微分方程
\[\begin{gathered}
\gdef \d {\mathrm d}
\frac{\d y}{\d x} = \frac{2 x y}{x ^ 2 + y ^ 2} \\
\frac{x \d u}{\d x} u = \frac{2 u}{1 + u ^ 2} \\
\frac{x \d u}{\d x} = \frac{u - u ^ 3}{1 + u ^ 2} \\
\frac{1 + u ^ 2}{u (1 - u ^ 2)} \d u = \frac{\d x}{x} \\
\left(\frac{1}{u} + \frac{1}{1 - u} - \frac{1}{1 + u}\right) \d u = \frac{\d x}{x} \\
\ln u - \ln(1 - u) - \ln(1 + u) - \ln(x) = C \\
C u = x (1 + u) (1 - u) \\
C y = x ^ 2 - y ^ 2
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
\frac{\d y}{\d x} = \frac{y}{x}(1 + \ln y - \ln x) \\
\frac{x \d u}{\d x} + u = u (1 + \ln u) \\
\frac{\d u}{u \ln u} = \frac{\d x}{x} \\
\int \frac{\d u}{u \ln u} = \int \frac{\d v}{v} = \ln \ln u + C \\
\ln \ln u = \ln x + C \implies \ln u = C x \implies y = x e ^ {C x}
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
y ^ 2 + x ^ 2 \frac{\d y}{\d x} = x y \frac{\d y}{\d x} \\
u ^ 2 + u + x \frac{\d u}{\d x} = u ^ 2 + u x \frac{\d u}{\d x} \\
\frac{\d x}{x} = \left(1 - \frac{1}{u}\right) \d u \\
\ln x = u - \ln u + C \\
\ln x = \frac{y}{x} - \ln y + \ln x + C \\
y = C e ^ \frac{y}{x} \\
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
(y + x) \d y = (y - x) \d x \\
(u + 1) (u \d x + x \d u) = (u - 1) \d x \\
(u ^ 2 + 1) \d x = -x (u + 1) \d u \\
\frac{\d x}{x} = -\frac{u + 1}{u ^ 2 + 1} \d u \\
\ln x + C = -\frac{1}{2} \ln(u ^ 2 + 1) - \arctan u \\
\ln(\sqrt{x ^ 2 + y ^ 2}) = -\arctan \frac{y}{x} + C \\
\sqrt{x ^ 2 + y ^ 2} = C e ^ {-\arctan \frac{y}{x}} \iff r = C e ^ {-\theta}
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
\left(x - y \cos \frac{y}{x}\right) \d x + x \cos \frac{y}{x} \d y = 0 \\
(1 - u \cos u)\d x + \cos u(u \d x + x \d u) = 0 \\
\frac{\d x}{x} = -\cos u \d u \\
\ln x = -\sin\frac{y}{x} + C \\
y = x \arcsin(-\ln x + C)
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
\tan\left(\arctan \frac{y}{x} - \arctan \frac{\d y}{\d x}\right) = \tan \alpha \\
\left(\frac{y}{x} - \frac{\d y}{\d x}\right) {\Large/} \left(1 + \frac{y \d y}{x \d x}\right) = \tan \alpha \\
\frac{y}{x} - \frac{\d y}{\d x} = \tan \alpha + \frac{y \d y}{x \d x} \tan \alpha \\
\frac{y}{x} - \tan \alpha = \left(1 + \frac{y}{x} \tan \alpha\right) \frac{\d y}{\d x} \\
\frac{\d y}{\d x} = \frac{y - x \tan \alpha}{x + y \tan \alpha} \\
\frac{x \d u}{\d x} + u = \frac{u - \tan \alpha}{1 + u \tan \alpha} \\
\frac{x \d u}{\d x} = -\frac{(1 + u ^ 2) \tan \alpha}{1 + u \tan \alpha} \\
\frac{\d x}{x} = -\frac{1 + u \tan \alpha}{(1 + u ^ 2) \tan \alpha} \d u \\
\frac{\d x}{x} = -\frac{1}{(1 + u ^ 2) \tan \alpha} \d u - \frac{u}{1 + u ^ 2} \d u \\
\ln x + \frac{\arctan u}{\tan \alpha} + \frac{1}{2} \ln(1 + u ^ 2) = C \\
\frac{\arctan y / x}{\tan \alpha} + \ln \sqrt{x ^ 2 + y ^ 2} = C \\
r = C e ^ {\theta / \tan \alpha}
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
\tan \left(2 \arctan \frac{\d y}{\d x}\right) = \frac{y}{x} \\
\frac{2 \d y}{\d x} = \frac{y}{x} \left(1 - \left(\frac{\d y}{\d x}\right) ^ 2\right) \\
\left(\frac{\d y}{\d x}\right) ^ 2 + \frac{2 x \d y}{y \d x} - 1 = 0 \\
\frac{\d y}{\d x} = -\frac{x}{y} \pm \sqrt{\frac{x ^ 2}{y ^ 2} + 1} \\
u + \frac{x \d u}{\d x} = -\frac{1 \pm \sqrt{u ^ 2 + 1}}{u} \\
\frac{\d x}{x} = -\frac{u \d u}{u ^ 2 + 1 \pm \sqrt{u ^ 2 + 1}} \\
\frac{\d x}{x} = -\frac{v \d v}{2(v ^ 2 \pm v)} \\
2 \ln x + \ln(v \pm 1) = C \\
\sqrt{x ^ 2 + y ^ 2} \pm x = C \\
y ^ 2 = 2 x C + C ^ 2
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
\frac{1}{b} \frac{\d z}{\d x} - \frac{a}{b} = \frac{z + c}{\lambda z + c_1} \\
(\lambda z + c_1) \d z - a(\lambda z + c_1) \d x = b (z + c) \d x \\
\frac{\lambda z + c_1}{b (z + c) + a(\lambda z + c_1)} \d z = \d x \\
\frac{\lambda z + c_1}{(a \lambda + b)z + a c_1 + b c} \d z = \d x \\
p = (a \lambda + b), q = a c_1 + b c \\
p = 0 \implies \frac{\lambda z ^ 2 + 2 c_1 z}{2 q} = \d x \\
\frac{\lambda}{p} \d z + \left(c_1 - \frac{\lambda q}{p}\right) \frac{1}{p z + q} \d z = \d x \\
\frac{\lambda z}{p} + \left(c_1 - \frac{\lambda q}{p}\right) \frac{\ln(p z + q)}{p} = x + C \\
\lambda p z + b(c_1 - \lambda c) \ln(p z + q) = p ^ 2 x + C \\
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
3 y - 7 x + 7 = (3 x - 7 y - 3) \frac{\d y}{\d x} \\
x = \xi + h, y = \eta + k, 3 k - 7 h + 7 = 3 h - 7 k - 3 = 0 \\
h = 0, k = 1, \frac{3 \eta - 7 \xi}{3 \xi - 7 \eta} = \frac{\d \eta}{\d \xi}, u = \frac{\eta}{\xi} \\
\frac{3 u - 7}{3 - 7 u} = u + \frac{\xi \d u}{\d \xi} \\
-\frac{7 (u + 1) (u - 1)}{(7 u - 3) \d u} = \frac{\xi}{\d \xi} \\
\frac{2 \d u}{7 (u - 1)} + \frac{5 \d u}{7 (u + 1)} = -\frac{\d \xi}{\xi} \\
\frac{2 \ln(u - 1)}{7} + \frac{5 \ln(u + 1)}{7} + ln \xi = C \\
2 \ln(u - 1) + 5 \ln(u + 1) + 7 \ln \xi = C \\
2 \ln \left(\frac{y - x - 1}{x}\right) + 5\ln \left(\frac{y + x - 1}{x}\right) + 7 ln x = C \\
2 \ln(y - x - 1) + 5 \ln(y + x - 1) = C \\
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
(x + 2 y + 1) \frac{\d y}{\d x} = 2 x + 4 y + 3 \\
\frac{1}{2} (u + 1) \left(\frac{\d u}{\d x} - 1\right) = 2 u + 3 \\
\frac{\d u}{\d x} = \frac{4 u + 6}{u + 1} + 1 \\
\frac{(u + 1) \d u}{5 u + 7} = \d x \\
\left(\frac{1}{5} - \frac{2}{5 (5 u + 7)}\right) \d u = \d x \\
\frac{u}{5} - \frac{2 \ln(5 u + 7)}{25} = x + C \\
5 (x + 2y) - 2 \ln(5 (x + 2 y) + 7) = 25 x + C \\
\ln(5 x + 10 y + 7) = 5 y - 10 x + C
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
\frac{\d y}{\d x} = 2 \left(\frac{y + 2}{x + y - 1}\right) ^ 2 \\
x = \xi + h, y = \eta + k, k + 2 = h + k - 1 = 0 \\
h = 3, k = -2, \frac{\d \eta}{\d \xi} = 2 \left(\frac{\eta}{\xi + \eta}\right) ^ 2 \\
\frac{\xi \d u}{\d \xi} + u = 2 \left(\frac{u}{1 + u}\right) ^ 2 \\
\frac{\xi \d u}{\d \xi} = \frac{2 u ^ 2 - u (1 + u) ^ 2}{(1 + u) ^ 2} \\
\frac{(1 + u) ^ 2 \d u}{u ^ 3 + u} + \frac{\d \xi}{\xi} = 0 \\
\left(\frac{1}{u} + \frac{2}{u ^ 2 + 1}\right) \d u + \frac{\d \xi}{\xi} = 0 \\
\ln u + 2 \arctan u + \ln \xi = C \\
\ln \frac{y + 2}{x - 3} + 2 \arctan \frac{y + 2}{x - 3} + \ln(x - 3) = C \\
\ln(y + 2) + 2 \arctan \frac{y + 2}{x - 3} = C
\end{gathered}\]
\[\begin{gathered}
\def \d {\mathrm d}
(x + y) ^ 2 \frac{\d y}{\d x} = a ^ 2 \\
u ^ 2 \frac{\d u}{\d x} - u ^ 2 = a ^ 2 \\
\frac{u ^ 2 \d u}{a ^ 2 + u ^ 2} = \d x \\
\left(1 - \frac{1}{1 + (u / a) ^ 2}\right) \d u = \d x \\
u - a \arctan \frac{u}{a} = x + C \\
x = a \tan \frac{y + C} a - y
\end{gathered}\]
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