2x^3y^'=y(2x^2-y^2)

Lösung

2 x^{3} y^{^{'}} = y (2 x^{2} - y^{2})

y = \frac{x}{ln ( x ) + c _{1}}, y = - \frac{x}{ln ( x ) + c _{1}}

Schritte zur Lösung

2 x^{3} y^{'} (x) = y (2 x^{2} - y^{2})

Solve Bernoulli ODE:

y = \frac{x}{ln ( x ) + c _{1}}, y = - \frac{x}{ln ( x ) + c _{1}}

y = \frac{x}{ln ( x ) + c _{1}}, y = - \frac{x}{ln ( x ) + c _{1}}

t an g e n t f (x) = 12 (x - \frac{1}{x})^{2}, at x = 2

\int \frac{( e ^{x} + e ^{- x} )}{2} d x

x \to 0 lim (\frac{2 x}{x})

x \to - 4 - lim (\frac{2}{x ^{2} - 16})

\frac{d y}{d x} = \frac{( e ^{x} - e ^{- x} )}{3 + y}