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

Lösung

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

y = 1 + t + c_{1} + 1, y = 1 - t + c_{1} + 1

Schritte zur Lösung

2 y y^{'} (t) = 1 + (y^{'} (t)) 2

Solve separable ODE:

y = 1 + t + c_{1} + 1, y = 1 - t + c_{1} + 1

y = 1 + t + c_{1} + 1, y = 1 - t + c_{1} + 1

x y^{^{'}} = y^{2} - 1

\frac{x ^{3} + 1}{x ^{2}} \frac{d y}{d x} = \frac{1}{y}

y y^{^{'}} = - 5 x

\frac{d y}{d x} - \frac{2 x}{x ^{2} + 1} y = - \frac{2 x}{x ^{2} + 1}

(y^{2} + 3) d x = y sec^{2} (x) d y