y^'+a*y^2=0

Lösung

y^{^{'}} + a \cdot y^{2} = 0

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

Schritte zur Lösung

y^{'} (t) + a \cdot y^{2} = 0

Solve separable ODE:

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

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

\frac{d y}{d x} = - \frac{4}{6} x^{3} y^{5}

\frac{d y}{d x} = (x^{3} y^{4})

4 x y^{2} d y = 2 y^{3} + 2 y d x

y^{2} d y = cos (5 x) d x

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