臺灣綜合大學系統107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:D09

解答:(a)L{cos(5t)}=ss2+52=ss2+25(b)L−1{1s−1}=et(c)L{y′}−L{y}=2L{cos(5t)}⇒sY(s)−y(0)−Y(s)=2ss2+25⇒Y(s)=2s(s2+25)(s−1)=−113s+2513s2+25+1/13s−1=−113⋅ss2+25+2513⋅1s2+25+113⋅1s−1⇒y(t)=−113L−1{ss2+25}+513L−1{5s2+25}+113L−1{1s−1}=−113cos(5t)+513sin(5t)+113et⇒y(t)=−113cos(5t)+513sin(5t)+113et

解答:(a)A=[010100001]⇒det

解答:\iint_R(2x-3x^2)dA = \int_{-1}^1 \int_0^1 (2x-3x^2)\,dydx = \int_{-1}^1 (2x-3x^2)\,dx = \left.\left[ x^2-x^3 \right]\right|_{-1}^1 =-2\\ \cases{C_1=\{(2t-1,0)\mid t\in[0,1]\} \\ C_2=\{(1,t)\mid t\in [0.1]\} \\ C_3=\{(-2t+1,1)\mid t\in[0,1] \} \\ C_3=\{(-1,1-t)\mid t\in[0,1]\}} \Rightarrow \oint_C 3x^2ydx +(x^2-5y)dy =\oint_{C_1} 3x^2ydx +(x^2-5y)dy \\+ \oint_{C_2} 3x^2ydx +(x^2-5y)dy + \oint_{C_3} 3x^2ydx +(x^2-5y)dy+ \oint_{C_4} 3x^2ydx +(x^2-5y)dy \\=0 +\int_0^1(1-5t)dt +\int_0^1 -6(-2t+1)^2\,dt +\int_0^1 5(1-t)-1\,dt \\=\int_0^1 -24t^2+14t-1\,dt =\left.\left[ -8t^3+7t^2-t \right]\right|_0^1 =-2 \\ 因此左式=右式=-2,\bbox[red,2pt]{故得證}
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