WEBVTT

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c 小题，假设 y 等于 f(x) 是微分方程

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在初始条件 f(2) = 3 时的特解。

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那么 f 在 x 等于 2 处是局部极大值？

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还是局部极小值？或者都不是？

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验证你的结论。

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好的，若要考虑

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局部极小或极大值，

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可以看看这一点的导数。

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如果等于 0，那么这里就有可能是

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有可能是局部极大或极小值，

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如果不等于 0，那么都不是。

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而如果确实等于 0，

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我们还要判断是极大值还是极小值，

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看看二阶导数算出来是正是负，就能判断。

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我们回到这一题，

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我们要计算 f'

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我们要计算的是 f'——

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f'(2) 等于什么。

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那么我们知道 f'(x)

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f'(x)，也就是 dy/dx，

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它等于 2x 减 y。

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从上一题就能看到。

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所以 f'(2)，我这样写

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f'(2) 就等于

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2 乘以 2，

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2 乘以 2 减去

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x 等于 2 时 y 的值。

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我们知道 x 等于 2 时 y 的值吗？

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当然，这里已经给出了。

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y 等于 f(x)，

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当 x 等于 2，

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当 x 等于 2 时，

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y 等于 3，

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所以是 2 乘以 2 减 3。

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那么它就等于 4 减 3，等于 1

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由于在 2 点的导数不等于 0，

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所以它就不是极小值，局部极小值，

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也不是局部极大值，

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所以可以说，由于 f'(2)

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f'(2) 不等于 0，

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这个，我们说 f，
我这样写，

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f 在 x 等于 2 处，既没达到极小值，

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局部极小值，这样说更好，

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局部极小，

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也没有达到局部极大值。

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好的，下一题。

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计算常数 m 和 b 的值，

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使得 y 等于 mx 加 b 是微分方程的一个解

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这道题有趣

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我们来，这样吧，
我们先把所有已知条件都写下来

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然后再考虑 y 等于 mx 加 b

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是微分方程的一个解这个条件。

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我们已经知道，

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dy/dx 等于 2x 减 y，

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是已知条件。

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我们也知道二阶导数，

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y 对 x 的二阶导数，等于

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2 减 dy/dx，

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这是我们在 b 小题中得出的结论。

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那么，我们可以将它表示成

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我们看，可以写为

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2 减 2x 加 y，通过代换

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就是把它代换进来

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那么这是 2 减 2x 加 y，

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我这么写，

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