國立成功大學113學年度碩士班招生考試
系所:資源工程學系
科目:工程數學
解答:(a)dydx=y+x2⇒y′−y=x2⇒e−xy′−ye−x=x2e−x⇒(e−xy)′=x2e−x⇒e−xy=∫x2e−xdx=−e−x(x2+2x+2)+c1⇒y=−(x2+2x+2)+c1ex⇒y(0)=−2+c1=1⇒c1=3⇒y=3ex−(x2+2x+2)(b)y″+y=0⇒yh=c1cosx+c2sinx⇒yp=Axcosx+Bxsinx⇒y′p=Acosx+Bsinx−Axsinx+Bxcosx⇒y″p=−2Asinx+2Bcosx−Axcosx−Bxsinx⇒y″p+yp=−2Asinx+2Bcosx=cos(x)⇒{A=0B=1/2⇒yp=12xsinx⇒y=yh+yp=c1cosx+c2sinx+12xsinx⇒y′=(12−c1)sinx+c2cosx+12xcosx⇒{y(0)=c1=0y′(0)=c2=0⇒y=12xsinx解答:(a)L{exp(3t−2)}=∫∞0e3t−2e−stdt=e−2∫∞0e(3−s)tdt=e−2[13−se(3−s)t]|∞0=e−2s−3(b)L−1{s+2(s+1)s}=L−1{2s−1s+1}=L−1{2s}−L−1{1s+1}=2−e−t
解答:(a-1)A=[1112112112112111]R2−R1→R2,R3−R1→R3,R4−2R1→R4→[1112001−1010−10−1−1−3]R1−R3→R1,R1+R4→R4→[1013001−1010−100−1−4]R4+R2→R4→[1013001−1010−1000−5]R2↔R3,R4=R4/(−5)→[1013010−1001−10001]R1−R3→R1,R2+R4→[10040100001−10001]R1−4R4→R1,R3+R4→R3→[1000010000100001]=I4⇒rref(A)=[1000010000100001](a-2)rref(A)=I4⇒rank(A)=4⇒nullity(A)=4−rank(A)=0(b)B=[1111]⇒det
解答:\textbf{(a)}\;\vec F=3x^2y^3\vec i-2y^3z^2\vec j+x^3z^2\vec k \Rightarrow \text{curl }\vec F=\begin{vmatrix}\vec i & \vec j& \vec k \\{\partial \over \partial x} & {\partial \over \partial y} &{\partial \over \partial z} \\\vec F_1 & \vec F_2 & \vec F_3\end{vmatrix} =\begin{vmatrix}\vec i & \vec j& \vec k \\{\partial \over \partial x} & {\partial \over \partial y} &{\partial \over \partial z} \\3x^2y^3 & -2y^3z^2 & x^3z^2\end{vmatrix} \\ =0\vec i+ 0\vec j+0\vec k -9x^2y^2\vec k-3x^2z^2 \vec j+4y^3z \vec i \Rightarrow \text{curl }\vec F(2,-1,1) =\bbox[red, 2pt]{-4\vec i-12\vec j-36\vec k} \\\textbf{(b)}\; \cases{f(x,y, z)={xy^2\over z^3} \Rightarrow \nabla f =({y^2\over z^3},{2xy\over z^3}, -{3xy^2\over z^4} ) \\ \vec n= 2\vec i+2\vec j+ 1\vec k \Rightarrow{\vec n\over |\vec n|}={2\over 3}\vec i+ {2\over 3}\vec j+{1\over 3}\vec k} \\ \Rightarrow \nabla f\cdot {\vec n\over |\vec n|} ={2\over3}\cdot {y^2\over z^3} +{2\over 3}\cdot {2xy\over z^3 } -{1\over3}\cdot {3xy^2\over z^4}\\ \Rightarrow \nabla f(3,2,1)\cdot {\vec n\over |\vec n|} ={8\over 3}+8-12 = \bbox[red, 2pt]{-{4 \over 3}}
\textbf{(c)}\;\text{Let }M(x,y) \text{ and }N(x,y) \text{ be continuous and have continuous partial derivatives } \\\text{in a region }R \text{ and on its boundary }C. \text{ Then, for both simply-connected and }\\\text{ multiply-connected }R.\\ \oint_C (Mdx+ Ndy)= \displaystyle \iint_R \left( {\partial N\over \partial x} -{\partial M\over \partial y}\right)\,dxdy
(d) Let S be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation. Also let \vec F be a vector field then,\int_C \vec F\cdot d\vec r= \iint_S \text{curl }\vec F\cdot d\vec S
解答:\textbf{(a)}\;a_0= {1\over 2}\int_{-1}^1 f(x) \,dx = {1\over 2}\int_{0}^1 1\,dx ={1\over 2} \\ a_n= \int_0^1\cos(n\pi x)\,dx =\left.\left[{1\over n\pi} \sin(n\pi x) \right] \right|_0^1=0\\ b_n= \int_0^1 \sin(n\pi x)\,dx =\left.\left[-{1\over n\pi} \cos(n\pi x) \right] \right|_0^1= {1\over n\pi}(1-(-1)^n) \\ \Rightarrow f(x)=\bbox[red, 2pt]{{1\over 2}+\sum_{n=1}^\infty {1\over n\pi}(1-(-1)^n) \sin(n\pi x)} \\ \textbf{(b)}\;\mathcal F(\omega) =\int_{-\infty}^\infty f(t)e^{-j\omega t}\,dt =\int_0^\infty ke^{-at} e^{-j\omega t}\,dt = k\int_0^\infty e^{-(a+j\omega) t}\,dt =k \left. \left[-{1\over a+j\omega} e^{-(a+j\omega) t}\right] \right|_0^\infty \\= \bbox[red, 2pt]{k\over a+j\omega}
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解題僅供參考,其他碩士班歷年試題及詳解
第三題的(b)eigenvalues已求錯,故eigenvectors 也不對.
回覆刪除第四題的(b)方向導數不對,因為那單位向量是(2/3,2/3,1/3)
謝謝告知,已修訂
刪除