f'(x)=f(1−1x)
| 1 | ||
y' = (1− | ) | |
| x |
| dy | 1 | ||
= (1− | ) | ||
| dx | x |
| 1 | ||
dy = (1− | )dx | |
| x |
| 1 | ||
∫dy = ∫(1− | )dx | |
| x |
| 1 | ||
y' = y(1− | ) | |
| x |
| dy | 1 | ||
= 1− | |||
| dx*y | x |
| dy | 1 | ||
= (1− | )dx | ||
| y | x |
| 1 | 1 | |||
Czy to jest rzeczywiście f'(x) = f(1− | ), czy może f'(x) = f(x) * (1− | ). Jeżeli to | ||
| x | x |
Jeżeli to drugie, to rozwiązanie jest takie
jak proponujesz o 19:38.