y^'=2x(y^2+1),y(0)=0

Lösung

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

y = tan (x^{2})

Schritte zur Lösung

y^{'} (x) = 2 x (y^{2} + 1)

Solve separable ODE:

y = tan (x^{2})

y = tan (x^{2})

\frac{d x}{d t} = e^{t + x}

\frac{d y}{d x} = - \frac{x ^{2}}{y ^{2}}

x y^{^{'}} + 2 y = 3, x y (1) = 5

2 y \frac{d y}{d x} = - x^{2}

y + 5 + (x - 3) \frac{d y}{d x} = 0