(e^y+1)^2e^{-y}dx+(e^x+1)^5e^{-x}dy=0

Lösung

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{5} e^{- x} d y = 0

y = ln (- \frac{4 ( e ^{x} + 1 ) ^{4}}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}} - \frac{4 c _{1} ( e ^{x} + 1 ) ^{4}}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}} - \frac{1}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}})

Schritte zur Lösung

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{5} e^{- x} d y = 0

Solve separable ODE:

y = ln (- \frac{4 ( e ^{x} + 1 ) ^{4}}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}} - \frac{4 c _{1} ( e ^{x} + 1 ) ^{4}}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}} - \frac{1}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}})

y = ln (- \frac{4 ( e ^{x} + 1 ) ^{4}}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}} - \frac{4 c _{1} ( e ^{x} + 1 ) ^{4}}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}} - \frac{1}{1 + 4 c _{1} ( e ^{x} + 1 ) ^{4}})

\int \frac{sec ( x )}{sec ( x ) - tan ( x )} d x

x \to 0 - lim (arcsin (e^{x}))

a re a y = 6 x^{2} - 2 x, y = x^{3} - 6 x + 5

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

\int_{1}^{e^{2}} \frac{2}{x} d x