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

Lösung

y = - \frac{1}{x} + c_{1}, y = - - \frac{1}{x} + c_{1}

Schritte zur Lösung

y = \frac{1}{2 y ^{'} ( x ) x ^{2}}

Solve separable ODE:

y = - \frac{1}{x} + c_{1}, y = - - \frac{1}{x} + c_{1}

y = - \frac{1}{x} + c_{1}, y = - - \frac{1}{x} + c_{1}

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

y^{^{'}} - 6 y + 11 y^{^{'}} - 6 y = 0

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

y^{^{'}} = \frac{1}{e ^{y} + 2}

f^{^{'}} (x) [2 - x] = f (x)