dx+(x+y+1)dy=0

Lösung

d x + (x + y + 1) d y = 0

y = W (- \frac{1}{e ^{- c_{1} - x}}) - x

Schritte zur Lösung

d x + (x + y + 1) d y = 0

Solve using substitution:

y = W (- \frac{1}{e ^{- c_{1} - x}}) - x

y = W (- \frac{1}{e ^{- c_{1} - x}}) - x

f (x) = x^{3} - x - 1

t an g e n t f (x) = \frac{x + 1}{x - 1}, at x = 0

5 t \cdot \frac{d y}{d t} + 4 y = t

\int_{2}^{\infty} \frac{2}{t ^{2} - 1} d t

d er i v a t i v e (9 - x^{2}) (\frac{1}{3}) π x^{2}