y^'=(2-e^x)/(3+2y)

Lösung

y^{^{'}} = \frac{2 - e ^{x}}{3 + 2 y}

y = \frac{- 3 + 9 - 4 ( - c _{1} + e ^{x} - 2 x )}{2}, y = \frac{- 3 - 9 - 4 ( - c _{1} + e ^{x} - 2 x )}{2}

Schritte zur Lösung

y^{'} (x) = \frac{2 - e ^{x}}{3 + 2 y}

Solve separable ODE:

y = \frac{- 3 + 9 - 4 ( - c _{1} + e ^{x} - 2 x )}{2}, y = \frac{- 3 - 9 - 4 ( - c _{1} + e ^{x} - 2 x )}{2}

y = \frac{- 3 + 9 - 4 ( - c _{1} + e ^{x} - 2 x )}{2}, y = \frac{- 3 - 9 - 4 ( - c _{1} + e ^{x} - 2 x )}{2}

\int e^{\frac{- x}{10}} d x

y^{^{'}} = \frac{y}{x} + tan (\frac{y}{x})

4 y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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

\frac{\partial}{\partial x} (- sin (4 x))