x^'+0.09*x=0,x(0)=1

Lösung

x^{^{'}} + 0.09 \cdot x = 0, x (0) = 1

x = e^{- 0.09 t}

Schritte zur Lösung

x^{'} (t) + 0.09 \cdot x = 0

Solve separable ODE:

x = e^{- 0.09 t}

x = e^{- 0.09 t}

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

y^{2} \frac{d y}{d x} = x^{2}

\frac{d x}{d t} = 50 - \frac{7 x}{100}

(- y) d x + (x) d y = 0

y^{^{'}} + \frac{1}{2} y = 9