y^'=-y/2+4

Lösung

y^{^{'}} = - \frac{y}{2} + 4

y = e^{\frac{- t - c _{1}}{2}} + 8

Schritte zur Lösung

y^{'} (t) = - \frac{y}{2} + 4

Solve separable ODE:

y = e^{\frac{- t - c _{1}}{2}} + 8

y = e^{\frac{- t - c _{1}}{2}} + 8

\frac{d y}{d t} y = (4 t - 1) (3 t - 6)^{- 1}

t y^{^{'}} 2 y = t^{2} - t^{1}

y^{^{'}} - \frac{y}{( ln ( y ) ) ^{2}} = 0, y (0) = 1

\frac{d y}{d x} = y^{2} - 9 y + 20

y^{^{'}} = - \frac{y}{2} - 2