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2023年12月29日 星期五

111年雲科大電子系碩士班-工程數學詳解

 




解答:(a)(x3y4)dx+(x4y3+2y)dy=0{P(x,y)=x3y4Q(x,y)=x4y3+2yPy=4x3y3=QxExactΦ(x,y)=x3y4dx=x4y3+2ydyΦ(x,y)=14x4y4+ϕ(y)=14x4y4+y2+ρ(x)Φ(x,y)=14x4y4+y2+C=0(b)(14x4y4+y2+C)=x3y4+x4y3y+2yy=0x3y4+(x4y3+2y)dydx=0



解答:(a)y+9y+20y=0λ2+9λ+20=0(λ+4)(λ+5)=0λ=4,5yh=c1e4t+c2e5t(b)yp=Aetyp=Aetyp=Aetyp+9yp+20yp=12Aet=36etA=3yp=3et(c)y=yh+yp=c1e4t+c2e5t3ety=4c1e4t5c2e5t+3et{y(0)=c1+c23=0y(0)=4c15c2+3=0{c1=12c2=9y=12e4t9e5t3et

解答:(a)L{e2t}=0e2testdt=0e(2s)tdt=[12se(2s)t]|0=1s2(b)y=[2213]y+[42]e2t,y(0)=0{y1=2y12y2+4e2t,y1(0)=0y2=y1+3y2+2e2t,y2(0)=0{L{y1}=L{2y12y2+4e2t}L{y2}=L{y1+3y2+2e2t}{sY1(s)=2Y1(s)2Y2(s)+4s2(1)sY2(s)=Y1(s)+3Y2(s)+2s2(2)(1)Y2(s)=2s2Y1(s)+2s2(2)Y1(s)=4(s1)(s2)Y2=2s22s1y1(t)=L1{Y1(s)}=4L1{1s21s1}=4(e2tet)y2(t)=L1{Y2(s)}=2L1{1s21s1}=2(e2tet){y1(t)=4(e2tet)y2(t)=2(e2tet)(c){limty1(t)>>y1(0)limty2(t)>>y2(0)unstable

解答:(a){A=[101213]B=[210034]A+B=[311247](A+B)T=[321417](b){2AB=[012412]C+D=[132422](2AB)(C+D)=[68212](c)AB=[111221]X=32(AB)=[3/23/23/2333/2]


解答:{0x+4y2z=26x2y+z=294x+8y4z=24[04226212948424]R3/4[0422621291216]6R3+R2R2[0422014771216]R2/14[0422011/21/21216]4R2+R1R1,2R2+R3R3[0000011/21/21005]{x=5yz/2=1/2(x,y,z)=(5,k,2k1),kR

解答:(a){A=(1,3,2)B=(3,2,3)AB=1×3+(3)×(2)+2×3=15(b)A×B=|ijk132323|=5i+3j+7k(c)(ABB)BB=1522(322,222,322)=(4522,3022,4522)=4522i1511j+4522k

解答:A=[201010203]det

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解題僅供參考,其他歷年試題及詳解

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