xy^'+2y-2x^2y^2=0

Lösung

x y^{^{'}} + 2 y - 2 x^{2} y^{2} = 0

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

y = \frac{1}{x ^{2} ( - 2 ln ( x ) + c _{1} )}

\int \frac{1}{x ^{2} + 6 x + 13} d x

\int_{1}^{4} 3 e^{x} d x

\frac{d}{d x} (tan (π))

a re a x = 1, y = 6 x (2 - x^{2})^{2}

x \to - 4 - lim (\frac{- 4 x ^{3}}{4 + x})