y*y^'+2x-1=0

Lösung

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

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

Schritte zur Lösung

y \cdot y^{'} (x) + 2 x - 1 = 0

Solve separable ODE:

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

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

y^{^{'}} = x y + x - 2 y - 2, y (0) = 1

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

(x^{2} + 4) d y + (8 x y - 2 x) d x = 0

y^{^{'}} + 2 y - 5 = 0

\frac{d y}{d x} y = \frac{6 x ^{2} + 4 x ^{2}}{3 y ^{6}}