-xy^'+y=y^{-1}

Lösung

- x y^{^{'}} + y = y^{- 1}

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

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

y^{^{'}} + 3.2 x^{4} \cdot y = 6.4 x^{4}

\frac{d y}{d x} = \frac{x}{y ( 5 x + 1 )}

\frac{( x ^{2} - 9 )}{x} \cdot \frac{d y}{d x} + y = 0, y (5) = 1

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