(e^y+1)^2e^{-y}dx+(e^x+1)^3e^{-x}dy=0

Lösung

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

y = ln (- \frac{2 ( e ^{x} + 1 ) ^{2}}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{2 c _{1} ( e ^{x} + 1 ) ^{2}}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{1}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}})

Schritte zur Lösung

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

Solve separable ODE:

y = ln (- \frac{2 ( e ^{x} + 1 ) ^{2}}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{2 c _{1} ( e ^{x} + 1 ) ^{2}}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{1}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}})

y = ln (- \frac{2 ( e ^{x} + 1 ) ^{2}}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{2 c _{1} ( e ^{x} + 1 ) ^{2}}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{1}{1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}})

\frac{\partial}{\partial y} (x + e^{z} + y)

t an g e n t \frac{x}{x + 4}, at 1, 0.2

t an g e n t 3 x + 4, at x = 7

x \to 0 - lim (x ln (- x))

\frac{d}{d x} (x + sin^{2} (\frac{x}{3}) - 8)