{nullity}(A)$ must be different from $\text{nullity}(A^T)$.}$$
解答:$$ \mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = \mathcal{L}\{\delta(t-\pi)\} - \mathcal{L}\{\delta(t-2\pi)\} \Rightarrow [s^2Y(s) - sy(0) - y'(0)] + 4Y(s) = e^{-\pi s} - e^{-2\pi s} \\ \Rightarrow (s^2+4)Y(s)=e^{-\pi s}-e^{-2\pi s} \Rightarrow Y(s) = \frac{e^{-\pi s}}{s^2 + 4} - \frac{e^{-2\pi s}}{s^2 + 4} \\ \Rightarrow y(t) = \mathcal{L}^{-1}\left\{Y(s)\right\}= \mathcal{L}^{-1}\left\{\frac{e^{-\pi s}}{s^2 + 4}\right\} - \mathcal{L}^{-1}\left\{\frac{e^{-2\pi s}}{s^2 + 4}\right\}\\ ={1\over 2}\sin(2(t-\pi))u(t-\pi)-{1\over 2} \sin(2(t-2\pi)) u(t-2\pi)\\ \Rightarrow \bbox[red, 2pt]{y(t) = \frac{1}{2}\sin(2t)u(t-\pi) - \frac{1}{2}\sin(2t)u(t-2\pi) }$$
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