求解一道简单的常微分方程,dy/dx=(x+y)^2

学习时间:2026-05-29 23:54:13 阅读：2435

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?

拼搏的蜻蜓

2026-05-29 23:54:13

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dy/dx = (x + y)²令t = x + y,dt/dx = 1 + dy/dxdt/dx - 1 = t²dt/dx = (1 + t²)dt/(1 + t²) = dxarctan(t) = x + C₁x + y = tan(x + C₁)y = tan(x + C₁) - x