(1+2y)dx+(4-x^2)dy=0

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d y}{d x} = - k y^{2}

\frac{y ^{^{'}}}{x} + y = 1

sin (y) + \frac{1}{2} x cos (y) \frac{d y}{d x} = 3

y^{^{'}} = e^{x} e^{- y}

\frac{d P}{d t} = 4 Pt, P (1) = 5