求解一道简单的常微分方程,dy/dx=(x+y)^2dy/dx=(x+y)^2怎么作适当变换来解?_数学解答

求解一道简单的常微分方程,dy/dx=(x+y)^2
dy/dx=(x+y)^2
怎么作适当变换来解?

人气：499 ℃　时间：2019-10-10 05:53:51

解答

dy/dx = (x + y)²
令t = x + y,dt/dx = 1 + dy/dx
dt/dx - 1 = t²
dt/dx = (1 + t²)
dt/(1 + t²) = dx
arctan(t) = x + C₁
x + y = tan(x + C₁)
y = tan(x + C₁) - x