x+xy+y^'=0

Lösung

x + x y + y^{^{'}} = 0

y = e^{- \frac{x ^{2}}{2} + c_{1}} - 1

Schritte zur Lösung

x + x y + y^{'} (x) = 0

Solve separable ODE:

y = e^{- \frac{x ^{2}}{2} + c_{1}} - 1

y = e^{- \frac{x ^{2}}{2} + c_{1}} - 1

\frac{d y}{d x} = 2 e^{x - y}, y (0) = ln (4)

\frac{d y}{d x} = \frac{4 y}{1 + x}

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

\frac{d y}{d x} = 2 \cdot (x^{2}) \cdot (y^{2})

d x - \frac{d y}{1 + y ^{2}} = 0