题目

求dy/dx=x+xy^/y+yx^满足初始条件y｜(下面是x=0)=2的特解．

提问时间：2020-12-16

答案

dy/dx = (x+xy²)/(y+yx²)
(y+yx²)dy = (x+xy²)dx
ydy + yx²dy - xdx - xy²dx = 0
ydy - xdx + 1/2 * (x²dy² - y²dx²) = 0
dy² - dx² + x²dy² - y²dx² = 0
(1+x²)dy² = (1+y²)dx²
dy²/(1+y²) = dx²/(1+x²)
ln(1+y²) = ln(1+x²) + C
x=0时y=2
则
ln5 = C
所求特解为
ln(1+y²) = ln(1+x²) + ln5
即
y = √(4+5x²)