4y^'-6y^'+7y=0

Lösung

4 y^{^{'}} - 6 y^{^{'}} + 7 y = 0

y = e^{\frac{7 t + c _{1}}{2}}

Schritte zur Lösung

4 y^{'} (t) - 6 y^{'} (t) + 7 y = 0

Solve separable ODE:

y = e^{\frac{7 t + c _{1}}{2}}

y = e^{\frac{7 t + c _{1}}{2}}

\frac{d y}{d t} = k (y + 2)

\frac{1}{8} \frac{d y}{d x} = y 4 x^{2} - 1

\frac{y ^{^{'}}}{x} = - y

y e^{- y} d y = - e^{x} x^{2} d x

\frac{d y}{d x} = 3 y, y (0) = 5