國立成功大學111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)y″+y=0⇒yh=c1cosx+c2sinxyp=Acos(2x)+Bsin(2x)⇒y′p=−2Asin(2x)+2Bcos(2x)⇒y″p=−4Acos(2x)−4Bsin(2x)⇒y″p+yp=−3Acos(2x)−3Bsin(2x)=sin(2x)⇒{A=0B=−1/3⇒yp=−13sin(2x)⇒y=yh+yp⇒y=c1cosx+c2sinx−13sin(2x)⇒y′=−c1sinx+c2cosx−23cos(2x)⇒{y(0)=c1=0y′(0)=c2−2/3=0⇒c2=23⇒y=23sinx−23sin(2x)(b)y′=y+ex⇒y′−y=ex⇒ integration factor I(x)=e∫−1dx=e−x⇒e−xy′−e−xy=1⇒(e−xy)′=1⇒e−xy=x+c1⇒y=xex+c1ex⇒y(0)=c1=2⇒y=xex+2ex解答:(a)∫∞0e3t+2e−stdt=e2∫∞0e(3−s)tdt=e2[13−se(3−s)t]|∞0=e2s−3(b)L−1{s+6s2+4s+20}=L−1{(s+2)+4(s+2)2+42}=e−2t(cos4t+sin4t)解答:(a-1)[13422−11−312311301]R2−2R1→R2,R3−R1→R3,R4−R1→R4→[13420−7−7−70−1−1−100−4−1]R1+3R3→R1,R2−7R3→R2→[101−100000−1−1−100−4−1]−R3→R3,−R4/4→R4→[101−1000001110011/4]R1−R4→R1,R2↔R3→[100−5/4011100000011/4]R2−R4→R2,R3↔R4→[100−5/40103/40011/40000](a-2)由rref(A)可知rank(A)=3(a-3)rank(A)=1−1+3+1=4(b)B=[010100001]⇒det(B−λI)=−(λ−1)2(λ+1)=0⇒eigenvalues: 1,−1λ1=1⇒(B−λ1I)v=0⇒[−1101−10000][x1x2x3]=0⇒x1=x2⇒v=x1(110)+x3(001),取v1=(110),v2=(001)λ2=−1⇒(B−λ1I)v=0⇒[110110002][x1x2x3]=0⇒{x1+x2=0x3=0⇒v=x2(−110),取v3=(−110)⇒eigenvectors: (110),(001),(−110)

解答:(a)→F(x,y,z)=(3xy2,−2yz2,4x2z)⇒div →F=∂∂x3xy2+∂∂y(−2yz2)+∂∂z4x2z=3y2−2z2+4x2⇒div →F(−1,2,3)=12−18+4=−2(b){F=x+y2+z3→v=(2,2,1)⇒{(Fx,Fy,Fz)=(1,2y,3z2)→n=→v‖→v‖=(23,23,13)⇒D→nF(x,y,z)=(1,2y,3z2)⋅(23,23,13)=23+4y3+z2⇒D→nF(3,2,−1)=23+83+1=133(c)假設C為平滑曲線所組合而成的封閉曲線,所圍區域為R,而P(x,y)及Q(x,y)為連續且一次偏導數也連續,則∮CPdx+Qdy=∬
解答:\textbf{(a)} \; f(x)為奇函數 \Rightarrow a_n=0,而b_n= \int_{-1}^0 -\sin(n\pi x)\,dx + \int_0^1 \sin(n\pi x)\,dx ={2\over n\pi}(1-(-1)^n)\\\quad \Rightarrow f(x)= \bbox[red, 2pt]{\sum_{n=1}^\infty {2\over n\pi}(1-(-1)^n) \sin(n\pi x)} \\\textbf{(a)} \; f(x)=e^{-x}, x\gt 0 \Rightarrow A(\alpha)=\int_0^\infty e^{-x} \cos(\alpha x)\,dx ={1\over 1+\alpha^2} \\ \quad \Rightarrow \bbox[red, 2pt]{ f(x)={2\over \pi}\int_0^\infty {\cos (\alpha x)\over 1+\alpha^2} \,dx} \\\textbf{(c)} \; F(\omega) =\int_{-\infty}^\infty e^{-|x|} e^{-j\omega x}\,dx =2\int_0^\infty e^{-x} e^{-j\omega x}\,dx = 2\int_0^\infty e^{-(j\omega+1) x}\,dx =2 \left. \left[ -{1\over j\omega +1} e^{-(j\omega+1)x}\right] \right|_0^\infty\\ \qquad =\bbox[red, 2pt]{2\over j\omega +1}========================= END =========================
解題僅供參考,其他歷年試題及詳解
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