(2t-1)y^'-4y=0

Lösung

(2 t - 1) y^{^{'}} - 4 y = 0

y = 4 e^{c_{1}} t^{2} - 4 e^{c_{1}} t + e^{c_{1}}

Schritte zur Lösung

(2 t - 1) y^{'} (t) - 4 y = 0

Solve separable ODE:

y = 4 e^{c_{1}} t^{2} - 4 e^{c_{1}} t + e^{c_{1}}

y = 4 e^{c_{1}} t^{2} - 4 e^{c_{1}} t + e^{c_{1}}

f (x) f^{^{'}} (x) = x x^{3} - x x^{2} - 6 xx

f^{^{'}} (x) = f (x)^{- 1} (x)

x y^{2} y^{^{'}} = x + x^{3}

\frac{3 d y}{d t} = \frac{cos ( t )}{y}

e^{x + y} \cdot y^{^{'}} = x