國立高雄科技大學115學年度碩士班招生考試
系所別:電子工程系碩士班(建工校區)
組別: 電信與系統組
考科: 微分方程
Solve the following ODEs.
解答:$$\textbf{(1) } y' \sin 2\pi x = \pi y \cos 2\pi x \Rightarrow \frac{1}{y} dy = \pi \frac{\cos 2\pi x}{\sin 2\pi x} dx \Rightarrow \int \frac{1}{y} dy = \pi \int \frac{\cos 2\pi x}{\sin 2\pi x} dx \\ \quad \Rightarrow \ln\vert{}y\vert{} = \ln\sqrt{\vert{}\sin 2\pi x\vert{}} + C_1 \Rightarrow \bbox[red, 2pt]{y=C_2 \sqrt{|\sin 2\pi x|}} \\ \textbf{(2) }u=x+y-2 \Rightarrow {du\over dx}=1+{dy\over dx} \Rightarrow y'=u'-1 \Rightarrow u'-1=u^2 \Rightarrow {1\over u^2+1} \,du=dx\\ \quad \Rightarrow \int \frac{1}{u^2 + 1} du = \int 1 \, dx \Rightarrow \tan^{-1} u = x + C \Rightarrow u= x+y-2 =\tan(x+C)\\\quad \Rightarrow y=\tan(x+C)-x+2 \Rightarrow y(0)= \tan C+2=2\Rightarrow C=0 \Rightarrow \bbox[red, 2pt]{y= \tan x-x+2} \\ \textbf{(3) } y' = 3x^2 - \frac{y}{x} \Rightarrow y' + \frac{1}{x} y = 3x^2 \Rightarrow \text{ integrating factor }I(x)=x \Rightarrow xy'+y=3x^3\\ \quad (xy)'=3x^3 \Rightarrow xy=\int 3x^3\,dx ={3\over 4}x^4+C \Rightarrow y={3\over 4}x^3+{C \over x} \Rightarrow y(1)={3\over 4}+C=5 \Rightarrow C={17\over 4} \\\quad \Rightarrow \bbox[red, 2pt]{y={3\over 4}x^3+{17\over 4x}}$$
解答:$$\textbf{(4) } xe^{x^2}dx + (y^5 - 1)dy = 0 \Rightarrow \int xe^{x^2} dx + \int (y^5 - 1) dy = \int 0 \Rightarrow \bbox[red, 2pt]{\frac{1}{2} e^{x^2} + \frac{1}{6} y^6 - y = C} \\ \textbf{(5) } \frac{dy}{dx} = \frac{xy^2 - \cos x \cdot \sin x}{y(1 - x^2)} \Rightarrow (1 - x^2) \cdot y \frac{dy}{dx} - x y^2 = -\cos x \cdot \sin x \\\quad \Rightarrow y \frac{dy}{dx} - \frac{x}{1 - x^2} y^2 = \frac{-\cos x \cdot \sin x}{1 - x^2} \Rightarrow \frac{1}{2} \frac{dv}{dx} - \frac{x}{1 - x^2} v = \frac{-\cos x \cdot \sin x}{1 - x^2} \quad (取v=y^2) \\\quad \Rightarrow \frac{dv}{dx} - \frac{2x}{1 - x^2} v = \frac{-2\cos x \cdot \sin x}{1 - x^2} \Rightarrow 積分因子I(x)= e^{\int \frac{-2x}{1 - x^2} dx} =1-x^2 \\ \Rightarrow (1 - x^2) \frac{dv}{dx} - 2x v = -2\cos x \cdot \sin x \Rightarrow (v(1-x^2))'=-2\cos x\sin x \\\quad \Rightarrow v(1 - x^2) = \int -2\cos x \cdot \sin x \, dx = \cos^2 x+C \Rightarrow v= \bbox[red, 2pt]{y^2 ={\cos^2 x+C\over 1-x^2}}$$
解答:$$\textbf{(6) } (x + y)dx + x \cdot \ln x \, dy = 0 \Rightarrow \frac{dy}{dx} + \frac{1}{x \ln x} y = \frac{-1}{\ln x} \Rightarrow 積分因子I(x) = e^{\int \frac{1}{x \ln x} dx} =\ln x \\ \quad \Rightarrow \ln x \frac{dy}{dx} + \frac{1}{x} y = -1 \Rightarrow (y\ln x)'=-1 \Rightarrow y\ln x = \int -1\,dx =-x+C \Rightarrow \bbox[red, 2pt]{y = \frac{-x + C}{\ln x}} \\ \textbf{(7) } y^{(4)} + y''' + y'' = 0 \Rightarrow r^4+r^3+r^2=0 \Rightarrow r^2(r^2+r+1)=0 \Rightarrow r=0, {-1\pm \sqrt 3i\over 2} \\ \quad \Rightarrow \bbox[red, 2pt]{y = C_1 + C_2 x + e^{-\frac{1}{2}x} \left( C_3 \cos \left(\frac{\sqrt{3}}{2}x\right) + C_4 \sin \left(\frac{\sqrt{3}}{2}x\right) \right)} \\\textbf{(8) } y''-4y=0 \Rightarrow r^2-4=0 \Rightarrow r=\pm 2\Rightarrow y_h = C_1 e^{2x} + C_2 e^{-2x} \\ \quad \Rightarrow y_p = A\cos 3x+B\sin 3x+ C\cos 2x+ D\sin 2x \\\quad \Rightarrow y_p''-4y_p = -13A \cos 3x -13B \sin 3x -8C\cos 2x-8D\sin 2x =\cos 3x+ \sin 2x \\ \quad \Rightarrow \cases{A=-1/13\\ B=0\\ C=0\\ D=-1/8} \Rightarrow y_p =-{1\over 13}\cos 3x-{1\over 8}\sin 2x \Rightarrow y= y_h+ y_p \\ \quad \Rightarrow \bbox[red, 2pt]{ y = C_1 e^{2x} + C_2 e^{-2x} - \frac{1}{13} \cos 3x - \frac{1}{8} \sin 2x }$$
解答:$$\textbf{(a) } 1 \cdot (\sin^2 x) + 1 \cdot (\cos^2 x) - \frac{1}{2\pi} \cdot (2\pi) = 1 + 1 - 1 = 0 \Rightarrow \cases{c_1=c_2=1\\ c_3=-1/2\pi}不全為零\\ \quad \Rightarrow \bbox[red, 2pt]{\text{linearly dependent}} \\ \textbf{(b) }1\cdot e^x \ln x^2-1\cdot 2e^x \ln x=2e^x \ln x-2e^x\ln x=0 \Rightarrow \cases{c_1=1\\ c_2=-1}不全為零 \\ \quad \Rightarrow \bbox[red, 2pt]{\text{linearly dependent}}$$
解答:$$\textbf{(a) } 基底為 e^{-\sqrt{2}x}, xe^{-\sqrt{2}x} \Rightarrow 特徵方程式: (r-(-\sqrt 2))^2=0 \Rightarrow r^2+2\sqrt 2r+ 2=0 \\ \quad \Rightarrow \bbox[red, 2pt]{ y'' + 2\sqrt{2}y' + 2y = 0 } \\\textbf{(b) } 基底為\cos 2\pi x, \sin 2\pi x \Rightarrow 特徵方程式: (r - 2\pi i)(r + 2\pi i) = 0 \Rightarrow r^2+4\pi^2=0 \\ \quad \Rightarrow \bbox[red, 2pt]{ y'' + 4\pi^2y = 0 } $$
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解題僅供參考，碩士班歷年試題及詳解
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