國立臺北科技大學 112學年度碩士班招生考試
系所組別 :1111、 l112機械工程系機電整合碩士班甲組
第一節 工程數學 試題
解答:先求齊次解,y″−3y′+2y=0⇒λ2−3λ+2=0⇒(λ−2)(λ−1)=0⇒λ=2,1⇒yh=c1ex+c2e2x再利用參數變換法求特定解,令{y1=exy2=e2xr(x)=−e3xex+1⇒{y′1=exy′2=2e2x⇒W=|y1y2y′1y′2|=|exe2xex2e2x|=e3x⇒yp=−y1∫y2r(x)Wdx+y2∫y1r(x)Wdx=−ex∫−e2xex+1dx+e2x∫−exex+1dx=ex(ex−ln(ex+1))−e2xln(ex+1)=e2x−ex(1+ex)ln(ex+1)⇒y=yh+yp⇒y=c1ex+c2e2x−ex(1+ex)ln(ex+1)
解答:L{y″}+2L{y′}+10L{y}=L{t}⇒s2Y(s)−sy(0)−y′(0)+2(sY(s)−y(0))+10Y(s)=1s2⇒(s2+2s+10)Y(s)−s−3=1s2⇒Y(s)=s+3s2+2s+10+1s2(s2+2s+10)=s+1(s+1)2+32+2(s+1)2+32+150⋅s+1(s+1)2+32−225⋅1(s+1)2+32−150s+110s2=5150⋅s+1(s+1)2+32+4825⋅1(s+1)2+32−150s+110s2⇒y(t)=L−1{Y(s)}⇒y(t)=5150e−tcos(3t)+1625e−tsin(3t)−150+t10
解答:{x1+x2+x3=63x1−x2+x3=4−2x1+x2−2x3=−6⇒Ax=b,其中A=[1113−11−21−2],x=[x1x2x3],b=[64−6]A−1=[1612132301316−12−23]⇒x=A−1b=[1612132301316−12−23][64−6]=[123]⇒{x1=1x2=2x3=3
解答:令y(x)=∞∑n=0anxn⇒y′(x)=∞∑n=1nanxn−1⇒y″(x)=∞∑n=2n(n−1)anxn−2又ex=∞∑n=01n!xn,因此y″−xy′+exy=4⇒∞∑n=2n(n−1)anxn−2−∞∑n=1nanxn+(∞∑n=01n!xn)(∞∑n=0anxn)=4考慮xn的係數bn=(n+2)(n+1)an+2−nan+(an+an−1+12an−2+13!an−3+⋯+1n!a0)=(n+2)(n+1)an+2−nan+n∑k=01k!an−k⇒{b0=4bn=0,n>0又初始值{y(0)=2y′(0)=3⇒{a0=2a1=3⇒b0=2a2+a0=4⇒a2=1n>0⇒bn=(n+2)(n+1)an+2−nan+n∑k=01k!an−k=0⇒an+2=1(n+2)(n+1)(nan−n∑k=01k!an−k)⇒{a3=16(a1−a1−a0)=−13a4=112(2a2−a2−a1−12a0)=−14⋯⇒y(x)=2+3x+x2−13x3−14x4+⋯

解答:
A.A=[1−2101−102−1]⇒det(A−λI)=0⇒−(λ−1)(λ2+1)=0⇒λ=1,±iλ1=1:(A−λ1I)x=0⇒[0−2100−102−2][x1x2x3]=0⇒{x2=0x3=0⇒E1(A)={x∣x=t[100],t∈R}λ2=i:(A−λ2I)x=0⇒[1−i−2101−i−102−1−i][x1x2x3]=0⇒{x1=−1+i2x3x2=1+i2x3⇒Ei(A)={x∣x=t[(−1+i)/2(1+i)/21],t∈R}λ3=−i:(A−λ3I)x=0⇒[1+i−2101+i−102−1+i][x1x2x3]=0⇒{x1=−1−i2x3x2=1−i2x3⇒E−i(A)={x∣x=t[(−1−i)/2(1−i)/21],t∈R}特徵值為1,i,−i相對應的特徵向量分別為E1(A)={x∣x=t[100],t∈R},Ei(A)={x∣x=t[(−1+i)/2(1+i)/21],t∈R},及E−i(A)={x∣x=t[(−1−i)/2(1−i)/21],t∈R} B. A=[1−2101−102−1]=PDP−1=[1−1+i2−1−i201+i21−i2011][1000i000−i][1−110−i1+i20i1−i2]⇒A2000=PD2000P−1=[1−1+i2−1−i201+i21−i2011][12000000i2000000(−i)2000][1−110−i1+i20i1−i2]=[1−1+i2−1−i201+i21−i2011][100010001][1−110−i1+i20i1−i2]=PIP−1=PP−1=I=[100010001]
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