2y^'+1/x y=x^2y^{-1}

Lösung

2 y^{^{'}} + \frac{1}{x} y = x^{2} y^{- 1}

y = \frac{x ^{4} + c _{1}}{2 x}, y = - \frac{x ^{4} + c _{1}}{2 x}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = \frac{x ^{4} + c _{1}}{2 x}, y = - \frac{x ^{4} + c _{1}}{2 x}

y = \frac{x ^{4} + c _{1}}{2 x}, y = - \frac{x ^{4} + c _{1}}{2 x}

x \to 0 lim (\frac{( 2 ^{x} - 7 ^{x} )}{x})

in v erse l a pl a ce \frac{1}{s ( s ^{2} + 2 s + 5 )}

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

t an g e n t f (x) = x^{3} ln (2 x), at x = 1

\int_{1}^{9} y ln (y) d y