题目

求微分方程y=xdy/dx+(y^2)(sinx)^2的通解

提问时间：2021-01-16

答案

Ans：y = 2x/(x - sinxcosx + C)
y = x * dy/dx + y²sin²x
-x * dy/dx + y = y²sin²x
-(dy/dx)/y² + 1/(xy) = (sin²x)/x
令v = 1/y则dv/dx = dv/dy * dy/dx = d(1/y)/dy * dy/dx = -1/y² * dy/dx = -(dy/dx)/y²
=> dv/dx + v/x = sin²x/x
积分因子= e^∫ (1/x) dx = e^lnx = x,将x乘以方程两边得
x * dv/dx + v = sin²x
x * dv/dx + v * dx/dx = sin²x
d(x * v)/dx = sin²x
x * v = (1/2)∫ (1 - cos2x) dx = (1/2)(x - 1/2 * sin2x) + C = x/2 - 1/2 * sinxcosx + C
v = (x - sinxcosx + C)/(2x)
1/y = (x - sinxcosx + C)/(2x)
y = 2x/(x - sinxcosx + C)