Loading [MathJax]/jax/output/CommonHTML/jax.js

2023年5月19日 星期五

112年中山大學碩士班-工程數學詳解

國立中山大學112學年度碩士班暨碩士在職專班招生考試

科目名稱:工程數學【材光系碩士班選考,材料前瞻應材碩士班選考,材光聯合碩士班選考】

解答μ(x)=etanxdx=secxysecx+ysecxtanx=secxsin(2x)(ysecx)=2sinxysecx=2sinxdx=2cosx+Cy=2cos2x+Ccosxy(0)=12+C=1C=3y=2cos2x+3cosx
解答y+y=xy2yy+2y2=2xv+2v=2x,v(x)=y2(x)   μ=e2dx=e2xve2x+2ve2x=2xe2x(ve2x)=2xe2xve2x=2xe2xdx=12e2x(12x)+Cv=y2=12(12x)+Ce2xy(0)=1C=12y2=12(12x+e2x)y=12x+e2x2
解答xy+4y=0,y=xmy=mxm1(m+4)xm=0m=4yh=Cx4yp=Ax4yp=4Ax38Ax4=8x4A=1yp=x4y=yh+yp=Cx4+x4,y(1)=22=C+1C=1y=x4+1x4
解答y=xmy=mxm1y=m(m1)xm2y=m(m1)(m2)xm3x3y+3x2y6xy6y=(m(m1)(m2)+3m(m1)6m6)xm=0(m+1)(m+2)(m3)xm=0m=1,2,3y=C1x1+C2x2+C3x3


解答y3y+2y=0(λ2)(λ1)=0λ=1,2yh=C1ex+C2e2xyp=Axe2x+Bx2+Cx+Dyp=Ae2x+2Axe2x+2Bx+Cyp=4Ae2x+4Axe2x+2Byp3yp+2yp=Ae2x+2Bx2+(2C6B)x+2B3C+2D=3e2x+2x27{A=3B=1C=3D=0yp=3xe2x+x2+3xy=yh+yp=C1ex+C2e2x+3xe2x+x2+3xy=C1ex+(2C2+3)e2x+6xe2x+2x+3{y(0)=1=C1+C2y(0)=0=C1+2C2+6{C1=8C2=7y=8ex7e2x+3xe2x+x2+3x
解答{y1=y1y2y2=y1y2y=[1111]y=Aydet(AλI)=0λ2+2λ+2=0λ=1±iλ1=1+i(Aλ1I)x=0x1=ix2,v1=[i1]λ2=1i(Aλ2I)x=0x1=ix2,v2=[1i]=PDP1,P=[i11i],D=[1+i001i]y=PDP1yP1y=DP1yz=Dz,z=P1yz=[1+i001i][z1z2]=[(1+i)z1(1i)z2]{z1=C1e(1+i)xz2=C2e(1i)xy=Pz=[i11i][C1e(1+i)xC2e(1i)x]=[iC1e(1+i)x+C2e(1i)xC1e(1+i)x+iC2e(1i)x]{y1(0)=1y2(0)=0{iC1+C2=1C1+iC2=0{C1=i/2C2=1/2{y1=12ex(eix+eix)=excosxy2=12ex(ieix+ieix)=exsinx{y1=excosxy2=exsinx
解答(a)L1{2s4}L1{48s6}=13L1{3!s4}25L1{5!s6}=13t425t5(b)L1{6s+72s2+4s+10}=L1{3s+1(s+1)2+22+142(s+1)2+22}=3etcos(2t)+14etsin(2t)
解答L{y}14L{y}=s2Y(s)sy(0)y(0)14Y(s)=(s214)Y(s)12s=0Y(s)=12ss21/4=6s+1/2+6s1/2y(t)=L1{Y(s)}=6L1{1s+1/2}+6L1{1s1/2}=6(et/2+et/2)y(t)=6(et/2+et/2)

解答w(t)=y(t+2)y+2y3y=w+2w3w=0L{w}+2L{w}3L{w}=s2W(s)sw(0)w(0)+2(sW(s)w(0))3W(s)=(s2+2s3)W(s)+3s+11=0W(s)=3s+11s2+2s3=121s+3721s1w(t)=L1{W(s)}=12e3t72etw(t2)=y(t)=12e3(t2)72et2

沒有留言:

張貼留言