| dy | 3x−4 | ||
= | |||
| dx | xy(x−2) |
| 3x−4 | ||
ydy = | dx | |
| x(x−2) |
| 3x−4 | ||
∫ydy = ∫ | dx | |
| x(x−2) |
| y2 | 32(2x−83) | ||
= ∫ | dx | ||
| 2 | x2−2x |
| y2 | 2x−2−23 | ||
= 32∫ | dx /*2 | ||
| 2 | x2−2x |
| 2x−2 | 1 | |||
y2 = 3*[ ∫ | dx − 23∫ | dx ] | ||
| x2−2x | x(x−2) |
| 2x−2 | 1 | |||
y2 = 3∫ | dx − 2∫ | dx | ||
| x2−2x | x(x−2) |
niestety nie może mi wyjść...
| A | B | 1 | |||
+ | = | ||||
| x | x−2 | x(x−2) |
| 1 | ||
A = − | ||
| 2 |
| 1 | ||
B = | ||
| 2 |
| 3x−4 | ||
∫ | dx=... | |
| x(x−2) |
| 3x−4 | A | B | Ax−2A+Bx | ||||
= | + | = | |||||
| x(x−2) | x | x−2 | x(x−2) |
| 3x−4 | 2 | 1 | ||||
∫ | dx=∫ | dx+∫ | dx=2ln|x|+ln|x−2| | |||
| x(x−2) | x | x−2 |
| 1 | |
y2=2ln|x|+ln|x−2| | |
| 2 |