y*y^'+4x-2=0

Lösung

y \cdot y^{^{'}} + 4 x - 2 = 0

y = - 4 x^{2} + 4 x + c_{1}, y = - - 4 x^{2} + 4 x + c_{1}

Schritte zur Lösung

y \cdot y^{'} (x) + 4 x - 2 = 0

Solve separable ODE:

y = - 4 x^{2} + 4 x + c_{1}, y = - - 4 x^{2} + 4 x + c_{1}

y = - 4 x^{2} + 4 x + c_{1}, y = - - 4 x^{2} + 4 x + c_{1}

\frac{d y}{d t} = - 0.4 (y - 70)

- v b = \frac{d v}{d t} - c

3 y^{^{'}} = y cos (5 x), y (0) = 2

y ln (y) d x - d y = 0

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