國立臺北科技大學113學年度碩士班招生考試
系所組別: 車輛工程系碩士班
第一節 工程數學
解答:1.y′=−4xy⇒∫ydy=−∫4xdx⇒12y2=−2x2+c1⇒y=±√−4x2+c22.y(2)=3⇒3=√−16+c2⇒c2=25⇒y=√25−4x2
解答:1.y″+y=0⇒λ2+1=0⇒λ=±i⇒yh=c1cosx+c2sinx2.{y1=cosxy2=sinx⇒W=|cosxsinx−sinxcosx|=1. Applying variations of parameters,yp=−cosx∫sinx(cosx−sinx)dx+sinx∫cosx(cosx−sinx)dx=−cosx(−14cos(2x)−12x+14sin(2x))+sinx(14sin(2x)+12x+14cos(2x))=14(cosx−sinx)+12x(cosx+sinx)⇒yp=14(cosx−sinx)+12x(cosx+sinx)3.y=yh+yp⇒y=c3cosx+c4sinx+12x(cosx+sinx)
解答:1.L{f(t)}=L{cos(ωt+θ)}=L{cos(ωt)cosθ−sin(ωt)sinθ}=cosθL{cosωt}−sinθL{sin(ωt)}=cosθ⋅ss2+ω2−sinθ⋅ωs2+ω2=scosθ−ωsinθs2+ω22.F(s)=s+1s2+9=ss2+32+13⋅3s2+32⇒L−1{F(s)}=L−1{ss2+32}+13L−1{3s2+32}=cos(3t)+13sin(3t)

解答:1.[A∣I]=[−1121003−11010−134001]R2+3R1→R2,R3+R1→R3→[−112100027310022−101]R2/2→R2,R3/2→R3→[−112100017232120011−12012]R1−R2→R1,R3−R2→R3→[−10−32−12−12001723212000−52−2−1212]−(2/5)R3→R3→[−10−32−12−1200172321200014515−15]R1+(3/2)R3→R2,R2−(7/2)R3→R2→[−100710−15−310010−1310−157100014515−15]−R1→R1→[100−71015310010−1310−157100014515−15]⇒A−1=[−71015310−1310−157104515−15]2.A=[1−10−12−10−11]⇒det(A−λI)=−λ(λ−1)(λ−3)=0⇒λ=0,1,3λ1=0⇒(A−λ1I)v=0⇒[1−10−12−10−11][x1x2x3]=0⇒{x1=x3x2=x3⇒v=x3(111),choose v1=(111)λ2=1⇒(A−λ2I)v=0⇒[0−10−11−10−10][x1x2x3]=0⇒{x2=0x1+x3=0⇒v=x3(−101),choose v2=(−101)λ3=3⇒(A−λ3I)v=0⇒[−2−10−1−1−10−1−2][x1x2x3]=0⇒{x1=x3x2+2x3=0⇒v=x3(1−21),choose v3=(1−21)⇒eigenvalues: 0,1,3, and eigenvectors: (111),(−101),(1−21)

解答:F(e−ax)=∫∞0e−axe−jωxdx=∫∞0e−(a+jω)xdx=[−1a+jωe−(a+jω)x]|∞0=1a+jω=a−jωa2+ω2
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