國立中山大學110學年度碩士班招生考試
科目名稱:工程數學【海下所碩士班】
解答:limx→∞lnxxa+1=limx→∞ddxlnxddxxa+1=limx→∞1/x(a+1)xa=limx→∞1(a+1)xa+1=0解答:(1) v1=u1=<3,1>⇒e1=v1|v1|=<3√10,1√10>=<3√1010,√1010>v2=u2−(u2⋅e1)e1=<1,1>−2√105<3√1010,√1010>=<−15,35>⇒e2=v2|v2|=<−√1010,3√1010>⇒ an orthonormal basis: {e1,e2}={<3√1010,√1010>,<−√1010,3√1010>}(2) v1=u1=<1,1,1>⇒e1=v1|v1|=<√33,√33,√33>v2=u2−(u2⋅e1)e1=<1,2,2>−5√33<√33,√33,√33>=<−23,13,13>⇒e2=v2|v2|=<−√63,√66,√66>v3=u3−(u3⋅e1)e1−(u3⋅e2)e2=<0,12,−12>⇒e3=v3|v3|=<0,√22,−√22>⇒ an orthonormal basis: {e1,e2,e3}={<√33,√33,√33>,<−√63,√66,√66><0,√22,−√22>}
解答:A=[22−11−10010]⇒AT=[2102−11−100]⇒{B=12(A+AT)C=12(A−AT)⇒{B=[23/2−1/23/2−11/2−1/21/20]C=[01/2−1/2−1/20−1/21/21/20]
解答:dydx+y=x⇒exdydx+yex=xex⇒(exy)′=xex⇒exy=∫xexdx=xex−ex+c1⇒y=x−1+c1ex⇒y(0)=−1+c1=3⇒c1=4⇒y=x−1+4e−x
解答:(1) L{sin(2t)}=∫∞0e−stsin(2t)dt=12i∫∞0e−st(e2ti−e−2ti)dt=12i∫∞0(e(2i−s)t−e−(2i+s)t)dt=12i[12i−se(2i−s)t+12i+se−(2i+s)t]|∞0=−12i(12i−s+12i+s)=−12i(2i+s−4−s2+2i−s−4−s2)=12i⋅4is2+4=2s2+4(2) L{dydt}+3L{y}=13L{sin(2t)}⇒sY(s)−6+3Y(s)=13⋅2s2+4⇒(s+3)Y(s)=26s2+4+6⇒Y(s)=26(s2+4)(s+3)+6s+3⇒y(t)=L−1{Y(s)}=L−1{26(s2+4)(s+3)+6s+3}=L−1{−2s+6s2+4+8s+3}⇒y(t)=−2cos(2t)+3sin(2t)+8e−3t
解答:Bessel’s equation of order ν is given byx2y″+xy′+(x2−v2)y=0
解答:y″x2+(x2−81)y=−xy′⇒y″x2+xy′+(x2−92)y=0 is a Bessel's equation of order v=9⇒y=c1J9(x)+c2Y9(x), where J9(x)=∞∑n=0(−1)nn!Γ(n+10)(x2)2n+9,J−9(x)=∞∑n=0(−1)nn!Γ(n−8)(x2)2n−9 and Y9(x)=limv→9cosvπJv(x)−J−v(x)sinvπ
解答:Case I k=0:y″=0⇒y=c1x+c2⇒{y(0)=c2=0y(L)=c1L+c2=0⇒{c1=0c2=0⇒y=0Cases II k>0:y″+ky=0⇒y=c1cos(√kx)+c2sin(√kx)⇒y(0)=c1=0⇒y(L)=c2sin(√kL)=0⇒√kL=nπ⇒k=n2π2L2,n∈NCase III k<0:y″+ky=0⇒y=c1e√kx+c2e−√kx⇒y(0)=c1+c2=0⇒c2=−c1⇒y=c1e√kx−c1e−√kx⇒y(L)=c1e√kL−c1e−√kL=0⇒c1(e2√kL−1)=0⇒c1=0⇒c2=0⇒y=0Summary, eigenfunctions: yn(x)=cnsin(nπxL),n∈N
解答:f(t) with period 2⇒f(t)={10≤t≤1−11≤t≤2≡f(t)={10≤t≤1−1−1≤t≤0⇒cn=12∫1−1f(t)e−inπtdt=12(∫0−1−e−inπtdx+∫10e−inπtdt)=12([1inπe−inπt]|0−1+[−1inπe−inπt]|10)=12(2inπ−1inπ(einπ+e−inπ))=1inπ(1−(−1)n),n=0,±1,±2,⋯⇒f(t)=∞∑n=−∞1inπ(1−(−1)n)einπt
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解題僅供參考,碩士班歷年試題及詳解
第5題的(2),倒數第二行中,不是2/(s+3),要是8/(s+3),故最後答案後面應是,8e^(-3t)
回覆刪除謝謝,已修訂
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