[1+t^2](dy)/(dt)=1+y^2

Lösung

[1 + t^{2}] \frac{d y}{d t} = 1 + y^{2}

y = tan (arctan (t) + c_{1})

Schritte zur Lösung

[1 + t^{2}] \frac{d y}{d t} = 1 + y^{2}

Solve separable ODE:

y = tan (arctan (t) + c_{1})

y = tan (arctan (t) + c_{1})

y^{^{'}} = x^{2} y, y (0) = 1

y^{3} y^{^{'}} + 4 x y^{4} = 8 x

2 x y^{3} + 3 x^{2} y^{2} \frac{d y}{d x} = 0

(t^{3} - 12 t + 27) \frac{d y}{d t} = y

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