y^'+0.5y=y^3

Lösung

y^{^{'}} + 0.5 y = y^{3}

y = \frac{1.99999 \dots}{2 1 - e ^{t + c_{1}}}, y = - \frac{1.99999 \dots}{2 1 - e ^{t + c_{1}}}

Schritte zur Lösung

y^{'} (t) + 0.5 y = y^{3}

Solve separable ODE:

y = \frac{1.99999 \dots}{2 1 - e ^{t + c_{1}}}, y = - \frac{1.99999 \dots}{2 1 - e ^{t + c_{1}}}

y = \frac{1.99999 \dots}{2 1 - e ^{t + c_{1}}}, y = - \frac{1.99999 \dots}{2 1 - e ^{t + c_{1}}}

3 x d y - (y + 1) d x = 0

\frac{d y}{d x} = - 4 x^{5} e^{y}

(x^{2} + 1) \frac{d y}{d x} = 6 x - 3 x y

y^{^{'}} + x y = \frac{x}{y ^{3}}

y - y^{2} + x y^{^{'}} = 0