y^3y^'+4xy^4=8x

Lösung

y^{3} y^{^{'}} + 4 x y^{4} = 8 x

y = 4 e^{- 8 x^{2} + c_{1}} + 2, y = - 4 e^{- 8 x^{2} + c_{1}} + 2

Schritte zur Lösung

y^{3} y^{'} (x) + 4 x y^{4} = 8 x

Solve separable ODE:

y = 4 e^{- 8 x^{2} + c_{1}} + 2, y = - 4 e^{- 8 x^{2} + c_{1}} + 2

y = 4 e^{- 8 x^{2} + c_{1}} + 2, y = - 4 e^{- 8 x^{2} + c_{1}} + 2

2 x y^{3} + 3 x^{2} y^{2} \frac{d y}{d x} = 0

(t^{3} - 12 t + 27) \frac{d y}{d t} = y

\frac{d y}{d x} = y, y > 0

7 x y \frac{d y}{d x} = 2, y > 0

sin (y) y^{^{'}} = x cos (y)