國立臺北科技大學113學年度碩士班招生考試
系所組別: 能源與冷凍空調工程系碩士班甲、乙組
第一節 工程數學
解答:y″+3y′+2y=0⇒λ2+3λ+2=0⇒(λ+2)(λ+1)=0⇒λ=−1,−2⇒yh=c1e−x+c2e−2xyp=Ax2+Bx+C⇒y′p=2Ax+B⇒y″p=2A⇒y″p+3y′p+2yp=2Ax2+(6A+2B)x+(2A+3B+2C)=−2x2+3⇒{2A=−26A+2B=02A+3B+2C=3⇒{A=−1B=3C=−2⇒yp=−x2+3x−2⇒y=yh+yp⇒y=c1e−x+c2e−2x−x2+3x−2
解答:y=∞∑n=0anxn⇒y″=∞∑n=0n(n−1)anxn−2=∞∑n=0(n+2)(n+1)an+2xn⇒x2y=∞∑n=0anxn+2=∞∑n=0an−2xn(ak=0,k<0)⇒y″+x2y=∞∑n=0((n+2)(n+1)an+2+an−2)xn=0⇒(n+2)(n+1)an+2+an−2=0{n=−2⇒0⋅a0=0n=−1⇒0⋅a1=0n=0⇒2a2=0n=1⇒6a3=0⇒{a0arbitrarya1arbitrarya2=0a3=0 and an=−an−4n(n−1),n≥4⇒y=a0+a1x−∞∑n=4an−4n(n−1)xn,a2=a3=0⇒y=a0(1−112x4+1672x3+⋯)+a1(x−120x5+11440x9+⋯)
解答:L{y′}−4L{y}=L{1}⇒sY(s)−1−4Y(s)=1s⇒Y(s)=1s(s−4)+1s−4⇒y(t)=L−1{Y(s)}=L−1{1s(s−4)+1s−4}=54L−1{1s−4}−14L−1{1s}=54e4t−14⇒y=54e4t−14
解答:L−1{4s2+4s+20}=L−1{4(s+2)2+42}=e−2tsin(4t)
解答:f(t)=2t2+∫t0f(t−τ)e−τdτ⇒L{f(t)}=L{2t2}+L{f(t)}L{e−t}⇒F(s)=4s3+F(s)⋅1s+1⇒ss+1F(s)=4s3⇒F(s)=4(s+1)s4⇒f(t)=L−1{F(s)}=L−1{4(s+1)s4}=L−1{4ss4+4s4}=4⋅t22+4⋅t36⇒f(t)=2t2+23t3
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解題僅供參考,其他歷年試題及詳解
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