y^'[t]=0.3y[t]

Lösung

y^{^{'}} [t] = 0.3 y [t]

y = e^{0.3 t + c_{1}}

Schritte zur Lösung

y^{'} (t) [t] = 0.3 y [t]

Solve separable ODE:

y = e^{0.3 t + c_{1}}

y = e^{0.3 t + c_{1}}

32 = v^{2} + xv \frac{d v}{d x}

\frac{d y}{d x} = y^{2} + 2 y - 15

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

y^{^{'}} = e^{t} \cdot y^{2}

y^{^{'}} = 6 y - 5