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

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

t^{2} x^{^{'}} + x = 0

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

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

y^{^{'}} = \frac{e ^{x}}{e ^{y}}

y y^{^{'}} = [sin (x) - cos (x)] 1 - y^{2}