113年高科大電子工程碩士班-微分方程詳解

國立高雄科技大學113學年度碩士班招生考試

系所別: 電子工程系碩士班
組別:電信與系統組
考科:微分方程

解答:$$y=x^3(1+x)^9e^{6x} \Rightarrow {dy\over dx}=\left(x^3(1+x)^9 \right)' e^{6x} +x^3(1+x)^9\left(e^{6x} \right)' \\=\left( 3x^2(1+x)^9+ 9x^3(1+x)^8 \right)e^{6x} +6x^3 (1+x)^9 e^{6x} \\= 3x^2(1+x)^8 e^{6x} \left( 1+ x+ 3x+2x(1+x) \right) = \bbox[red, 2pt]{3x^2(1+x)^8 e^{6x}(1+6x+2x^2)}$$

解答:$$y=\sqrt{1+x^2} \sin^2 x \Rightarrow \frac{\text{d}y}{\text{d}x} ={ 2x\over 2\sqrt{1+x^2}} \sin^2x + 2\sqrt{1+x^2} \sin x\cos x \\= \bbox[red, 2pt]{{x\over \sqrt{1+x^2}} \sin^2x +\sqrt{1+x^2} \sin(2x)}$$

解答:$$y=x^2(1-x)^8 \Rightarrow \frac{\text{d}y}{\text{d}x} = 2x(1-x)^8 - 8x^2 (1-x)^7 =2x(1-x)^7(1-x-4x) \\ =\bbox[red, 2pt]{2x(1-x)^7(1-5x)}$$

解答:$$$$

解答:$$\cases{P(x,y)=2x+3y-2\\ Q(x,y)=3x-4y+1} \Rightarrow P_y=3=Q_x \Rightarrow \text{ Exact} \\ \Rightarrow \Phi(x,y)= \int (2x+3y-2)\,dx = \int (3x-4y+1)\,dy \\ \Rightarrow \Phi=x^2+3xy-2x+\phi(y)=3xy-2y^2+y+ \rho(x) \\ \Rightarrow \bbox[red, 2pt]{x^2+3xy+2y^2-2x-y=c_1}$$

解答:$$v=y^{-2} \Rightarrow v'=-2y^{-3}y' \Rightarrow y'=-{1\over 2}v'y^3 \Rightarrow -{1\over 2}v'y^3-{1\over x}y=xy^3 \\ \Rightarrow -{1\over 2}v'-{1\over x}v=x \Rightarrow v'+{2\over x}v=-2x \Rightarrow x^2v'+2xv=-2x^3 \\ \Rightarrow (x^2v)'=-2x^3 \Rightarrow x^2v = \int -2x^3\,dx = -{1\over 2}x^4+c_1 \\ \Rightarrow v={1\over y^2}=-{1\over 2}x^2+{c_1\over x^2} ={-x^4+2c_1\over 2x^2} \Rightarrow y=\pm \sqrt{2x^2 \over -x^4+2c_1} \\ \Rightarrow \bbox[red, 2pt]{y=\pm {\sqrt 2 x\over \sqrt{c_2-x^4}}}$$

解答:$$y= \int (3x^2+2x)\,dx = x^3+x^2+c_1 \Rightarrow y(1)=4 \Rightarrow 4=2+c_1 \Rightarrow c_1=2 \\ \Rightarrow \bbox[red, 2pt]{y=x^3+x^2+2}$$

解答:$$\lambda^2-6\lambda+9=0 \Rightarrow (\lambda-3)^2 \Rightarrow \lambda=3 \Rightarrow \bbox[red, 2pt]{y=c_1e^{3x}+ c_2xe^{3x}}$$

解答:$$y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda+2 =0 \Rightarrow (\lambda-2)(\lambda -1)=0 \Rightarrow \lambda=1,2 \\ \Rightarrow y_h=c_1e^x+ c_2e^{2x} \\ y_p=Ae^{3x} \Rightarrow y_p'=3Ae^{3x} \Rightarrow y_p''=9Ae^{3x} \Rightarrow y_p''-3y_p'+2y_p= 2Ae^{3x}=e^{3x} \\ \Rightarrow A={1\over 2} \Rightarrow y_p={1\over 2}e^{3x} \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^x+ c_2 e^{2x}+ {1\over 2}e^{3x}}$$

解答:$$y=x^m \Rightarrow \cases{y'=mx^{m-1} \\ y''= m(m-1)x^{m-2}} \Rightarrow x^2 y''-2xy'+2y = (m^2-m-2m+2)x^m \\=(m^2-3m+2)x^m=0 \Rightarrow m=2,1 \Rightarrow y_h= c_1x+ c_2x^2 \\ y_p =Ax^3+Bx^2+Cx+D \Rightarrow y_p'=3Ax^2+2Bx+C \Rightarrow y_p''=6Ax+2B \\ \Rightarrow x^2y_p''-2xy_p'+2 y_p = 6Ax^3+2Bx^2-6Ax^3-4Bx^2-2Cx+2Ax^3+2Bx^2+2Cx+2D \\ =2Ax^3+2D=x^3 \Rightarrow \cases{A=1/2\\ D=0} \Rightarrow y_p={1\over 2}x^3+Bx^2+Cx \Rightarrow y=y_h+y_p \\ \Rightarrow \bbox[red, 2pt]{y=c_3x+c_4x^2+{1\over 2}x^3}$$
 

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解題僅供參考,其他歷年試題及詳解