臺灣綜合大學系統107學年度學士班轉學生聯合招生考試
科目名稱:工程數學
類組代碼:D36
解答:f(x)=ex⇒f[n](x)=ex⇒f[n](0)=1,n≥0⇒泰勒級數=∞∑n=01n!xnf[n](0)=∞∑n=01n!xn=1+x+12x2+13!x3+⋯+1n!xn+⋯
解答:ex≈∞∑n=01n!xn=1+x+12x2+13!x3+⋯+1n!xn+⋯⇒e−x≈1−x+12x2−13!x3+⋯+1n!(−1)nxn+⋯⇒∫101−exxdx≈∫101−12x+13!x2−⋯+1n!(−1)n−1xn−1+⋯dx=[x−14x2+118x3−196x4+1600x5−⋯]|10≈1−14+118−196+1600≈0.797
解答:y′+4y=20⇒∫120−4ydy=∫1dx⇒−14ln(20−4y)=x+C1⇒ln(20−4y)=−4x−4C1將y(0)=2代入⇒ln12=−4C1⇒ln(20−4y)=−4x+ln12⇒20−4y=12e−4x⇒y=5−3e−4x
解答:取u=1y⇒u′=−1y2y′=−u2y′⇒y′=−1u2u′代回原式⇒−xu2u′−3u+x4u2=0⇒xu′+3u=x4⇒u′+3xu=x3⇒積分因子I(x)=e∫(3/x)dx=x3⇒x3u′+3x2u=x6⇒(x3u)′=x6⇒x3u=∫x6dx=17x7+C⇒x3y=17x7+C將y(1)=1代入上式⇒1=17+C⇒C=67⇒y=7x3x7+6
解答:y‴−5y″+8y′−4y=0⇒λ3−5λ2+8λ−4=0⇒(λ−1)(λ−2)2=0⇒y=C1ex+C2e2x+C3xe2x⇒y′=C1ex+(2C2+C3)e2x+2C3xe2x⇒y″=C1ex+(4C2+4C3)e2x+4C3xe2x⇒{y(0)=0=C1+C2y′(0)=1=C1+2C2+C3y″(0)=−1=C1+4C2+4C3⇒{C1=−5C2=5C3=−4⇒y=−5ex+5e2x−4xe2x

解答:dωdt+3CD2(2G+1)dω2=2(G−1)g2G+1≡ω′+Aω2=B⇒∫1B−Aω2dω=∫1dt⇒tanh−1(√ABω)√AB=t+C,由初始值ω(0)=0⇒C=0⇒√ABω=tanh(t√AB)⇒ω=√BAtanh(t√AB)=√4(G−1)gd3CDtanh(t√3CD(G−1)g(2G+1)2d)⇒ω(t)=√4(G−1)gd/(3CD)⋅tanh(t√3CD(G−1)g/(2G+1)2d)

解答:u(x,t)=X(x)T(t)⇒uxx−3ut=X″T−3XT′=0⇒X″X=3T′T=k,k為常數⇒X″-kX=0,以下就k值分別討論:Case I: k=0⇒X″=0⇒X=C1x+C2,由於u(0,t)=u(2,t)=0,即X(0)=X(2)=0⇒{C2=02C1+C2=0⇒X=0為明顯解,不討論Cases II: k>0⇒X″−kX=0⇒X=C1e√kx+C2e−√kx⇒{X(0)=C1+C2=0X(2)=C1e2√k+C2e−2√k=0⇒{C2=−C1C1e4√k+C2=0⇒C1e4√k−C1=0⇒C1(e4√k−1)=0⇒C1=0(k>0⇒e4√k≠1)⇒C1=C2=0⇒X=0為明顯解,不討論Cases III:k<0⇒k=−t2,t≠0,t∈R⇒X″+t2X=0⇒X=C1costx+C2sintx⇒{X(0)=0=C1X(2)=0=C1cos2t+C2sin2t⇒C2sin2t=0⇒{C2=0⇒C1=C2=0⇒X=0為明顯解2t=mπ,m∈Z⇒t=mπ/2⇒X=C2sinmπx2;又t=mπ/2⇒k=−m2π2/4⇒T′−k3T=0⇒T=C3ekt/3=C3e−m2π2t/12因此u(x,t)=X(x)T(t)=∞∑n=1Ane−n2π2t/12sin(nπx/2),再由u(x,0)=2sin(3πx)⇒{A6=2An=0,n≠6⇒u(x,t)=2e3π2tsin(3πx)
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