y^'=y*ln(y)

Lösung

y^{^{'}} = y \cdot ln (y)

y = e^{e^{t + c_{1}}}

Schritte zur Lösung

y^{'} (t) = y \cdot ln (y)

Solve separable ODE:

y = e^{e^{t + c_{1}}}

y = e^{e^{t + c_{1}}}

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

d y = (\frac{y}{x} - cos (e) c^{2} \frac{y}{x}) d x

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

y^{^{'}} + 4 t y = - 2 t

x y^{^{'}} - y = 4 x^{9} y, y (1) = 10