e^x=yy^'

Lösung

e^{x} = y y^{^{'}}

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

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

\frac{d y}{d x} = e^{2 y} sin (x), y (0) = 0

(x + 1) y^{^{'}} - (2 x + 3) y = 0

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

\frac{d Q}{d t} = 4 Q^{2} ln (t)