2y^'+y^3=0

Lösung

2 y^{^{'}} + y^{3} = 0

y = - \frac{1}{- t + c _{1}}, y = - - \frac{1}{- t + c _{1}}

Schritte zur Lösung

2 y^{'} (t) + y^{3} = 0

Solve separable ODE:

y = - \frac{1}{- t + c _{1}}, y = - - \frac{1}{- t + c _{1}}

y = - \frac{1}{- t + c _{1}}, y = - - \frac{1}{- t + c _{1}}

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

x^{3} y^{^{'''}} - x^{2} y^{^{'}} = 0

5 x (x + 4 y) y^{^{'}} = 5 y (x - 4 y)

3 y^{^{'}} + 7 y = x^{3} - 2 x^{2} + 1

y^{^{'}} = x \cdot y - 3 \cdot x + 3 \cdot y + 3 \cdot y - 9