Question

求微分方程(xdy/dx - y ) arctany/x=x，求通解

Answer 1

(xdy/dx-y)arctan(y/x) = x, 定义域， x ≠ 0.
两边同除于 x， 得 (dy/dx-y/x)arctan(y/x) = 1, 是齐次方程。
令 y/x = u，则 y = xu，dy/dx = u+xdu/dx, 齐次方程化为
（xdu/dx)arctanu = 1，分离变量为 (arctanu)du = dx/x
两边分别积分，得 uarctanu - ∫udu/√(1+u^2) = lnx
uarctanu - √(1+u^2) = lnx + lnC
即 Cx = e^[uarctanu-√(1+u^2)]
原微分方程的通解是 Cx = e^[(y/x)arctan(y/x)-√(1+y^2/x^2)]

Answer 2

y/x = t => y = x * t => dy = x dt + t dx => dy/dx = t + x dt/dx 代入原方程得：t + x dt/dx = t + tan t => x * dt/dx = tan t => cot t dt = 1/x dx 积分=> ln ...