xy+y^'=0

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

y^{^{'}} = x^{2} y^{2} - 2 y^{2} + x^{2} - 2

\frac{d y}{d x} = x y + 2 x - 3 y - 6

f^{^{'}} (x) = 5 f (x)

y^{^{'}} - y = 2, cos (5 t) y (0) = 0

\frac{d y}{d x} = - \frac{4}{4} 4 x^{4} y^{2}