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

Lösung

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

y = - \frac{1}{ln ( 1 + x ) + c _{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{1}{ln ( 1 + x ) + c _{1}}

y = - \frac{1}{ln ( 1 + x ) + c _{1}}

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

x^{^{'}} = 4 x + 2 e^{t} \cdot x

y^{^{'}} = - 4 x y^{2}

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

y^{^{'}} + sin (x) y = 5 sin (x)