Question

求微分方程dy/dx-y/x-√(y/x)=0的通解。急！！！！！真的没有金币了，哪位

求微分方程dy/dx-y/x-√(y/x)=0的通解。急！！！！！真的没有金币了，哪位大神能来帮帮忙

Answer 1

答：(2/3)(y/x)^(3/2) + y/x + √(y/x) - ln(√(y/x) - 1) = ln(x) + C

设y = ux，dy/dx = u + x*du/dx
u + x*du/dx - u + √u = 0
x*du/dx = (u - √u)/u
u/(u - √u) du = 1/x dx
左边的积分，令√u = s，则u = s²，du = 2s ds

∫ u/(u - √u) du
= ∫ s³/(s² - s) * 2s ds
= 2∫ s³/(s - 1) ds
= 2∫ [(s³ - 1) + 1]/(s - 1) ds
= 2∫ [(s - 1)(s² + s + 1) - 1]/(s - 1) ds
= 2∫ [s² + s + 1 - 1/(s - 1)] ds
= 2[ s³/3 + s²/2 + s - ln(s - 1) ] + C
= (2/3)u^(3/2) + u + √u - ln(√u - 1) + C
于是原式等价
(2/3)u^(3/2) + u + √u - ln(√u - 1) = ln(x) + C
将y = ux回代，得
(2/3)(y/x)^(3/2) + y/x + √(y/x) - ln(√(y/x) - 1) = ln(x) + C