2x+2yy^'=0

Lösung

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

y = - x^{2} + c_{1}, y = - - x^{2} + c_{1}

Schritte zur Lösung

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

Solve separable ODE:

y = - x^{2} + c_{1}, y = - - x^{2} + c_{1}

y = - x^{2} + c_{1}, y = - - x^{2} + c_{1}

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

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

y^{^{'}} y = x - 3 - 6 cos (x)

\frac{e ^{- 1}}{2} \cdot y^{3} + x^{2} \cdot y^{2} \cdot y^{^{'}} = 0

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