2024年4月13日 星期六

113年臺北科技大學製造所碩士班-微分方程詳解

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=y2ex1y2dy=exdx1y=ex+c1y=1exc1



解答: {P(x,y)=cosx2xyQ(x,y)=eyx2Py=2x=QxExactΦ(x,y)=(cosx2xy)dx=(eyx2)dysinxx2y+ϕ(y)=eyx2y+ρ(x)Φ(x,y)=sinxx2y+ey+c1=0
解答: y+1xy=1x3y4/3x3y4/3y+x2y7/3=2(x2y7/32)dx+x3y4/3dy=0Let {P(x,y)=x2y7/32Q(x,y)=x3y4/3{Py=73x2y4/3Qx=3x2y4/3QxPyQ=23x (depend on x)u=23xuintegrating factor u(x)=x2/3{uP=x4/3y7/32x2/3uQ=x7/3y4/3(uP)y=73x4/3y4/3=(uQ)x exactΦ(x,y)=uPdx=uQdyΦ=(x4/3y7/32x2/3)dx=x7/3y4/3dyΦ=37x7/3y7/36x1/3+ϕ(y)=37x7/3y7/3+ρ(x)Φ=37x7/3y7/36x1/3=c1y=((7/3)(6x1/3+c1)x7/3)3/7y=(14x1/3+c2x7/3)3/7

解答: y+2y+2y=0λ2+2λ+2=0λ=1±iyh=ex(c1cosx+c2sinx)yp=Acos(3x)+Bsin(3x)yp=3Asin(3x)+3Bcos(3x)yp=9Acos(3x)9Bsin(3x)yp+2yp+2yp=(7A+6B)cos(3x)(6A+7B)sin(3x)=3.5sin(3x)3cos(3x){7A+6B=36A+7B=3.5{A0B=1/2yp=12sin(3x)y=yh+yp=ex(c1cosx+c2sinx)12sin(3x)y=ex((c2c1)cosx(c1+c2)sinx)32cos(3x){y(0)=c1=0y(0)=c23/2=1/2c2=2y=2exsinx12sin(3x)

解答: L{y}+4L{y}=L{f(t)}=L{tu(t3)}s2Y(s)+4Y(s)=e3s(3s+1s2)Y(s)=e3s(3s(s2+4)+1s2(s2+4))y(t)=L1{Y(s)}=L1{e3s(3s(s2+4)+1s2(s2+4))}=u(t3)L1{3s(s2+4)+1s2(s2+4)}(t3)y(t)=u(t3)(3434cos(2(t3))+t3418sin(2(t3)))

解答: Let u(x,t)=X(x)T(t),then ut=k2ux2XT=kXTTT=kXXBC.{u(0,t)=X(0)T(t)=0u(L,t)=X(L)T(t)=0{X(0)=0X(L)=0Suppose that TkT=XX=λCases I. λ=0X=0X=c1x+c2BC.{X(0)=c2=0X(L)=c1L+c2=0c1=c2=0X=0u=0Case II. λ>0λ=ρ2(ρ0)Xρ2X=0X=c1eρx+c2eρxBC.{X(0)=c1+c2=0X(L)=c1eρL+c2eρL=0c2=c1c1eρLc1eρL=0c1(e2ρL1)=0c1=0c2=0X=0u=0Case III. λ<0λ=ρ2(ρ>0)X+ρ2X=0X=c1cos(ρx)+c2sin(ρx)BC.{X(0)=c1=0X(L)=c2sin(ρL)=0sin(ρL)=0ρ=nπLX=sinnπxL,nNT=λkTT=eλkt=en2π2kt/L2un(x,t)=en2π2kt/L2sinnπxLu(x,t)=n=1anen2π2kt/L2sinnπxLu(x,0)=n=1ansinnπxL=Aan=2LL0AsinnπxLdxu(x,t)=n=1anen2π2kt/L2sinnπxL,an=2LL0AsinnπxLdx

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解題僅供參考,其他歷年試題及詳解

2 則留言:

  1. 請問第四題2e^-x*sinx 是不是少個"-"號

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