| y | 2(y/x) | |||
⇔ y'[(y/x)2 + 1] = 2* | ⇔ y' = | |||
| x | (y/x)2 +1 |
| y | ||
... i podstaw: | = u | |
| x |
| y | |
=u | |
| x |
| 2u | ||
u'x+u= | ||
| u2+1 |
| 2u−u3−u | ||
u'x= | ||
| u2+1 |
| u2+1 | 1 | ||
du= | dx Poradzisz już sobie? | ||
| u−u3 | x |
| y | ||
u = | ⇔ y = ux ⇒ y' = u + u'x | |
| x |
| 2u | ||
masz równanie: u + u'x = | ..... i rozdzielasz zmienne | |
| u2 + 1 |
