(4x^5-16x^3-2x)(1-y^2)-2y(x^2-4)y^'=0

Lösung

(4 x^{5} - 16 x^{3} - 2 x) (1 - y^{2}) - 2 y (x^{2} - 4) y^{^{'}} = 0

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

Schritte zur Lösung

(4 x^{5} - 16 x^{3} - 2 x) (1 - y^{2}) - 2 y (x^{2} - 4) y^{'} (x) = 0

Solve separable ODE:

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

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

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = 1

y^{^{'}} = \frac{1 + x}{x y}, y (1) = - 4

\frac{d v}{d t} = 0.4 v + 1000

\frac{d y}{y} = - \frac{d x}{x}

y^{^{'}} = \frac{4 \cdot y}{x} - \frac{4}{x}