國立雲林科技大學111學年度碩士班招生考試
系所:電機系
科目:工程數學
解答:1.y′=5sin(3x)⇒y=∫5sin(3x)dx=−53cos(3x)+c1⇒y=−53cos(3x)+c12.積分因子I(x)=e∫5x4dx=ex5⇒y′I(x)+5x4yI(x)=3x2e−x5⇒y′ex5+5x4yex5=3x2⇒(yex5)′=3x2⇒yex5=x3+c1⇒y=x3e−x5+c1e−x53.x2y′=y2+xy⇒y′−yx=y2x2為白努利方程式因此取v=1y⇒v′=−y′y2⇒y′=−y2v′代入原式⇒−y2v′−yx=y2x2⇒v′+1xy=−1x2⇒v′+vx=−1x2⇒xv′+v=−1x⇒(xv)′=−1x⇒xv=−lnx+c1⇒xy=−lnx+c1⇒y=x−lnx+c1解答:先求齊次解,y″+9y=0⇒λ2+9=0⇒λ=±3i⇒yh=c1cos(3x)+c2sin(3x)yp=Acosx+Bsinx⇒y′p=−Asinx+Bcosx⇒y″p=−Acosx−Bsinx⇒y″p+9yp=8Acosx+8Bsinx=16sinx⇒{A=0B=2⇒yp=2sinx⇒y=yh+yp=c1cos(3x)+c2sin(3x)+2sinx⇒y′=−3c1sin(3x)+3c2cos(3x)+2cosx⇒{y(π)=−c1=0y′(0)=3c2+2=0⇒{c1=0c2=−23⇒y=−23sin(3x)+2sinx
解答:(1)F(s)=L{t2}+L{e−6t}+L{sin(6t)}=2s3+1s+6+6s2+36(2)f(t)=L−1{8s(s−2)2}=L−1{2s−2s−2+4(s−2)2}=2−2e2t+4e2tt(3)L{y′}+L{∫t0y(τ)dτ}=2L{1}−2L{sint}⇒sY(s)+Y(s)=2s−2s2+1⇒Y(s)=2s(s+1)−2(s2+1)(s+1)⇒y(t)=L−1{2s(s+1)}−L−1{2(s2+1)(s+1)}=2L−1{1s−1s+1}−L−1{1s2+1−ss2+1+1s+1}=2−2e−t−(sint−cost+e−t)⇒y(t)=2−3e−t−sint+cost
解答:y′−3y=te3tsin(t)⇒L{y′}−3L{y}=L{te3tsin(t)}⇒sY(s)−3Y(s)=−ddsL{e3tsin(t)}=−dds1(s−3)2+1=2s−6(s2−6s+10)2⇒Y(s)=2(s2−6s+10)2⇒y(t)=L−1{2(s2−6s+10)2}=L−1{2((s−3)2+1)2}=e3tL−1{2(s2+1)2}=e3t(sint−tcost)⇒y(t)=e3t(sint−tcost)
解答:(a)A=[003021201]⇒det\mathbf{(b)}\; X=[v_1 v_2 v_3], D=\begin{bmatrix}\lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix} \\\Rightarrow \bbox[red, 2pt]{A=\begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}-2 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} -3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}^{-1}} \\ \Rightarrow A^5= \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}(-2)^5 & 0 & 0\\ 0 & 2^5 & 0 \\ 0 & 0 & 3^5 \end{bmatrix} \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}^{-1} \\= \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}-32 & 0 & 0\\ 0 & 32 & 0 \\ 0 & 0 & 243 \end{bmatrix} \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}^{-1}\Rightarrow \bbox[red, 2pt]{A^5 = \left[ \begin{matrix} 78 & 0 & 165 \\78 & 32 & 133 \\110 & 0 & 133\end{matrix}\right]}


解答:A=\left[\begin{matrix}1 & 4 & 2 & 1\\0 & 1 & 1 & -1\\-2 & -8 & -4 & -2 \end{matrix}\right] \Rightarrow rref(A)= \left[ \begin{matrix}1 & 0 & -2 & 5\\0 & 1 & 1 & -1\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow \bbox[red, 2pt]{Rank(A)=2} \\ 又\left[ \begin{matrix}1 & 0 & -2 & 5\\0 & 1 & 1 & -1\\0 & 0 & 0 & 0\end{matrix}\right] \begin{bmatrix} x_1\\ x_2\\ x_3\\x_4\end{bmatrix} =0 \Rightarrow \cases{x_1-2x_3+5x_4=0\\ x_2+x_3-x_4=0} \\ \Rightarrow \begin{bmatrix} x_1\\ x_2\\ x_3\\x_4\end{bmatrix} = \begin{bmatrix} 2x_3-5x_4\\ -x_3+x_4\\ x_3\\x_4\end{bmatrix} = x_3 \begin{bmatrix} 2\\ -1\\ 1\\0\end{bmatrix}+ x_4 \begin{bmatrix} -5\\ 1\\ 0 \\1 \end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{ Null(A)=Span\left\{ \begin{bmatrix} 2\\ -1\\ 1\\0\end{bmatrix}, \begin{bmatrix} -5\\ 1\\ 0 \\1 \end{bmatrix} \right\}, Nulity(A)=2}
解答:A=\left[\begin{matrix}1 & 0 & 1\\7 & 7 & 8\\1 & 2 & 1\\7 & 7 & 6\end{matrix}\right] \Rightarrow rref(A)=\left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0 \\0 & 0 & 1\\0 & 0 & 0\end{matrix}\right] \Rightarrow \text{a basis for column space of }A=\{a_1,a_2, a_3\},\\ \text{where }a_1=\begin{bmatrix} 1\\ 7\\1\\7 \end{bmatrix}, a_2=\begin{bmatrix} 0 \\ 7 \\ 2\\7 \end{bmatrix}, a_3= \begin{bmatrix}1\\ 8\\ 1\\ 6\end{bmatrix} \\ b_1=a_1 \Rightarrow e_1={b_1\over \Vert b_1\Vert}=\begin{bmatrix} 1/10 \\ 7/10 \\ 1/10\\7/10 \end{bmatrix}\\ b_2=a_2-(a_2\cdot e_1)e_1 =\begin{bmatrix} 0 \\ 7 \\ 2\\7 \end{bmatrix}-\begin{bmatrix} 1\\ 7\\1\\7 \end{bmatrix}= \begin{bmatrix} -1\\ 0\\1\\0 \end{bmatrix} \Rightarrow e_2={b_2\over \Vert b_2 \Vert} = \begin{bmatrix} -1/\sqrt 2\\ 0\\1/\sqrt 2\\0 \end{bmatrix} \\ b_3=a_3-(a_3\cdot e_1)e_1-(a_3\cdot e_2)e_2 =\begin{bmatrix}1\\ 8\\ 1\\ 6\end{bmatrix}-\begin{bmatrix} 1\\ 7\\1\\7 \end{bmatrix} -0=\begin{bmatrix} 0\\ 1\\0\\-1 \end{bmatrix}\\ \qquad \Rightarrow e_3={b_3\over \Vert b_3 \Vert}=\begin{bmatrix} 0\\ 1/\sqrt 2\\0\\-1/\sqrt 2 \end{bmatrix}\\ \Rightarrow \text{an orthonomal basis for the column space of }A= \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1/10 \\ 7/10 \\ 1/10\\7/10 \end{bmatrix}, \begin{bmatrix} -1/\sqrt 2\\ 0\\1/\sqrt 2\\0 \end{bmatrix} , \begin{bmatrix} 0\\ 1/\sqrt 2\\0\\-1/\sqrt 2 \end{bmatrix} \right\}}
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解題僅供參考,其他歷年試題及詳解
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