111年北科大製造科技碩士班-微分方程詳解

國立臺北科技大學111學年度碩士班招生考試

系所組別 :1201製 造科技研究所 
第一節 微分方程 試題 (選考)

解答：$$令\cases{P(x,y)=4y\\ Q(x,y)= x+12xy} \Rightarrow \cases{P_y=4\\ Q_x=1+12y} \Rightarrow P_y\ne Q_x \Rightarrow \text{Not exact}\\ 取u'=-{P_y-Q_x\over P}u = (3-{3\over 4y})u \Rightarrow u=y^{-3/4}e^{3y} \Rightarrow \cases{uP=4y^{1/4}e^{3y} \\ uQ=xy^{-3/4}e^{3y} +12xy^{1/4}e^{3y}} \\ \Rightarrow (uP)_y=(uQ)_x= y^{-3/4}e^{3y} +12y^{1/4}e^{3y} \\ \Phi(x,y)=\int 4y^{1/4}e^{3y}\,dx = \int xy^{-3/4}e^{3y} +12xy^{1/4}e^{3y}\,dy \\ \Rightarrow  4x y^{1/4}e^{3y} +\phi(y)= 4xy^{1/4}e^{3y}+ \rho(x) \Rightarrow \bbox[red, 2pt] {4xy^{1/4}e^{3y}+c_1=0} $$

解答：$$1+2x\tan y+(x^2-x \tan y)y'=0 \Rightarrow (1+2x\tan y)dx+(x^2-x \tan y)dy=0 \\ 令\cases{P(x,y) =1+2x\tan y\\ Q(x,y)= x^2-x\tan y} \Rightarrow P_y\ne Q_x \Rightarrow \text{ Not Exact} \\ \Rightarrow u'={Q_x-P_y\over P}u = -\tan y u \Rightarrow u= -c_1 \cos y \Rightarrow \cases{uP= -c_1\cos y(1+2x\tan y) \\ uQ = -c_1\cos y(x^2-x\tan y)} \\ \Rightarrow \Phi(x,y) =\int \cos(y)(1+2x\tan y)\,dx = \int \cos(y)(x^2-x\tan y)\,dy \\ \Rightarrow \int \cos y+2x\sin y\,dx = \int x^2\cos y-x\sin y\,dy \\\Rightarrow x\cos y+x^2\sin y +\phi(y)=x^2 \sin y+ x\cos y + \rho(x) \Rightarrow x\cos y+x^2 \sin y+c_1=0\\ y(-1)=\pi \Rightarrow 1+c_1=0 \Rightarrow c_1=-1 \Rightarrow \bbox[red, 2pt]{x\cos y+x^2 \sin y=1}$$

解答：$$y'+y=y^2 \Rightarrow {dy\over y^2-y}=dx \Rightarrow \ln( y-1)-\ln y= x+c_1 \Rightarrow {y-1\over y}=c_2e^x \\ \Rightarrow y={1\over 1-c_2e^x} \Rightarrow y(0)={1\over 1-c_2}=-{1\over 3} \Rightarrow c_2=4 \Rightarrow \bbox[red, 2pt]{y={1\over 1-4e^x}}$$

解答：$$先求齊次解，即y'''+3y''+3y'+y=0 \Rightarrow \lambda^3+3\lambda^2 +3\lambda +1=0 \Rightarrow (\lambda+1)^3=0 \Rightarrow \lambda=-1\\ \Rightarrow y_h= e^{-x}(c_1+ c_2x+ c_3x^2  )\\ 特解y_p = Ax^3e^{-x} \Rightarrow y_p'=3Ax^2e^{-x}-Ax^3e^{-x} \Rightarrow y_p''= 6Axe^{-x}-6Ax^2e^{-x} +Ax^3e^{-x} \\ \Rightarrow y_p'''=6Ae^{-x}-18Axe^{-x} +9Ax^2e^{-x}   -Ax^3e^{-x} \\ \Rightarrow y_p'''+3y_p''+ 3y_p'+y_p= 6Ae^{-x} =30e^{-x} \Rightarrow A=5 \Rightarrow y_p=5x^3e^{-x} \\ \Rightarrow y=y_h +y_p =e^{-x}(c_1+ c_2x+ c_3x^2+ 5x^3)  \Rightarrow y(0)=c_1=3\\\Rightarrow y'=e^{-x}(c_2-3+ (2c_3-c_2)x+ (15-c_3)x^2- 5x^3) \Rightarrow y'(0)=c_2-3=-3 \Rightarrow c_2=0 \\ \Rightarrow y''=e^{-x}(2c_3+3+ (30-4c_3)x +(c_3-30)x^2 +5x^3) \Rightarrow y''(0)=2c_3+ 3=-47 \Rightarrow c_3=-25\\ \Rightarrow \bbox[red, 2pt]{y= e^{-x}(3-25c^2+ 5x^3)}$$

解答：$$L\{ y''''\}+3L\{ y''\}-4L\{y\} = (s^4Y(s)-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0)) \\\qquad + 3(s^2Y(s)-sy(0)-y'(0))-4Y(s)=0 \\ \Rightarrow (s^4+3s^2-4)Y(s)=10-10s^2 \Rightarrow Y(s)={-10(s^2-1) \over (s^2+4)(s^2-1)} \\ \Rightarrow Y(s)=-{10\over s^2+4} \Rightarrow y(t)= L^{-1}\{Y(s)\} =-5L^{-1}\left\{ {2\over s^2+2^2} \right\}=-5\sin(2t) \\ \Rightarrow \bbox[red, 2pt]{y=-5\sin(2t)}$$
 

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解題僅供參考,其他歷年試題及詳解