y^'=x(2y+1)

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

y (6 x^{2} - x - 1) d x + 2 x d y = 0

y^{^{'}} - y = 0, y (0) = c e^{x}

y^{^{'}} = \frac{1}{1 - 3 y}

\frac{d y}{y} = 35 x^{4} d x

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