l*******y
发帖数: 4006 | | k*****y
发帖数: 744 | 2 f(t) = k * exp( c*t ), where k*c != 0.
【在 l*******y 的大作中提到】
: f/f'=f'/f'';
: 通解哦
| A**u
发帖数: 2458 | 3 猜 sin,cos,exp
【在 l*******y 的大作中提到】
: f/f'=f'/f'';
: 通解哦
| m*******t
发帖数: 6 | 4 actually, df'/df=df'/dt*dt/df=f"/f', so back to your equation, we have:
f/f'=df/df' then df'/f'=df/f, which easily yiled to kinecty's anwser: f=k*
exp(ct), where kc!=0, | s********r
发帖数: 529 | | l******n
发帖数: 9344 | 6 自己试一下
【在 s********r 的大作中提到】
: 多项式不可以吗？
| l******i
发帖数: 1404 | 7 kinecty and mathquant are correct,
here is my detailed solution:
Notice that $ df'/df = df'/dt \times dt/df = f"/f' $;
Taking it back to the problem, we have: $ f/f' = df/df' $,
i.e. $ df'/f'=df/f $;
Integrating on both sides, then we get:
$ \log(|f'|) = \log(|f|) + C_1 $ for some constant C_1;
Taking exponential on both sides, then we get:
$ |f'| = C_2 \times |f| $ for some constant C_2>0;
Now $ |f'/f| = C_2 $ with C_2>0,
i.e. $ f'/f = C_3 $ with some constant C_3.
Integrating on both sides, then we get:
$ \log(|f|) = C_3 \times t+C_4 $ for some constant C_3 and C_4;
(here t is the variable)
Finally taking exponential on both sides, then we get:
$ f = L \times \exp(Kt) $ for some constants K,L;
Note that f' and f" cannot be zero, thus f cannot be constant,
i.e. K !=0 and L!0, i.e. KL !=0. |
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