臺灣綜合大學系統111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:先求齊次解,即y‴+y″=0⇒λ3+λ2=0⇒λ2(λ+1)=0⇒λ=0,−1⇒yh=C1+C2x+C3e−x;接著由3ex+4x2,假設yp=aex+bx4+cx3+dx2⇒y′p=aex+4bx3+3cx2+2dx⇒y″p=aex+12bx2+6cx+2d⇒y‴p=aex+24bx+6c⇒y‴p+y″p=2aex+12bx2+(24b+6c)x+6c+2d=3ex+4x2⇒{2a=312b=424b+6c=06c+2d=0⇒{a=3/2b=1/3c=−4/3d=4⇒yp=32ex+13x4−43x3+4x2⇒y=C1+C2x+C3e−x+32ex+13x4−43x3+4x2
解答:令A=[940−6−10643],依題意y′=Ay;⇒det(A−λI)=(λ−3)2(λ−5)⇒{λ1=3λ2=5Case I: λ1=3⇒(A−λ1I)x=[640−6−40640][x1x2x3]=0⇒3x1+2x2=0⇒取v1=(0,0,1)T,v2=(2,−3,0)TCase II: λ2=5⇒(A−λ2I)x=[440−6−6064−2][x1x2x3]=0⇒{x1+x2=03x2+2x2=x3⇒取v3=(1,−1,1)T⇒y=C1[001]e3x+C2[2−30]xe3x+C3[1−11]e5x
解答:先考慮f(t)={0,if −π<t≤0t,if 0≤t<π⇒F(t)=a0+∞∑n=1(ancosnt+bnsinnt)其{a0=12π∫π−πf(t)dt=12π∫π0tdt=π4an=1π∫π−πf(t)cosntdt=1π∫π0tcosntdt=1n2π((−1)n−1)bn=1π∫π−πf(t)sinntdt=1π∫π0tsinntdt=−1n(−1)n再考慮{limt→π−f(t)=πlimt→π+f(t)=0⇒F(π)=π2,同理F(nπ)=π2,∀n∈N因此F(t)={π2,if t=nππ4+∑∞n=1(1n2π((−1)n−1)cosnt−1n(−1)nsinnt),otherwise
解答:先求f(t)的Fourier series,由於f(t)為奇函數,所以an=0,只需要求bn;bn=∫f(t)sin(nt)dt=∫π/2−π/2tsin(nt)dt+∫3π/2π/2(π−t)sin(nt)dt={4/n2,n=1,5,9,…−4/n2,n=3,7,11,…0,n=2,4,6,8,…令y=Ancos(nt)+Bnsin(nt)⇒y′=−nAnsin(nt)+nBncos(nt)⇒y″=−n2Ancos(nt)−n2Bnsin(nt)⇒y″+ky′+y=(An+nkBn−n2An)cos(nt)+(−n2Bn−nkAn+Bn)sin(nt)=bnsin(nt)⇒{(1−n2)An+nkBn=0(1−n2)Bn−nkAn=bn⇒{An=−nk(1−n2)2+n2k2bnBn=1−n2(1−n2)2+n2k2bn⇒y=Ancos(nt)+Bnsin(nt),n∈N,其中{An=−nk(1−n2)2+n2k2bnBn=1−n2(1−n2)2+n2k2bn及bn={4/n2,n=1,5,9,…−4/n2,n=3,7,11,…0,n=2,4,6,8,…
==================== END ===========================
未公告答案,解題僅供參考
沒有留言:
張貼留言