國立中山大學111學年度碩士班暨碩士在職專班招生考試
科目名稱:工程數學(資工系碩士班乙組)
解答:[A∣I]=[1−10100213010021001]R2−2R1→R2,(1/2)R3→R3→[1−10100033−21001120012]R1+R3→R1,R2−3R3→R2→[101210120032−21−3201120012]R2↔R3→[10121012011200120032−21−32](2/3)R3→R3→[1012101201120012001−4323−1]R1−(1/2)R3→R1,R2−(1/2)R3→R2→[10053−13101023−131001−4323−1]⇒A−1=[53−13123−131−4323−1]解答:2.1A=[−√32−1212−√32]=[cos5π6−sin5π6sin5π6cos5π6]⇒A is a rotation matrix which counterclock angle θ=5π62.25π6×2011=837×2π+11π6⇒A2011=[cos11π6−sin11π6sin11π6cos11π6]=[√3/21/2−1/2√3/2]
解答:3.1y′=yex−2ex+y−2=y(ex+1)−2(ex+1)=(y−2)(ex+1)⇒1y−2dy=(ex+1)dx⇒ln(y−2)=ex+x+c1⇒y=eex+x+c1+2⇒y=c2eex+x+23.2ty′=t4−2y⇒y′+2ty=t3⇒ integrating factor I(x)=e∫(2/t)dt=t2⇒t2y′+2ty=t5⇒(t2y)′=t5⇒t2y=16t6+c1⇒y=16t4+c1t2y(1)=0⇒0=16+c1⇒c1=−16⇒y=16t4−16t2
解答:L{y″
解答:y''-2y'+y=0 \Rightarrow \lambda^2-2\lambda+1=0 \Rightarrow (\lambda-1)^2=0 \Rightarrow \lambda=1 \Rightarrow y_h=c_1e^t +c_2xe^t\\ \text{Apply variation of parameters, let }\cases{y_1=e^t \\ y_2=te^t\\ r(t) =e^t} \\\Rightarrow W= \begin{vmatrix}y_1 & y_2 \\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix}e^t & te^t \\ e^t & e^t+te^t \end{vmatrix}= e^{2t} \Rightarrow y_p = -y_1\int {y_2\cdot r(t)\over W}\,dt +y_2 \int {y_1\cdot r(t)\over W}\,dt \\ \Rightarrow y_p = -e^t \int {te^{2t}\over e^{2t}} \,dt + te^t \int 1\,dt =-{1\over 2}e^t t^2+ t^2e^t= {1\over 2}t^2e^t \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y(t)=c_1e^t+ c_2te^t+ {1\over 2}t^2e^t}
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解題僅供參考,其他歷年試題及詳解
老師您好:能否請問一題2023AMC upper primary 第24題:有4個相異正整數的和=98, 將4數各自分別:加6,減6,乘6,除6, 結果新4數與原4數一樣, 求最大的2數和=? Ans:90,謝謝!
回覆刪除a=39,b=45,c=2,d=12
刪除a'=a+6=45, b'=45-6=39, b'=cx6=12,d'=12/6=2
以上四數應該符合要求, 但答案不是90
不曉得是否我理解錯誤?
謝謝老師的解答.
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