臺灣聯合大學系統109學年度學士班轉學生聯合招生考試
科目名稱:工程數學
類組別:A5
解答:(a)y′+xy=xy−1⇒ydydx+xy2=x⇒ydydx=x(1−y2)⇒∫y1−y2dy=∫xdx⇒−12ln|1−y2|=12x2+C1⇒ln|1−y2|=−x2+C2⇒1−y2=C3e−x2⇒y=±√1−C3e−x2,將y(0)=3代入⇒C3=−8⇒y=√1+8e−x2(b)令y=xm⇒y′=mxm−1⇒y″=m(m−1)xm−2⇒x2y″−3xy′+3y=0⇒m(m−1)xm−3mxm+3x3=0⇒(m2−4m+3)xm=0⇒(m−3)(m−1)=0⇒m=1,3⇒yh=C1x+C2x3⇒y′h=C1+3C2x,將{y(1)=0y′(0)=1代入yh⇒{0=C1+C2C1=1⇒{C1=1C2=−1⇒yh=x−x3令yp=Alnx⇒y′p=Ax⇒y″p=−Ax2⇒x2y″p−3xy′p+3yp=−A−3A+3Alnx=3lnx−4⇒A=1⇒yp=lnx⇒y=yh+yp⇒y=x−x3+lnx解答:y″−y=t⇒L{y″}−L{y}=L{t}⇒s2Y(s)−sy(0)−y′(0)−Y(s)=1s⇒s2Y(s)−s−1−Y(s)=1s⇒(s2−1)Y(s)=s+1+1s⇒Y(s)=s2+s+1s(s2−1)=−1s+12⋅1s+1+32⋅1s−1⇒y(t)=L−1{Y}=−L{1s}+12L{1s+1}+32L{1s−1}=−1+12e−t+32et⇒y(t)=−1+12e−t+32et
解答:{m=1k=1F(t)=1−t2/π2代入−ky+F(t)=my″⇒y″+y=1−t2/π2⇒y=C1cost+C2sint−t2π2+1+2π2⇒y′=−C1sint+C2cost−2tπ2將初始值{y(0)=0y′(0)=0代入⇒{C1+1+2π2=0C2=0⇒{C1=−1−2/π2C2=0⇒y″+y=1−t2/π2,其解為y(t)=(−1−2/π2)sint−t2/π2+1+2/π2
解答:A=[−2−λ2−321−λ−6−1−20−λ]⇒det(A−λI)=|−2−λ2−321−λ−6−1−20−λ|=−(λ−5)(λ+3)2=0⇒λ1=5,λ2=−3λ1=5⇒(A−λ1I)x=[−72−32−4−6−1−2−5][x1x2x3]=0⇒{x1+x3=0x2+2x3=0,取v1=[12−1]λ2=−3⇒(A−λ2I)x=[12−324−6−1−23][x1x2x3]=0⇒x1+2x2=3x3,取v2=[2−10],v3=[301]⇒特徵值為5及−3,相對應的特徵向量為[12−1],[2−10]及[301]
解答:(a){x(t)=acos3ty(t)=asin3t⇒{x′(t)=−3acos2tsinty′(t)=3asin2tcost⇒{(x′(t))2=9a2cos4tsin2t(y′(t))2=9a2sin4tcos2t⇒∫π/20√(x′(t))2+(y′(t))2dt=∫π/20√9a2cos2tsin2t(cos2t+sin2t)dt=∫π/20√9a2cos2tsin2tdt=∫π/203acostsintdt=∫π/2032asin(2t)dt=[−34acos(2t)]|π/20=34a+34a=32a(b)f(x,y)=16x2−y2=399⇒∇f=(fx,fy)=(32x,−2y)⇒∇f(1/8,1)=(4,−2)⇒(4,−2)√42+22=(2√5,−1√5)註:P不在surface上,題目有誤?!(c)f=x2+y2+z2⇒∇f=(fx,fy,fz)=(2x,2y,2z)⇒∇f(2,2,−1)=(4,4,−2)⇒∇f(2,2,−1)⋅→a=(4,4,−2)⋅(1,1,3)=4+4−6=2
解答:{A→x=→e1A→y=→en⇒A→z=2→e1+3→en=2A→x+3A→y⇒→z=2→x+3→y
解答:∫S→u⋅→ndA=∫V∇⋅→udV=∫1/2−1/2∫1/2−1/2∫1/2−1/2(∂∂x,∂∂y,∂∂z)⋅(x,y,z)dxdydz=∫1/2−1/2∫1/2−1/2∫1/2−1/23dxdydz=3
解答:依d'Alembert's solution,utt=c2uxx的通解為u(x,t)=F(x+ct)+G(x−ct)=F(x+t)+G(x−t)=12(f(x+t)+f(x−t))+12∫x+tx−tg(s)ds=12(sin(π(x+t))+3sin(3π(x+t))+sin(π(x−t))+3sin(3π(x−t))+0⇒u(x,t)=12(sin(π(x+t))+3sin(3π(x+t))+sin(π(x−t))+3sin(3π(x−t))
解答:令u(x,t)=X(x)T(t)⇒{ut=XT′uxx=X″T⇒ut=uxx−u⇒XT′=X″T−XT⇒T′T=X″−XX=k⇒k為一常數⇒X″−(k+1)X=0原邊界條件{u(0,t)=X(0)T(t)=0u(1,t)=X(1)T(t)=0,若T(t)=0⇒u(x,t)=0為明顯解,不討論;因此邊界條件變為X(1)=X(0)=0Case I: k=−1⇒X″=0⇒X=ax+b代入邊介條件⇒{X(1)=a+b=0X(0)=b=0⇒X=0⇒u(x,t)=0為明顯解,不討論Case II: k>−1⇒X=C1e√k+1x+C2e−√k+1x代入邊介條件⇒{X(1)=C1e√k+1+C2e−√k+1=0X(0)=C1+C2=0⇒C1=C2=0⇒X=0⇒u(x,t)=0為明顯解,不討論Case III: k<−1,假設k+1=−μ2⇒X=Acosμx+Bsinμx代入邊界條件⇒{Acosμ+Bsinμ=0A=0⇒{A=B=0⇒明顯解,不討論A=0且sinμ=0⇒sinu=0⇒u=nπ⇒Xn(x)=sinnπx,n∈N接著求T′T=k⇒T′−kT=0⇒T′+(u2+1)T=0⇒T(t)=Ae−(u2+1)t⇒Tn(t)=Ane−(n2π2+1)t⇒u(x,t)=Ane−(n2π2+1)tsin(nπx);再將初始條件u(x,0)=sin(πx)⇒Ansin(nπx)=sin(πx)⇒{A1=1Am=0,m≥2⇒u(x,t)=e−(π2+1)tsin(πx)
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