y^'=(e^{-x}-e^x)/(3+4y)

Lösung

y^{^{'}} = \frac{e ^{- x} - e ^{x}}{3 + 4 y}

y = \frac{- 3 + 9 - 8 ( - c _{1} + e ^{- x} + e ^{x} )}{4}, y = \frac{- 3 - 9 - 8 ( - c _{1} + e ^{- x} + e ^{x} )}{4}

Schritte zur Lösung

y^{'} (x) = \frac{e ^{- x} - e ^{x}}{3 + 4 y}

Solve separable ODE:

y = \frac{- 3 + 9 - 8 ( - c _{1} + e ^{- x} + e ^{x} )}{4}, y = \frac{- 3 - 9 - 8 ( - c _{1} + e ^{- x} + e ^{x} )}{4}

y = \frac{- 3 + 9 - 8 ( - c _{1} + e ^{- x} + e ^{x} )}{4}, y = \frac{- 3 - 9 - 8 ( - c _{1} + e ^{- x} + e ^{x} )}{4}

y + 2 y^{^{'}} - 3 y = 0

4 (1 + y^{2}) d x - x y d y = 0, y (2) = 15

2 x y d x = (2 - x^{2}) d y, y (4) = 3

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

y^{^{'}} = \frac{x + 2}{y ^{2}}