112年高科大碩士班-微分方程詳解

解答： $$y'''＝e^{-0.2x} =e^{-x/5} \Rightarrow y有e^{-x/5}項,只有選項(C)符合,且y＝-125e^{-x/5}+c_1x^2 +c_2x+c_3\\ \Rightarrow y'=25e^{-x/5}+2c_1x+c_2 \Rightarrow y''=-5x^{-x/5}+2c_1 \Rightarrow y'''=x^{-x/5},故選\bbox[red, 2pt]{(C)}$$

解答： $$y'=1+y^2 \Rightarrow {1\over 1+y^2}dy =dx \Rightarrow \tan^{-1}y=x+c \Rightarrow y=\tan(x+c),故選\bbox[red, 2pt]{(A)}$$

解答： $$(e^{x+y}+ye^y)dx +(xe^y-1)dy=0 \Rightarrow \cases{P=e^{x+y}+ye^y\\ Q=xe^y-1} \Rightarrow \cases{P_y=e^{x+y}+e^y+ye^y \\ Q_x=e^y} \Rightarrow Not\; Exact\\ But {P_y-Q_x\over P}={e^{x+y}+ye^y \over e^{x+y}+ye^y}=1  \Rightarrow Non-Exact，故選\bbox[red, 2pt]{(C)}$$

解答： $$u=e^{\ln y}=y \Rightarrow uydx+u(2x+4)dy = y^2dx+(2xy+4y)dy=0 \\ \Rightarrow \cases{P=y^2\\ Q=2xy+4y} \Rightarrow \cases{P_y=2y\\ Q_x=2y} \Rightarrow P_y=Q_x \Rightarrow e^{\ln y}為積分因子，故選\bbox[red, 2pt]{(A)}$$

解答： $$y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda=\pm i \Rightarrow y=c_1\cos x+c_2\sin x\\ \Rightarrow y'=-c_1\sin x+c_2\cos x \Rightarrow \cases{y(0)=c_1=3\\ y'(0)=c_2=-0.5 \ne 0.5} \Rightarrow y=3\cos x-0.5\sin x \\ \Rightarrow \cases{y(\pi/2)=-0.5\\ y(-\pi)=-3} ，故選\bbox[red, 2pt]{(B)}$$

解答： $$\mathbf{(a)}\; \text{Wronskians: }W(x)=\begin{vmatrix}p& q & r\\ p'& q'& r'\\ p''& q''& r''\end{vmatrix} \\=\begin{vmatrix}1& e^{-2x}\cos x & e^{-2x}\sin x\\ 0& -2e^{-2x}\cos x-e^{-2x}\sin x& -2e^{-2x}\sin x+e^{-2x}\cos x\\ 0 & 3e^{-2x}\cos x+4e^{-2x}\sin x&  -4e^{-2x}\cos x+ 3e^{-2x}\sin x\end{vmatrix} =\bbox[red, 2pt]{5e^{-4x}} \\\mathbf{(b)}\; W(x)\ne 0, \exists x\in \mathbb R \Rightarrow \bbox[red,2pt]{\text{linearly independent}}$$

解答：

(a) $$\frac{d^5 y}{dx^5}+ 2\frac{d^3 y}{dx^3}+ \frac{d y}{dx}=0 \Rightarrow \lambda^5+2\lambda^3+\lambda=0 \Rightarrow \lambda(\lambda^2+1)^2=0\\ \Rightarrow \lambda =0, \pm i \Rightarrow \bbox[red, 2pt]{y=c_1+c_2\cos x+c_3\sin x+ c_4x\cos x+c_5x\sin x}$$ (b) $$y''+9y=0 \Rightarrow y=c_1\cos(3x)+c_2\sin(3x)\\ 令\cases{y_1=\cos(3x) \\y_2=\sin(3x)} \Rightarrow W=\begin{vmatrix}y_1&y_2 \\ y_1'& y_2' \end{vmatrix} =3 \Rightarrow y_p=-y_1\int{y_2r\over W}\,dx +y_2\int{y_1r\over W}\,dx, r(x)=\csc(3x) \\ \Rightarrow y_p=-\cos(3x)\int{1\over 3}\,dx +\sin(3x)\int{\cos(3x)\over 3\sin(3x)}\,dx = -{1\over 3}x\cos(3x)+{1\over 9}\sin(3x)\ln(\sin(3x)) \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1 \cos(3x)+ c_2 \sin(3x)-{1\over 3}x\cos(3x)+{1\over 9}\sin(3x)\ln(\sin(3x))}$$ (c) $$v=y^2 \Rightarrow v'=2yy' \Rightarrow y'={v'\over 2y} \Rightarrow {v'\over 2y}+y=-{x\over y} \Rightarrow v'+2v=-2x \\ \Rightarrow 積分因子=e^{2x} \Rightarrow v'e^{2x}+2ve^{2x}=-2xe^{2x} \Rightarrow (ve^{2x})'=-2xe^{2x} \\ \Rightarrow ve^{2x}=-2\int xe^{2x}\,dx =-xe^{2x}+{1\over 2}e^{2x}+c_1 \Rightarrow v=y^2=-x+{1\over 2}+c_1e^{-2x} \\ \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{-x+{1\over 2}+c_1e^{-2x}}}$$
 

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