xy^'+y=8xy^2

Lösung

x y^{^{'}} + y = 8 x y^{2}

y = \frac{1}{x ( - 8 ln ( x ) + c _{1} )}

Schritte zur Lösung

x y^{'} (x) + y = 8 x y^{2}

Solve Bernoulli ODE:

y = \frac{1}{x ( - 8 ln ( x ) + c _{1} )}

y = \frac{1}{x ( - 8 ln ( x ) + c _{1} )}

d er i v a t i v e \frac{3 x}{8 - cot ( x )}

d er i v a t i v e f (x) = 5^{- \frac{x}{2}} sin (2 x)

\int_{1}^{3} r^{3} ln (r) d r

d er i v a t i v e 4^{s i n^{2} (x) + c o s (3 x)}

x \to 0 lim (\frac{4 x}{4 + x - 4 - x})