2xy^'ln(x)=-2y

Lösung

2 x y^{^{'}} ln (x) = - 2 y

y = \frac{c _{1}}{ln ( x )}

Schritte zur Lösung

2 x y^{'} (x) ln (x) = - 2 y

Solve separable ODE:

y = \frac{c _{1}}{ln ( x )}

y = \frac{c _{1}}{ln ( x )}

x \cdot \frac{d y}{d x} + x^{3} y = 0

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

\frac{d y}{d t} = \frac{3 + y}{2}

y^{^{'}} = (1 + y^{2}) \cdot cos (t)

y^{^{'}} = x y + 4 x + 3 y + 12