y^'+2xy=0

Lösung

y^{^{'}} + 2 x y = 0

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

Schritte zur Lösung

y^{'} (x) + 2 x y = 0

Solve separable ODE:

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

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

\int sec^{4} (1) d x

t an g e n t f (x) = \frac{4}{1 + x ^{2}}

l a pl a ce t r an s f or m cos (- 2 t)

\int 23 x d x

y^{^{'}} = \frac{y}{t}