y^'+a*y^2=b

Lösung

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

y = - \frac{b e ^{2 a b t + 2 c_{1} a b}}{a ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{b}{a ( 1 - e ^{2 a b (t + c_{1})} )}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{b e ^{2 a b t + 2 c_{1} a b}}{a ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{b}{a ( 1 - e ^{2 a b (t + c_{1})} )}

y = - \frac{b e ^{2 a b t + 2 c_{1} a b}}{a ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{b}{a ( 1 - e ^{2 a b (t + c_{1})} )}

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

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

4 y^{^{'}} - 2 e^{3 x} y = 0, y (0) = 6

cos^{2} (x) \frac{d y}{d x} = y + 3

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