國立中山大學113學年度碩士班招生考試
科目名稱:工程數學【光電系碩士班】
解答:y′+2y=4e−2x⇒e2xy′+2e2xy=4⇒(e2xy)′=4⇒e2xy=4x+c1⇒y=4xe−2x+c1e−2x
解答:cos(x+y)dx+(3y2+2y+cos(x+y))dy=0⇒{P(x,y)=cos(x+y)Q(x,y)=3y2+2y+cos(x+y)⇒Py=−sin(x+y)=Qx⇒exact⇒Φ(x,y)=∫Pdx=∫Qdy⇒Φ=∫cos(x+y)dx=∫(3y2+2y+cos(x+y))dy⇒Φ=sin(x+y)+ϕ(y)=y3+y2+sin(x+y)+ρ(x)⇒Φ=sin(x+y)+y2+y2+c1=0解答:y″+3y+2=0⇒λ2+3λ+2=0⇒(λ+2)(λ+1)=0⇒λ=−1,−2⇒y=c1e−x+c2e−2x
解答:dydx=(x+1)e−xy2⇒1y2dy=(x+1)e−xdx⇒−1y=−xe−x−2e−x+c1⇒y=1xe−x+2e−x+c2

解答:cos(wt)=12(eiwt+e−iwt)⇒L{f(t)}=∫∞0eatcos(wt)e−stdt=12∫∞0(e(iw+a−s)t+e(−iw+a−s)t)dt=12[1iw+a−se(−iw+a−s)t+1−iw+a−se(−iw+a−s)t]|∞0=12[−1iw+a−s−1−iw+a−s]=s−a(a−s)2+w2
解答:{x1−x2+x3=0−x1+x2−x3=010x2+25x3=9020x1+10x2=80=[1−11−11−10102520100][x1x2x3]=[009080]augmented matrix:[1−110−11−1001025902010080]R1+R2→R2,R4−20R1→R4→[1−11000000102590030−2080]R3/10→R3→[1−110000001529030−2080]R1+R3→R1,R4−30R3→R4→[1072900000152900−95−190]R4/(−95)→R4→[107290000015290012]R1−(7/2)R4→R2,R3−(5/2)R4→R3→[1002000001040012]⇒{x1=2x2=4x3=2

解答:[A∣I]=[−1121003−11010−134001]−R1→R1→[1−1−2−1003−11010−134001]R2−3R1→R2,R1+R3→R3→[1−1−2−100027310022−101]R2/2→R2→[1−1−2−100017232120022−101]R1+R2→R1,R3−2R2→R2→[10321212001723212000−5−4−11]R3/(−5)→R3→[1032121200172321200014515−15]R1−1.5R3→R1,R2−3.5R3→R2→[100−71015310010−1310−157100014515−15]⇒A−1=[−71015310−1310−157104515−15]
解答:F(f(x))=∫∞−∞f(x)e−iωtdt=∫1−1e−iωtdt=[1−iωe−iωt]|1−1=1−iω(e−iω−eiω)=1−iω(−2isinω)=2sinωω解答:ex.digitalsignalprocessing...
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解題僅供參考,其他歷年試題及詳解
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