2y^'=y+1

Lösung

2 y^{^{'}} = y + 1

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

Schritte zur Lösung

2 y^{'} (t) = y + 1

Solve separable ODE:

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

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

y^{^{'}} = x y - y e^{5 x}

(1 + x) y + (1 - y) x y^{^{'}} = 0

y^{^{'}} = \frac{2 y ^{2} - 9}{( 3 x y )}

\frac{d y}{d x} - x y = 0, x = 0

\frac{d x}{d t} = 3 - x