2xy^'+x^2y=0

Lösung

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

y = e^{- \frac{x ^{2}}{4} + c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{- \frac{x ^{2}}{4} + c_{1}}

y = e^{- \frac{x ^{2}}{4} + c_{1}}

y^{^{'}} = \frac{x + x y ^{2}}{4 x}

\frac{d y}{d x} = \frac{sin ( 5 x )}{y}

\frac{d}{d x} ((y) \cdot 10) + 5 y = 0

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

\frac{d v}{d t} = 8 - 7 v