xy^'=yln(y)

Lösung

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

y = e^{c_{1} x}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{c_{1} x}

y = e^{c_{1} x}

y^{^{'}} = \frac{cos ^{2} ( y )}{x}

e^{y} d x + e^{- x} d y = 0

θ^{2} d θ = sin (t + 0.2) d t

\frac{d y}{d t} = (1 + 4 t) y

\frac{d y}{d x} + 2 x y = x, y (0) = 2