求方程dy⼀dx=y^2cosx满足初始条件x=0,y=1的特解

2025-06-28 00:55:45
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回答1:

这是分离变量的方程:dy/dx = y^2 * cosx =>1 / y^2 dy = cos x dx 积分=>-1/y = sin x + C =>y = -1 / (sin x + C).y(0) = 1 代入 => C = -1.故 y = -1 / (sin x - 1) = 1 / (1 - sin x).