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求方程dy/dx=y^2cosx满足初始条件x=0,y=1的特解
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求方程dy/dx=y^2cosx满足初始条件x=0,y=1的特解
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答案和解析
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).
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).
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