國立臺北科技大學113學年度碩士班招生考試
系所組別: 製造科技研究所
第一節 微分方程
解答: y′=y2e−x⇒1y2dy=e−xdx⇒−1y=−e−x+c1⇒y=1e−x−c1

解答: {P(x,y)=cosx−2xyQ(x,y)=ey−x2⇒Py=−2x=Qx⇒Exact⇒Φ(x,y)=∫(cosx−2xy)dx=∫(ey−x2)dy⇒sinx−x2y+ϕ(y)=ey−x2y+ρ(x)⇒Φ(x,y)=sinx−x2y+ey+c1=0
解答: y′+1xy=1x3y−4/3⇒x3y4/3y′+x2y7/3=2⇒(x2y7/3−2)dx+x3y4/3dy=0Let {P(x,y)=x2y7/3−2Q(x,y)=x3y4/3⇒{Py=73x2y4/3Qx=3x2y4/3⇒Qx−PyQ=23x (depend on x)⇒u′=23xu⇒integrating factor u(x)=x−2/3⇒{uP=x4/3y7/3−2x−2/3uQ=x7/3y4/3⇒(uP)y=73x4/3y4/3=(uQ)x⇒ exact⇒Φ(x,y)=∫uPdx=∫uQdy⇒Φ=∫(x4/3y7/3−2x−2/3)dx=∫x7/3y4/3dy⇒Φ=37x7/3y7/3−6x1/3+ϕ(y)=37x7/3y7/3+ρ(x)⇒Φ=37x7/3y7/3−6x1/3=c1⇒y=((7/3)(6x1/3+c1)x7/3)3/7⇒y=(14x1/3+c2x7/3)3/7

解答: y″+2y′+2y=0⇒λ2+2λ+2=0⇒λ=−1±i⇒yh=e−x(c1cosx+c2sinx)yp=Acos(3x)+Bsin(3x)⇒y′p=−3Asin(3x)+3Bcos(3x)⇒y″p=−9Acos(3x)−9Bsin(3x)⇒y″p+2y′p+2yp=(−7A+6B)cos(3x)−(6A+7B)sin(3x)=3.5sin(3x)−3cos(3x)⇒{−7A+6B=−36A+7B=−3.5⇒{A=0B=−1/2⇒yp=−12sin(3x)⇒y=yh+yp=e−x(c1cosx+c2sinx)−12sin(3x)⇒y′=e−x((c2−c1)cosx−(c1+c2)sinx)−32cos(3x)⇒{y(0)=c1=0y′(0)=c2−3/2=1/2⇒c2=2⇒y=2e−xsinx−12sin(3x)

解答: L{y″}+4L{y}=L{f(t)}=L{tu(t−3)}⇒s2Y(s)+4Y(s)=e−3s(3s+1s2)⇒Y(s)=e−3s(3s(s2+4)+1s2(s2+4))⇒y(t)=L−1{Y(s)}=L−1{e−3s(3s(s2+4)+1s2(s2+4))}=u(t−3)L−1{3s(s2+4)+1s2(s2+4)}(t−3)⇒y(t)=u(t−3)(34−34cos(2(t−3))+t−34−18sin(2(t−3)))

解答: Let u(x,t)=X(x)T(t),then ∂u∂t=k∂2u∂x2⇒XT′=kX″T⇒T′T=kX″XBC.{u(0,t)=X(0)T(t)=0u(L,t)=X(L)T(t)=0⇒{X(0)=0X(L)=0Suppose that T′kT=X″X=λCases I. λ=0⇒X″=0⇒X=c1x+c2⇒BC.{X(0)=c2=0X(L)=c1L+c2=0⇒c1=c2=0⇒X=0⇒u=0Case II. λ>0⇒λ=ρ2(ρ≠0)⇒X″−ρ2X=0⇒X=c1eρx+c2e−ρxBC.{X(0)=c1+c2=0X(L)=c1eρL+c2e−ρL=0⇒c2=−c1⇒c1eρL−c1e−ρL=0⇒c1(e2ρL−1)=0⇒c1=0⇒c2=0⇒X=0⇒u=0Case III. λ<0⇒λ=−ρ2(ρ>0)⇒X″+ρ2X=0⇒X=c1cos(ρx)+c2sin(ρx)⇒BC.{X(0)=c1=0X(L)=c2sin(ρL)=0⇒sin(ρL)=0⇒ρ=nπL⇒X=sinnπxL,n∈NT′=λkT⇒T=eλkt=e−n2π2kt/L2⇒un(x,t)=e−n2π2kt/L2sinnπxL⇒u(x,t)=∞∑n=1ane−n2π2kt/L2sinnπxL⇒u(x,0)=∞∑n=1ansinnπxL=A⇒an=2L∫L0AsinnπxLdx⇒u(x,t)=∞∑n=1ane−n2π2kt/L2sinnπxL,an=2L∫L0AsinnπxLdx
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解題僅供參考,其他歷年試題及詳解
請問第四題2e^-x*sinx 是不是少個"-"號
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