國立臺北科技大學110學年度碩士班招生考試
系所組別:電機工程碩士班丙組
科目:工程數學
解答:v(x)=1y⇒v′=−y′y2⇒y′=−y2v′=−v′v2⇒−v′v2−21v=−1v2e3x⇒v′+2v=e3x⇒e2xv′+2e2xv=e5x⇒(e2xv)′=e5x⇒e2xv=15e5x+c1⇒v=1y=15e3x+c1e2x⇒y=115e3x+c1e2x⇒y=5e2xe5x+c2解答:y=xm⇒y′=mxm−1⇒m(m−1)xm−2⇒x2y″−5xy′+9y=m(m−1)xm−5mxm+9xm=(m2−6m+9)xm=0⇒m2−6m+9=(m−3)2=0⇒m=3⇒yh=c1x3+c2x3lnxApplying variation of parameters, let {y1=x3y2=x3lnxr(x)=2x+lnx/x2⇒W=|y1y2y′1y′2|=|x3x3lnx3x23x2lnx+x2|=x5⇒yp=−x3∫x3lnx⋅(2x+lnx/x2)x5dx+x3lnx∫x3(2x+lnx/x2)x5dx=−x3((lnx)2−(lnx)23x3−2lnx9x3−227x3)+x3lnx(2lnx−lnx3x3−19x3)=x3(lnx)2+19lnx+227⇒y=hh+yp⇒y=c1x3+c2x3lnx+x3(lnx)2+19lnx+227
解答:f(t)={0t<12t−31≤t<30t≥3⇒f(t)=(u(t−1)−u(t−3))(2t−3)⇒L{f(t)}=−3se−3s−2s2e−3s−1se−s+2s2e−sL{y″(t)}+2L{y′(t)}+2L{y(t)}=s2Y(s)−1+2sY(s)+2Y(s)=(s2+2s+2)Y(s)−1⇒=(s2+2s+2)Y(s)−1=−3se−3s−2s2e−3s−1se−s+2s2e−s⇒Y(s)=1(s+1)2+12(1−3se−3s−2s2e−3s−1se−s+2s2e−s)⇒y(t)=L−1{Y(s)}=e−tsin(t)−u(t−3)(t−12e−t(3sint+cost)−52)+u(t−1)(t+12e−t(sint+3cost)−52)⇒y(t)=e−tsin(t)+(u(t−1)−u(t−3))(t−52)+12e−t(u(t−1)(sint+3cost)+u(t−3)(3sint+cost))
解答:[1234a624b][x1x2x3]=[12c]≡Ax=b(1)det(A)=(a−8)(b−6)≠0⇒a≠8,b≠6⇒unique solution: {a≠8b≠6c∈R(2)many solution: {12=24=3b=1c42=a4=6b=2c⇒{(a∈R,b=6,c=2)(a=8,b=3,c=1)(3)no solution: {12=24=3b≠1c42=a4=6b≠2c⇒{(a∈R,b=6,c≠2)(a=8,b=3,c≠1
解答:A=[11−11]⇒A−1=[12−121212]⇒A−2=[0−12120]⇒A−4=[−1400−14]=−14I2⇒A−8=124I2⇒A−16=128I2⇒A−64=1232I2⇒A−86=A−64⋅A−16⋅A−4⋅A−2=−1242I2⋅A−2=[01/243−1/2430]
解答:Given a 2x2 matrix [x1x2x3x4],there exists a,b,c,d such thata[11−11]+b[1101]+c[1011]+d[100−1]=[x1x2x3x4]⇒{a+b+c+d=x1a+b=x2−a+c=x3a+b+c−d=x4⇒[11111100−1010111−1][abcd]=[x1x2x3x4]⇒[abcd]=[11111100−1010111−1]−1[x1x2x3x4]=[12−1−112−1221−1212−10121200−12][x1x2x3x4]⇒{a=x1/2−x2−x3+x4/2b=−x1/2+2x2+x3−x4/2c=x1/2−x2+x4/2d=x1/2−x4/2⇒Any 2x2 matrix can be the combination of vectors in V⇒Yes!V form a basis.
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解題僅供參考,其他歷年試題及詳解
第三題,想討論一下
回覆刪除1.u(t-3)那部份,sin&cos裡面都應該是(t-3),以及乘上的e^(-t)也應該改成e^-(t-3)?
2.u(t-1)也是同理,sin&cos裡面應該是(t-1),以及乘上的e^(-t)也應該改成e^-(t-1)?